Math Practice Fourth Grade

Introduction to math practice fourth grade:

Here we are going to see the problems on math practice fourth grade; it covers some topics of the grade 4 Number sense ,Addition ,Subtraction ,Multiplication, Division ,Mixed operation ,Algebra ,Functions, statistics .In these topics we are going to see the practice problem with answers


Number sense math practice fourth grade


1) How do write this number   using digits?

Seventy nine

A) 7.90

B) 790

C) 0.79

D) 79

Addition- math practice fourth grade:

2) Add 4890and 567

Using the addition operation to do the above problem

A) 4812

B) 5457

C) 4702

D) 5902

subtraction- math practice fourth grade:

3) Subtract 6892and 5263

Using the subtraction  operation to do the above problem

A) 2529

B) 7890

C) 1629

D) 1529

Multiplication- math practice fourth grade:

4) Multiplication facts to 18

Multiply

18*12

A) 158

B) 216

C) 138

D) 197

Division - math practice fourth grade:

5) Fill the mixing number

5X /25=5

A) 25

B) 35

C) 125

D) 45

Algebra- math practice fourth grade:

6) Write an expression for 67 divided by 8

A) `67*8`

B) `67/8`

C) 67-8

D) 67+8

Mixed operation - math practice fourth grades:

7)John and his friend jolly went together to the department store for buy the chocolate, john bought 15 chocolates and jolly bought 4 chocolates ,about how many pieces of chocolates did they buy in all ?choose the better estimation.

A) 20

B) 15

C) 4

D) 11

Statistics - math practice fourth grades:

8) Find the median for the following numbers 7, 9,6,4,5,2,11

A) 4

B) 5

C) 6

D) 11



Geometry - math practice fourth grades:

9) What is the perimeter of the square whose side is 6?

A) 36

B) 12

C) 6

D) 24

10) What is the dimension of the line?

A) One

B) Two

C) three

D) Zero

what is the dimension of  the square?

A) One

B) Two

C) three

D) Zero


Answer key:


1) D

2) B

3) C

4) B

5) A

6) B

7) A

8) C

9) D

10) A

11) B

8th Grade Math Practice Learning

Introduction to 8th grade math practice learning:

Mathematics is a study of basic math operations and maths functions. Mathematics is one of a language and logic thing of a science. It is an essential tool for science, medicine and the geography fields. In mathematics concept a fundamental concept is arithmetic operations. In 8th grade mathematics includes algebra, number system, measurements, geometry, etc. In this article we shall discuss for 8th grade math practice learning.

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Sample problem for 8th grade math practice learning:


8th grade math practice learning problem 1:

Solve the given algebra sample equation and find out m and n value of the equation.

4m + 6n – 10 = 0

-4m + 18n – 14 = 0

Solution:

We are going to find out the m and n value of the given algebra 1 linear equation.

4m + 6n = 10

-4m + 18n = 14

In the first we are going to add equation (1) and (2). We get

4m + 6n = 10

-4m + 18n = 14

0m + 24n = 24

n = 1

Now, we get the n value as 1. In the equation (2) we substitute n = 1, we get

-4m + 18(1) = 14

-4m + 18 = 14

-4m = -4

M = 1

Now, we get the m value as 1. So, the given system of linear equation values are m = 1, n = 1.

8th grade math practice learning problem 2:

Solve the given algebra equation and find out the ‘p’ value 3 - 2(4p - 5) = 5

Solution:

Given:

3 - 2(4p - 5) = 5

We are going to find ‘p’ value of the given equation. Subtract 3 on both side of the equation, we get

3 - 2(4p - 5) - 3 = 5 - 3

-8p + 10 = 2

In the next step grouping the terms, we get

-8p = 2 - 10

-8p = -8

p = 1

8th grade math practice learning problem 3:

Simplify a given algebraic expression: 4p + 6q - 8 + 5p + 6q + 7.

Solution:

We are going to find a value of the given algebraic expression.

4p and 5p are like terms, and can be combined to give 9p,

6q and 6q combine to give 12q, and

-8 and 7 combine to give -1.

In the next we add all the terms, we get 9p+ 12q - 1.

8th grade math practice learning problem 4:

Find a diameter value from a given circle radius. A given circle radius is 5 cm.

Solution:

Given:

Radius of the circle is r = 5 cm

Diameter of the circle is d = 2r

d = 2 * 5

d = 10

So, the diameter of the circle is 10 cm2.

Practice problem for 8th grade math practice learning:

Find the value of the given equation 4m + 26 = 10

Answer: m = -4

Solve the given algebraic equation and find the ‘n’ value 2n -3 = 7.

Answer: n = 5

Find a diameter value from a given circle radius. A given circle radius is 3 cm.

Answer: d = 6 cm2

Algebra 1 Final Exam Practice

Introduction to algebra 1 final exam practice:

Let us see about algebra connections equations. The algebra is a mathematics that can be studied in words of mathematics numbers. In this we learn about the algebra connections equations with a few concepts. The algebra is the branch in which we learned the arithmetical functions, polynomials, factors and some equations. Let us see some examples in  algebra 1 final exam practice.

Important Concepts of Algebra

The followings are some of the algebra concepts.

1. Polynomial

In algebra connections, a polynomial of degree one is known as a linear polynomial.

2. Quadratic equation

In algebra connections, a quadratic equations in a variable of the x .An equation in the form ax2 + bx + c = 0.

3. Algebraic identities

The form (a + b)2 = a2 + 2ab + b2 is called as the algebraic identities.

4. Algebraic expressions

It is a division of polynomial can be denoted by a letters and exponent.


Examples to understand algebra 1 final exam practice:


Algebra Connections Problems

1)Solve the following equation 49x2 – 36 = 0.

Solution:

49x2 – 36 = 0 or

(7x + 6) (7x – 6) = 0 or (by simplify the equations)

7x2- 62 =0

7x2- 36 = 0

x2= 36/7

The solution set = `6 / sqrt 7`

2) Conclude whether (x–3) is a cause of polynomial p(x) = x³ – 3x² + 5x – 15.

Solution:

For (x–3) is a factor of p(x),

p(3) must be zero by the factor theorem.

Now p(3) = 3³ – 3(3) ² + 5(3) – 15

= 27 – 27 + 15 – 15 (by equating the known polynomial)

= 0

Hence (x–3) is a factor of the given polynomial.

3) Calculate 105 * 104 without multiplying directly.

Solution:

105 * 106 = (100 + 5) * (100 + 4)

= (100)² + (4 + 4) (100) + (5 * 4) (by using identity)

= 10000 + 800 + 20

= 10820.

4) Expand: (3a + 2b – 3c) ²

Solution:

Comparing the given expression (3a + 2b – 3c)² with (a + b + c) ²,

From the expression we have,

a = 3a

b = 2b

c = -3c

(3a + 2b – 3c) ² = (3a) ²+ (2b) ² - (3c) ²+ 2(3a) (2b) + 2(2b)(-3c) + 2(-3c)(3a)

Therefore, the equation becomes as,

= 9a² + 4b² + 9c² + 12ab² -12bc -18ca.

These are the examples of algebra connections equations.

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Exam questions to practice for algebra 1 final exam practice:

Problem 1:

Solve the equation 6x + 8 = 38 for x.

Answer: 5

Problem 2:

Solve the equation 8x + (40 ÷ 8) = 53 for x.

Answer: x = 6

Problem 3:

Solve the equation for x2 + 8= 89 for x.

Answer: x = 9

Problem 4:

Solve the equation 5x2 + 15x + 10 = 0 by using quadratic formula.

Answer: x1 = -1 and x2 = -2

Problem 5:

Solve the equation 5x2 + 15x + 10 = 0 by using factorizing method.

Answer: x1 = -1 and x2 = -2

Problem 6:

Solve the equation 3x2 + 4x + 1 = 0 by using quadratic formula.

Answer: x1 = -0.3 and x2 = -1

Problem 7:

Solve the equation 3x2 + 4x + 1 = 0 by using factorizing method.

Answer: x1 = -0.3 and x2 = -1

These are algebra 1 final exam practice as for exam preparation.

Plane Geometry Practice

Introduction for plane geometry practice:

In this branch of mathematics,we study to measure a line segment, find the area of a plane surface and perimeter and the volume of solids.Proper units are chosen for different measurements. In this chapter we will derive standard formula for calculating area of the rectangle, the perimeter of a rectangle,the area of a square of a geometry practice problem.


Area and perimeter of plane geometry practice:

Area of a rectangle = length × breadth

A = l × b

perimeter of a rectangle = 2 lengths + 2 breadths

P = 2l + 2b

Area of a square = side × side

A = a2

Perimeter of a square = 4 × side

P = 4a




Examples for plane geometry practice problems:


Example 1:

Find (a) area and (b) perimeter of a square whose sides is 15 cm.

Solution :

(a) Side of the square, a = 15 cm

Area of the square, A = a2

= a × a

= 15 × 15

= 225 sq.cm (or 225 cm2)

(b) Perimeter of the square, P = 4a

= 4 × 15 cm

= 60 cm

Example 2:

Find the area and perimeter of a rectangle whose length is 4m and breadth is 50 cm.

Solution :

Length of the rectangle, l = 4 m or 400 cm

Breadth of the rectangle, b = 50 cm

(a) Area of the rectangle, A = l × b

= 400 cm × 50 cm

? Area = 20,000 sq.cm.

(b) Perimeter of the rectangle, P = 2l + 2b

= 2 × 400 cm + 2 × 50 cm

= 800 cm + 100 cm

? Perimeter = 1,000 cm

Example 3 :

The perimeter of the floor of a rectangular hall is 24 m. Its rectangle length is 9 meter. Find its area of rectangle.

Note : To find area we should know the measurements of length and breadths. Here rectangle  length is given.

Hence we should find the breadths from the perimeter.

Solution :

2 length + 2 breadth = perimeter

2 × 9 + 2b = 24

18 + 2b = 24

2b = 24 – 18

2 b = 6

b =6/2 = 3

Now the area of the rectangle A = l b

= 9 m × 3 m

= 21 sq.m.

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Practice problem for plane geometry :


Find the cost of fencing a rectangular park of lengths 170 m and breadth 100 m at the rate of Rs.9 per metre.

Answer:Rs.2800

Probability Maths

Introduction to probability maths:

Probability is a word that is been used commonly in our day today life without going into the details of its actual meaning. We come across statements like:

(i) Probably it may rain to day.

(ii) He may possibly join politics.

(iii) Argentina team has good chance of winning world cup.

(iv) She is probably right.

In these kinds of statements, we generally use the terms: possible, probable, chance, likely etc. All these terms convey the same meaning that the event is not certain to take place or, or other words there is uncertainty about the occurrence of the event in question.

Now let us see few problems of this kind..


Example problems on probability maths:


1. Find the probability that a leap year selected at random will contain 53 Sundays.

Soln: We know that in a leap year there are 366 days.

They can be written as, 366 days = 52 weeks and 2 days.

Thus a leap year has always 52 Sundays.

The remaining two days cane as follows:

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday.

Clearly, there are seven events are associated with this random experiment. Let X be the event that a leap year has 53 Sundays. This is possible if the last two days are either Sunday and Monday or Saturday and Sunday.

Therefore the required probability is `2/7` .

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More example problems on probability maths


2. Let X and Y throw a pair of dice. If X throws 9, find Y’s chance of throwing a higher number.

Soln: To throw a value greater than 9, we have the following possibilities:

{(4,6),(5,5),(5,6),(6,4),(6,6),(6,6)}. Therefore, the number of chances is 6.

When two dice are thrown, the total possibility is 36.

Hence the required probability is `6/36 = 1/6` .

3. A letter is chosen at random from the letters of the word “ASSASSINATION”. Find the probability that the letter chosen is a consonant.

Soln: The word “ASSASSINATION” has 13 words.

Here the consonants are {s, s, s, s, n, t, n}. Hence we have 7 consonants.

Therefore the probability to get a consonant is `7/13` .

Decimal Skill Practice

Introduction of decimal skill practice:
Decimal is one of the types of the numbers system. The point is called decimals. Here we are going to discuss about adding decimals, subtracting decimals, multiplication decimals and dividing decimals. And we are going to learn how to improve our decimal skills and practice problems here .Decimals are also fractional numbers. For example 0.5 is the same as the fraction 5/10.


Decimal skill practice:


There are 4 different types of decimal skills.

That is ,

Adding and subtracting decimal skills.
Multiplication decimal skills.
Division decimal skills.


Addition and subtracting decimal skill practice:

Its commonly used to find the total of two numbers. Generally addition and subtraction of decimals is totally based on adding and subtracting whole numbers. But here we need to concentrate to place the decimals

Example of adding decimals:

Example 1:

Add the decimal value: 12.74, 15.11

Solution:

4 2 . 15

1 3.  1 1    (+)

-------------

5 5  .  2 6



Example of subtracting decimals:

Example 2:

Subtract the decimal value 243.87, 23.56.

Solution:

3 2 1 . 8 7

1 5 6 . 5 6   (-)

------------------

1 6  5 . 3  1


Example problem for multiplication and division decimal skill practice :


Example of multiplication decimals:

Multiplication is a product of two terms.

Example 1:

Multiply 7.9 * 1.3

Step 1: First remove the decimal point.

Step 2: And then perform multiplication operation.

Step 3: lastly add the decimal point in the answer.

Solution:

79

13   (*)

----------

2 3 7

7 9

-----------

1 0 2 7

Answer: Form the given question we have two decimal points. So we need to put the 2 decimal points to the answer So, the answer is 10.27.

Dividing decimals:

Its also the equal of whole number division.

Example 2:

10.5 divided by 0.5

Step1: Convert the decimal values in to the whole number.

Step 2: So, multiply 10 with both top and bottom.

Step 3: `(10.5 times 10)/(0.5 times 10)` .So we get `(105)/(5)`

Step 4: Here we can perform the division process.

Step 5: Therefore the answer is 21.

Practice Integers

Introduction to practice integers:

In mathematics, integers is one interesting topics in number representation. Integer has a set of numbers in which there are two types of integers that are non negative integers and negative integers. Integers have complete entity or unit.Integers perform different types of arithmetic operations such as addition, subtraction, multiplication and division. Let us see some example problems and practice problems.

Example integers:

7254, - 8564, 0 etc,


practice integers - Example problems:


Different types of example problems for integers are,

Example Problem for addition:

Perform addition for the given two integers

1087 + 5892

Solution:

Given two integer are

1087+ 5892

Here we add 1087 into 5892, and then we get the result

=1087 + 5892

=6979

Solution to the given two integers is 6979.

Example Problem for subtraction:

Perform arithmetic operation subtraction for the given two integers

854 and 789

Solution:

Given two integer numbers are

854 – 789

Here we subtract 854 into 789, and then we get the result

=854 – 789

=65

Solution to the given two integers is 65.

Example Problem for multiplication:

Perform arithmetic operation multiplication for the given two integers

264 × 147

Solution:

Given two integer numbers are

264 × 147

Here we multiply 246 into 147, and then we get the result

=264 × 147

=38808

Solution to the given two integers is 38808.

Example Problem for division:

Perform arithmetic operation division for the given two integer numbers

735/15

Solution:

Given two integer numbers are

735 / 15

Here we divide 735 by 15, and then we get the result

=735/ 15

=49

Solution to the given two integers is 49.

Example Problem for addition:

Perform arithmetic operation addition for the given two integer numbers

- 364 + 785

Solution:

Given two integer numbers are

- 364 + 785

- 364 is negative number and 785 is positive number

Here we add - 364 into 785, and then we get the result

= - 364 + 785

=461

Solution to the given two integers is 461.


practice integers - Practice problems:


Perform arithmetic operation for the given two integers are,

i). 786 – 225

ii). 621 + 106

iii). 45 × 21

iv). 1250 / 50

v). – 897 × 154

vi) – 456 + 264

Solution:

i). 561

ii). 727

iii). 945

iv). 25

v) – 138138

vi) – 192

Mixed Numbers Practice

Introduction for Mixed numbers:

In a fraction if numerator is greater than denominator then this kind of fraction is know improper fraction the improper fraction in standard form is known as mixed number.

A mixed number consists of

A whole number.

Proper fraction.

A `b/c`

Here A is whole number.

Now let see problems on mixed numbers operations.


Mixed numbers practice problems:


Practice problem 1.

Find the sum of two mixed number 6`1/3` and 7`1/3`

Solution:

The given mixed numbers are  6`1/3` and  7`1/3`

Initially to perform any operations on mixed numbers we must convert it to fraction

6`1/3` in fraction

Multiply 6 and 3 and add with 1

6`1/3` = `(19+1)/3`

=`20/3`

Now Convert the 7`1/3`

Multiply 7 and 3 and add with 1

7`1/3` =` (21+1)/3`

=`22/3`

6`1/3` +7`1/3` =`20/3` +`22/3`

=`(20+22)/3`

= `42/3`

This can be simplified has 14

Practice problem 2.

Find the difference of two mixed number 8`1/3` and 9`1/3`

Solution:

The given mixed numbers are 8`1/3` and 9`1/3`

Initially to perform any operations on mixed numbers we must convert it to fraction

8`1/3 ` in fraction

Multiply 8 and 3 and add with 1

8`1/3`  =` (24+1)/3`

=`25/3`

Now convert  9`1/3` in fraction

Multiply 9 and 3 and add with 1

9`1/3` = `(27+1)/3`

= `28/3`

8`1/3` - 9`1/3` = `25/3` -`28/3`

=`(25-28)/3`

=`-3/3`

This can be simplified has  -1

Practice problem 3.

Find the product of 5` 1/3` and 6`1/3`

Solution:

The given mixed numbers are 5`1/3` and 6`1/3`

Initially to perform any operations on mixed numbers we must convert it to fraction

5`1/3` in fraction

Multiply 5 and 3 and add with 1

5`1/3` =`(15+1)/3`

=`16/2`

Now convert the nextmixed numbers  6`1/3 `

Multiply 6 and 3 and add with 1

6`1/3`  =` (18+1)/3`

= `19/3`

5`1/3` `xx` 6`1/3` =`16/3` `xx` `19/3`

= `(16xx19)/(3xx3)`

= `304/9`

Understanding Completing the Square Formula is always challenging for me but thanks to all math help websites to help me out.

Some more mixed numbers practice problems:

Practice problem 4.

Find the sum of two mixed number 5`1/5` and 6`1/5`

Solution:

The given mixed numbers are 5`1/5` and 6`1/5`

Initially to perform any operations on mixed numbers we must convert it to fraction

5`1/5` in fraction

Multiply 5 and 5 and add with 1

5`1/5` = `(25+1)/4 ` =`26/4`

6`1/5`  `rArr` Multiply 6 and 5 and add with 1

6`1/5`  =` (30+1)/5`

= 3`1/5`

5`1/5` + 6`1/5` =`26/5` +3`1/5`

=`(26+31)/5`

= `57/5`

This can be simplified has 11.4

These are some examples of mixed numbers.

Multiplying Rational Expressions

Multiplying rational expressions I (math)

A rational expression  is an algebraic expression  which is in the  form P/Q, where P and Q are simpler expressions P and Q are usually polynomials.  The denominator Q is not zero. It is the quotient of two polynomials. It is important to remember that the divisor cannot be zero. A quotient is the answer to a division problem. Polynomials are expressions which has the sum of powers with at least one variable.  This variable is multiplied by coefficients.

Examples of rational expressions are :

2/5 ,    3x + 9    ,        x^2 + 6x + 3
X + 3               x^2 +x +2

Multiplication Rational Expressions

When  we multiply the rational-expressions the numerators  and the denominators  are to be multiplied.   If P,Q,R  and S polynomials   then

P / Q   .  R / S  =   PR / QS  When Q is not equal to zero and S is not equal to 0.

Here we multiply the  numerators and the denominators.   Before multiplying it has to be seen if they can be reduced.   If reduction is possible then  it has to be done before multiplication.

Multiplying Rational Expressions Examples

Multiply the rational expression given below

18a3       x     5b4

20b2                   6a

Before multiplying the above expression it can be simplified.  Here we can cancel the numerator with the denominator or denominator with the numerator


18 a3       x     5 b4

20 2                   6a

18 can be divided by 6 and we get 3 and   a3 divided by a  = a2   and  b4 /b2 =b2 .  Thus we have 3a2 b2 in the numerator and 4 in the denominator.



3a2        x   b2                                           3a2  b2
=                 4
4               1

Multiply 3a2 with b2 and 4 with 1.  We get 3a2b2  in the  numerator and 4 in the denominator.  In the above problem first the expression was simplified then it was multiplied.  It can also be done vice versa.  First the rational-expressions can be  multiplied  that is numerator with numerator 4xy^2  with 2x and denominator with denominator 3y with 4y.
Then it can  be simplified.

4xy^2         2x
x
3y           4y

4x y^2. 2 x      =                4 . 2x^2 y^2
3y. .4y                             4. 3 y^2

Understanding formula for quadratic equation is always challenging for me but thanks to all math help websites to help me out.

When multiplying the numerator we get 8x^2 y^2   and when multiplying the denominator we get 12 y^2


4.2 x^2 y^2

4.3y^2

The 4 in the numerator and denominator gets cancelled. The y^2 in the numerator and denominator also gets cancelled. And we are left out with

2 x^2
3

Which gives the final answer.

Practice Percentages

Introduction for practice percentage:

The word “percent” is consequent from Latin. It was at first “per centum”, which means “by the hundred”. Thus the statement is frequently complete that “percent means hundredths”.

Percentage deals with the collection of decimal fraction whose denominators are 100 – that is, fractions of two decimal spaces Since hundredths be used so regularly, the decimal position was drop and the symbol % be located after the number and understand “percent”. Thus, 0.25 and 25% represent the same value, 25/100.  The first is read “25 hundredths”, and the second is read “25 percent”. Both mean 25 parts out of 100.

Originally, percent is used in discussing relative values. For example, 25 percent may convey an idea of relative value or relationship. To say “35 percent of the crew is ashore” gives an idea of what part of the crew is gone, but it does not tell how many of discussing the percentage.

I like to share this formula percentage change with you all through my article.

Practice percentage for steps and example problems:


Steps for practice percentage:

STEP 1: Start the percentage x/100 = is/of. X is the percentage (over 100 of course), "is" refers to fraction, and "of" refers to entire.

STEP 2: In the question "80 is to 40 percent of what number? x=40, is=40 ("80 is"), and of = the unknown ("of what number"). Therefore write 40/100=80/x.

STEP 3: Cross multiply. You will have a constant value on one side and multiply a variable on the other side. Here it is 40x=8,000.

STEP 4: Solve for x. Here, x = 8,000/40 = 400, So x = 400.

Example problems for percentage:

Example problem 1:

The $169.99 poodle I purchased was on sale for 10% off, what did I pay for my poodle?

Solution:

$ 169.99 x 10 / 100

= $ 169.99 / 10

= $16.999

So Pay for poodle = $169.99 - $16.99

Answer = $153

Example problem 2:

I got 40% off when I purchased a rare comic book regularly priced at $74.50. How much did I pay?

Solution:

$74.50 x (40 / 100)

= $74.50 x (4 / 10)

= ($74.50 x 4) / 10

= $298 / 10 = $29.8

Pay for books = $74.50 - $29.8

= $44.7

Example problem 3:

Our take out dinner was $74.95 but we got 20% off because we picked it up. What did our meal end up costing us? $59.96

Solution:

$74.95 * (20 / 100)

= $74.95 * (2 / 10)

= ($74.95 * 2) / 10

=$149.9 /10 = $14.99

End of meal cost = $74.95 - $14.99

Answer = $ 59.96


Practice problem for percentage:


Practice problem 1:

My flight was $199.99 but I got 30% off because the plane wasn’t full. What did I pay?

Answer: $139.99

Practice problem 2:

$40.50 video games were on sale for 30% off, how much are they now?

Answer: $28.25

Practice problem 3:

My new cell phone cost me $190.00 but when I signed a 2 year plan, I got 20% off so it only cost me?

Answer: $152.00

Practice Problem 4:

It’s usually $16.50 to go to the show, but if you go on Tuesday, there’s a 40% discount, what does it cost to go to the show on Tuesday?

Answer: $9.90

Place Value Practice

Introduction for place value practice:

Place value is a second important concept in mathematics. It refers to the value of the digit, relative to its location in the number. The place value can be identified for any given digit in a written number, in the number 7403, the digit 4 is in the hundreds place and stands for 400. The 3 in 34 is not the same as the 3 in 23, even though both digits are 3. Is this topic Statistics Problems hard for you? Watch out for my coming posts.

This is because the 3 in 34 is in the tens place and therefore worth 30, while the 3 in 23 is in the ones place and worth only 3. Child is developing an understanding of place value up to at least the thousands. (This also means that she needs to practice skip counting starting with any number up to one thousand: 326,336,346)

Example: The number 3725 is read as three thousand seven hundred twenty- five. In it, there are 3 thousands, 7 hundreds, 2 tens, and 5 ones. A number with only ones has only one digit, one with tens two digits, and one with thousands has four digits.

Example problem for place value practice:

5 ones =             5

2 tens =            20

7 hundreds =     700

3 thousands =  3000

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Example problem for place value practice:


Read the numbers, and then tell the place and value of the underlined digit; Up to 10 millions.

1. 5,760,000

What place is the underlined digit in?

Ten thousands place

What is the value of the underlined digit?

60,000

2.   86,003,040

What place is the underlined digit in?

Millions place

What is the value of the underlined digit?

6,000,000

3.   4,954,560

What place is the underlined digit in?

Tens place

What is the value of the underlined digit?

60

4.    5,520,254

What place is the underlined digit in?

Millions place

What is the value of the underlined digit?

5,000,000

Practice Area Problems

Introduction

Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed curve. The surface area of a 3-dimensional solid is the total area of the exposed surface, such as the sum of the areas of the exposed sides of a polyhedron. (Source: From Wikipedia).

Here we are going to solve some practice problems, to find the area of basic regular shapes.

Practice problems to find the area

Square

The area of a square is given by the formula, A = a ^2 square units

Practice problem 1

Find the area of a square, with side lengths equal to 4 ft.

Solution

Area of the square = a ^2 square units

= 4^2

= 16 square feet

So teh area of the square = 16 square feet.

Rectangle

The area of a rectangle can be calculateed by the following formula,

Area of  a rectangle, A = lb square units

Where, l = length and b = breadth of the rectangle.

Practice problem 2

What is the area of a 4 in by 6 in rectangle.

Solution

Area of  a rectangle, A = lb square units

= 6 * 4 square inch

= 24 square inch

Answer: The area of the rectangle = 24 square inch

Is this topic solve equation by factoring hard for you? Watch out for my coming posts.

Few more practice problems to find the area


Circle

The area of a circle = `pi` r^2 square units

Here, r is the radius of the circle.

Practice problem 3

Calculate the area of a circle with radius 10 centimeter.

Solution

Area of a circle = `pi` r^2 square units

= 3.14 * 102 square centimeter

= 314 square centimeter

So the area of the circle is 314 square centimeter.

Triangle

Area of a triangle = `1/2` b h square units

Where, b = base of the triangle

h = perpendicular height of the triangle

Practice problem 4

Find the area of a triangle whose base length is 5 meter and height is 10 meter.

Solution

Area of a triangle =  `1/2` b h square units

= square meter

= 25 square meter

The area of the given triangle = 25 square meters

Solving Probability Practice

Introduction to Probability:
Numerical measure of the likelihood of an event to occur is called as Probability. The probability should be a range in between 0 and 1.For solving probability practice problems we must know the following formula,

Probability of an event   =    number of times an event occurs / total number of outcomes

Consider an observation with 'n' possible ways and out of them in 'm' ways if the event 'A' occurs,then the probability of occurrence of the event 'A' is given by P(A) = m/n.

Let us workout some practice problems on solving probability in the following sections.

I like to share this Probability Problems and Solutions with you all through my article.

Probability on a coin problem:


Here Tossing a coin is a random experiment. When you toss a coin, you may get head or tail.Let us see how to  solving the coin problem.

1: What is the probability of getting Head when you toss a coin?

We can have 2 out-comes for tossing a coin :  Head or tail.

Here the event is getting Head. So we have 1 head.

So probability of getting head is one out of two outcomes.

Probability of getting Head    = number of head event occurs / total number of outcomes

= 1/2

= 0.5

So the probability of this practice problem is 0.5


Probability on Card problem:


In 52 cards, there are 13 spades, 13 clovers, 13 diamonds and 13 hearts , 26 cards are red in color (diamonds and heart) and 26 are black (spade and clover).Let us see how to  solving the card problems.

(1)  What is the probability that you get a red card?

The event is getting a red card. There are 26 red cards.

Total sample spaces are 52(52 cards).

Probability of getting a red cards = number of red cards / sample space

= 26/52

=1/2

= 0.5

(2) What is the probability of getting a spade when you draw from a well shuffled pack of 52 cards?

Number of Spade cards = 13

Total Sample space       = 52

Probability of getting a spade card = number of spade / total sample space

= 13/52

= 1/4

= 0.25

(3) What is the probability of getting a black queen in a pack of 52 cards?

In a pack of 52 cards, there are 2 Black Queens (spade 1, clover 1)

Total Sample space = 52

Probability of getting a black queen = number of black queen / total sample space

= 2/52

= 1/26

= 0.04

I have recently faced lot of problem while learning Fraction Simplifier, But thank to online resources of math which helped me to learn myself easily on net.

Practice Problem:


The following practice problems on solving probability will helps for good understanding.

Practice problem 1: What is the probability of getting Tail when you toss a coin?

Practice problem 2: What is the probability of getting a diamond when you draw from a well shuffled pack of

52 cards ?

Answer key:

Practice problem 1:  0.5

Practice problem 2: 0.25

Study Practice Algebra Problems

Introduction of Study Practice Algebra Problems:

Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorial, and number theory, algebra is one of the main branches of pure mathematics. (Source: Wikipedia)

Looking out for more help on Discriminant Function in algebra by visiting listed websites.

Study Example Algebra Problems:


Study practice algebra problem 1:

Simplify the equation 5(a -3) + 4b - 7(a -b -3) + 5.

Solution:

5(a -3) + 4b - 7(a -b -3) + 5

Multiply factors.

5a - 15 + 4b -7a + 7b + 21 + 5

Group like terms.

-2a + 11b + 11

Answer:   - 2a + 11b + 11

Study practice algebra problem 2:

To find the 4x - 8y = 9 equation on x-intercept.

Given:

4x - 8y = 9

Solution:

4x - 8y = 9

To find the x intercept of y = 0 and solve for x.

4x - 0 = 9

Solve the value of x.

x = 9 / 4

The x intercept is at the point (9/4, 0).

Answer: x intercept is at the point (9/4, 0).

Study practice algebra problem 3:

16x + 2y = 32.

y + 8 = 16x

Find the x and y value.

Solution:

16x + 2y = 32 (equation 1)

y + 8 = 16x   (equation 2)

Step 1: To choose the equation where the coefficient of the variable is 1.Choose equation 2 to isolate the variable y

y = 16x – 8 (equation 3)

Step 2: From equation 3, we know that y-value is the same as the 16x – 8. We can to substitute the variable y-values in the equation 1 with 16x – 8.

16x + 2 (16x – 8) = 32

Step 3: Remove the brackets by using the distributive property

16x + 32x – 16 = 32

Step 4: To combine the terms

48x – 16 = 32

Step 5: To Isolate the variable of x

48x = 48

x = 48 / 48

x = 1

Step 6: Substitute x = 1 in the equation 3 to get the y-value.

y = 16 (1) – 8

=16 – 8

y = 16 - 8

y = 8

Step 7: To check the answer with the equation (1)

16(1) + 2 (8) = 16+ 16 = 32

Answer: x = 1 and y = 8

I have recently faced lot of problem while learning Antiderivative Calculator, But thank to online resources of math which helped me to learn myself easily on net.

Study Practice Algebra Problems:


Practice Problem 1:

Find the x and y value 12x + 2y = 20, y + 8 = 12x

[Answer: x = 1 and y = 4]

Practice problem 2:

Solve for x and y for the following equations 5x + y = 8, 6x + y = 7

[Answer: x = -1 and y = 13]

Practice Problem 3:

Simplify the equation 7(x -3) + 5y - 3(x -y -3) + 5.

[Answer:   4x + 8y – 7]

Practice Algebra Class

Introduction to practice algebra class:

Algebra class deals with the Pre-algebra, Functions and Graphs, Integers, Rational Numbers, Equations in One Variable, Equations in Two Variables, Simultaneous Equations, Problem Solving etc. in arithmetic, we use digit like 1,2,3, etc to represent numbers. In algebra we use numbers as well as letters of the alphabet such as a, b, c, etc for any numerical values we choose. The basic thought of the algebra is balancing the algebraic equations on both sides of the give equations in the problems. Algebra will perform the four fundamental operations such as addition, subtraction, multiplication and division. In this article we shall discuss with practice algebra class


Sample problem on practice algebra class


Problems of algebra class on linear equations:

Example 1:

Solve the linear equation for algebra class

11 x - 15 = 19 x - 23

Solutions:

11 x - 15 = 19 x - 23

Subtract 19 x from both sides of the above equation, we get the below term

-8x – 15 = -23
Add 15 to both sides of the above equation, we get the below term

-8x – 15 + 15 = - 23 + 15

-8x = - 8
Divide both sides by -8 of the equation, we get

X = 1

The answer is x = 1

Test the solution for the given equation. Replace with in the above given equation for x. If the left side of the equation sum value is equal to the right side of the equation sum value after the replacement of x value, you have got the exact answer.

Left side of the equation:
11(1) - 15 = - 4

Right side of the equation:
19(1) - 23 = - 4

Practice word problems of algebra class

Example 2:

A bag contains ten, five and two rupees currencies. The total number of Currencies is 30 and the total value of money is Rs.135. If the second and third currencies are interchanged the value will decrease Rs.8. Find the number of currency in each sort.

Solution :

Let x, y, z be the number of Rs.10, Rs.5 and Rs.2 currencies respectively.

The total number of currencies is 30 => x + y + z = 30--------(1)

If the total value of money is Rs.135 => 10x + 5y + 2z = 136--------(2)

If the II and III of currencies are interchange the value will be decreased by Rs.8

=>10x + 2y + 5z = 135 – 8 = 127----------------(3)

10x + 2y + 5z = 127

10 x (1) – (2) =>     5y + 8z = 106------------(4)

(2) – (3)        =>     3y - 3z = 9

y – z = 3------------------(5)

Let us solve (4) and (5)

(4) =>      5y + 8z = 106

8 x (5)   ⇒ 8y – 8z = 24

13y = 130

y = 10

Substituting y = 10 in (5) we get

z = 7

Substituting z = 7, y = 10 in (1) we get

x + 10 + 7 = 30 => 30 – 10 – 7 or x = 13

Number of Rs.10, Rs.5 and Rs.2 currencies are respectively 13, 10 and 7.

Is this topic Fraction to Decimal Calculator hard for you? Watch out for my coming posts.

Practice problem for algebra class:

Problem 1:

2x + 11 = 20 - x

Answer: x = 3

Problem 2:

7x + 11 = 19 + 3x


Answer: x = 2

Statistics Practice Exam

Introduction to statistics practice exam:

Statistics is the science of making effective use of numerical data relating to groups of individuals or experiments. It deals with all aspects of this, including not only the collection, analysis and interpretation of such data, but also the planning of the collection of data, in terms of the design of surveys and experiments.

A statistician is someone who is particularly versed in the ways of thinking necessary for the successful application of statistical analysis. Often such people have gained this experience after starting work in any of a number of fields. There is also a discipline called mathematical statistics, which is concerned with the theoretical basis of the subject.

Let us see several statistics based solved problems and statistics practice exam. Statistics practice exam with solutions which would be helpful for students to evaluate their performance.

Concepts on statistics practice exam:


Before going for a statistics practice exam, let us recall how to find mean deviation. The steps to find the mean deviation are as follows.

Step 1: find the mean for given data
`barx` = (sum of all variables) /( given number of  variables).

Step 2: find the deviation of the respective observations from mean `barx` ,  in other words  xi -  `barx`
Step 3: find the modulation of deviation ,     | Xi - `barx` |
Step 4: mean deviation of the mean,
M.D.(`barx` ) = `sum` n i-1 | xi - x| / n

Example1:

Find the mean deviation about the mean for the following data:

12, 5, 16, 14, 11, 5, 10, 15

Solution:

We proceed step-wise and get the following:

Step 1 Mean of the given data is `barx`

`barx`  = `(12+5+16+14+11+5+10+15)/8` = `88/8` = 11

Step 2 The deviations of the respective observations from the mean x, i.e., xi – x are

12–11,5–11,16–11,14–11,11–11,5–11,10–11,15–11,

or 1,–6,5,3,0,–6,–1,4

Step 3 The absolute values of the deviations, i.e.,|xi − x |are

1,6,5,3,0,6,1,4

Step 4 The required mean deviation of the mean is

M.D. (`barx`   ) =`sum` 8 i-1 |xi-x| / 8

=`(1+6+5+3+0+6+1+4)/8` = `26/8` = 3.25.

Example2:

Find the mean deviation about the mean for the following data:

15, 5, 12, 16, 12, 4, 9, 16,7.

Solution:

We proceed step-wise and get the following:

Step 1 Mean of the given data is `barx`

`barx`  = `(15+5+12+16+12+7+9+16+7)/9` = `99/9` = 11

Step 2 The deviations of the respective observations from the mean x, i.e., xi – x are

15–11,5–11,12–11,16–11,12–11,7–11,9–11,16–11,7-11.

or 4,–6,1,5,1,–4,–2,5,-4

Step 3 The absolute values of the deviations, i.e.,|xi − x |are

4,6,1,5,1,4,2,5,4

Step 4 The required mean deviation of the mean is

M.D. (`barx`   ) =`sum` 8 i-1 |xi-x| / 8

=`(4+6+1+5+1+4+2+5+4)/8` = `32/8` = 4.

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Practice on statistics practice exam:

Statistics Practice exam:

1. Find the mean deviation for the mean of the following data:
6, 7, 10, 12, 13, 4, 8, 12

2. Find the mean deviation for the mean of the following data :
12, 3, 18, 17, 4, 9, 17, 19, 20, 15, 8, 17, 2, 3, 16, 11, 3, 1, 0, 5

Answers for the above questions of statistics practice exam:

1.   2 75

2.   6.2

Practice Abstract Algebra Exams

Abstract for algebra:

In algebra the algebraic expressions represent mathematical ideas and operations in a short way using symbols and variables.The algebra equation states that one expression names the same number as another expression. The symbol in algebra which can be assigned different numerical values is called a variables. Algebra uses known quantities to find the unknown quantities. Here it is about the practice for abstract algebra exams. I like to share this algebra 2 problems and solutions with you all through my article.


practice abstract algebra exams - Examples:


practice abstract algebra exams - Problem 1: Simple equations:

solving simple equation: 7x + 13 = 4x + 43.

Solution:

since 7x is larger than 4x,

we decide to shift the letter terms to left and the number terms to right side.

7x + 13 = 4x + 43

we want to remove 4x from the right side,

we can do so by subtracting 4x from both sides.

7x + 13 – 4x = 4x + 43 – 4x

3x + 13 = 43

we want to remove 13 from both sides,

we do so by subtracting 13 from both sides.

3x + 13 – 13= 43 – 13

3x = 30

Dividing both sides by 3.

3x/3 = 30 / 3

x = 10

check to substituting the x in as:

left side: 7(10) + 13 = 70 + 13 = 83

right side: 4 (10) + 43 = 40 + 43 = 83

practice abstract algebra exams - Problem 2: Algebraic expressions:

Simplify the algebraic expression: 17m² – 9m + 5m – 3m² – 7m + 11

solution:

By rearranging the given terms as,

we have,

=17m² – 3m² + 5m – 9m – 7m + 11

By equating, we get

= (17 – 3) m² + (5 – 9 – 7) m + 11

To simplify the equation as

=14m² + (– 4 – 7) m + 11

=14m² + (–11) m + 11

=14m² – 11m + 11.

This is the equation for the given algebraic expression.

practice abstract algebra exams - Problem 3: Algebraic identities:

Evaluate 102 × 104 without directly multiplying the two given numbers.

Solution:

Taking 102 as (100 + 2) and 104 as (100 + 4)

we get the multiplier as

= 102 × 104

Taking the multiplier as

= (100 + 2) × (100 + 4)

By multiplying and adding the identities we get,

= 100² + (2 + 4) 100 + 2 × 4

= 10000 + 6 × 100 + 8

= 10000 + 600 + 8

= 10608

This is the solution for the given algebraic identities.

Understanding Derivative of Sin Squared x is always challenging for me but thanks to all math help websites to help me out.

practice abstract algebra exams - Problems for practice exam:


1. solving simple equation:18 – 5x = 3x – 6.

Answer: x = 3

2. Evaluate the algebraic identities 103 × 105 without directly multiplying the two given numbers.

Answer: = 10815

3. Simplify the algebraic expression: 16m² – 9m + 5m – 3m² – 7m + 12.

Answer: =13m² – 11m + 12.

Study Maths Advantage

Introduction to study maths advantage:

The study of maths is important nowadays in everyday life.  For developing the skills of the individual persons, maths is very important. Every industry needs mathematics today. Generally math is invented for the purpose of real time application for calculation purpose. In this article,  we are going to see about the maths advantage using real time application with some example problems.

Advantage to study maths

Advantage:

Employment
Science
Technology
Medicine
Example for the study of maths advantage:

In computing industry, complex programs are created using the mathematics.
In Cryptography, the encode and decode are done using the mathematics.
In Internet, credit and debit card are also done using the maths.
For calculating the area of land, we have formula in mathematics. By using this formula, it will be easy to calculate the area.
Speed can also be calculated using the formula in mathematics.
Distance can also be calculated using the formula in mathematics.

Example problems for study maths advantage:

Real Time Example:

Problem 1: Find the area of the given square, where, width = 24, height = 35.

Solution:

Step 1:Given:

Width = 24, Height = 35

Step 2: To Find:

Area of the square.

Step 3: Formula:

Area of the square = width `xx` Height

Step 4: Solve:

Area of the square = width `xx` Height

= 24  `xx` 35

= 840

Result: Area of the square = 840

Problem 2: Find the speed of the given car, where, distance = 345, Time = 30.

Solution:

Step 1:Given:

Distance = 345

Time = 30.

Step 2: To Find:

Speed of the car

Step 3: Formula:

Speed of the car = `(Distance)/(Time)`

Step 4: Solve:

Speed of the car = `(Distance)/(Time)`

= `345/30`

= 11.5

Result: Distance of the car = 11.5

Understanding Dilation Geometry is always challenging for me but thanks to all math help websites to help me out.

Practice problems for study maths advantage:


Problem 1: Find the speed of the given car, where, distance = 240, Time = 20.

Answer: 12

Problem 2:  Find the area of the given square, where, width = 34, height = 46.

Answer: 1564

Pre Algebra Practice Tests

Introduction to online pre algebra practice tests :

Algebra , most important topic in mathematics and it’s very easy to calculating, concerning the study of structure, quantity and relation. We can use arithmetic operation and the letter and symbols are referred to as Variables.  Algebra is derived from Arabic language Al-jabr. Algebra also performs addition, subtraction, multiplication, division. For example,   2+3=3+2 is same as p+q=q+p. Let us see about online pre algebra practice tests.


Problems for pre algebra:


Example 1 for online pre algebra practice tests problem:

Multiply -2a2bc, 2a2b and −1/4

Solution for online pre algebra practice tests problem:

=>( - 2a2bc ) ( 2a2b ) ( -1/4 )

=>( -4 a4b2c ) ( - 1/4 )

=>a4b2c = a2bc * a2b

Example 2 for online pre algebra practice tests problem:

Find factors of 3x2y

Solution for online pre algebra practice tests problem:

=>( 1 * 3x2y ) (3x2y ) ( 3x * xy )  (3xy * x ) ( x2 * 3y ) (y * 3x2 )

=>(1,3x2y), (3x2y), (3x,xy), (3xy, x),(x2 ,3y),(y,3x2 ).

Example 3 for online pre algebra practice tests problem:

Simplify : ( 2x + 2) ( 4x – 10 ) +  25

Solution for online pre algebra practice tests problem:

First of all remove the paranthesis by multiplying the terms inside the bracket.

=>8x2 – 20x + 8x – 20 + 25

=>8x2 – 12x + 5

Having problem with Area of a Kite keep reading my upcoming posts, i will try to help you.

More Example Problems


More Example Problems For Online Pre Algebra Practice Tests:

1.Evaluate (a) (2x + 3y)2 (b) (3 / 2x − 6y)2 (c) (x2 + 2y2 )(x2 − 2y2 )

Answer: (a)4x2 +12xy + 9y2 (b) 9/4x2 18x y 36y2 (c)x4 − 4y4

2.Factorize 6x3 + 8 x2y.

Answer:2x2×3x+2x2×4y = 2x2(3x+4y)

3.Find the greatest common factor of 2x5 and 12x2 and then factorize 2x5+12x2

Answer:2x2 (x3 + 6)

4.Factorize 4m2 - 4m +1.

Answer:(2m - 1)2 =(2m- 1) (2m-1)

5.Factorize x2- 2xy + y2 – 9

Answer:(x-y + 3) (x-y-3)

Practice Algebra i

Introduction to practice algebra i Introduction:

Algebra i follows the rules of arithmetic in which letters representing numbers. The numbers are the constants. Practice of Algebra  i includes matrices, real and complex numbers etc. When we move from Arithmetic to Algebra i we will look something like this:  Arithmetic: 7 +5 = 7 + 5 .in Algebra: 2x + 2y = 5y. Practice Algebra i includes addition, subtraction, multiplication, division operations with variables and numbers following rules according to arithmetic.

Important formulas for practice algebra i :


Commutative property of addition: a + b = b + a
Associative property of addition: (a + b) + c = a + (b + c)
Commutative property of multiplication: a × b = b × a
Distributive property: a(b + c) = ab + ac


Definitions of trigonometric functions:

Sine = `a/c`  Cosine = `b/c`  Tangent = `a/b`

Is this topic Acute Obtuse Angles hard for you? Watch out for my coming posts.

Examples for practice algebra i:


Pro 1:Solve the following system of equations
`-x/2 + y/3 = 0`
x + 6y = 16

Solution:We first multiply all elements of the first equation by the LCM of 2 and 3 which is 6.
6(`-x/2 + y/3 = 0` ) = 6(0)
x + 6y = 16
We then solve the following equivalent system of equations.
-3x + 2y = 0
x + 6y = 16
which gives the solution
x = `8/5 `  and y = `12/5 `

Pro 2: Solve 3x2 + 5x = -1 for x in algebra equation.

Solution:First determine a, b, and c.

3x2 + 5x + 1 = 0

a = 3

b = 5

c = 1

Plug the values you found for a, b, and c into the

Quadratic formula

x     = `( (-5) +- sqrt(5^2 - 4 *3*1))/(2*3)`
Perform any indicated operations.
x     = `( (-5) +- sqrt(25 -12))/(6)`   = x     = `( (-5) +- sqrt(13))/(6)`
The solutions are as follows:
x     = `( (-5) - sqrt(13))/(6)`    or x     = `( (-5) + sqrt(13))/(6)`

Practice Geometry Questions

Introduction :

The geometry is a branch of mathematics that investigates the relations, properties, and measurement of solids, surfaces, lines, and angles; the science which treats of the properties and relations of magnitudes; the science of the relations of space. Specially created windows classified as either Straight line Geometric such as rectangles, triangles, trapezoid, octagons, pentagons, etc. I like to share this Quadrilateral Properties with you all through my article.


Sample Questions :


1.The length of a certain rectangle is 3 cm bigger than its width. If the breaths were doubled and the lengths were decreased by 1 cm, the new rectangle would have the same perimeter as the original. Find the dimensions of the original rectangle.

2.The radius of a circle is 3 centimeters. What is the circle's circumference?

3.A cube has a surface area of fifty-four square centimeters. What is the volume of the cube?

4.A circle has an area of 49pi square units. What is the length of the circle's diameter?

5.In a quadrilateral two angles are equal. The third angle is equal to the sum of the two equal angles. The fourth angle is 60° less than twice the sum of the other three angles. Find the measures of the angles in the quadrilateral. Understanding online math help for free is always challenging for me but thanks to all math help websites to help me out.

6.The length and the breath of a rectangle are given by consecutive integers. The area of the rectangle is 90 cm squared . Find the length of a diagonal of the rectangle.

7.In a right triangle, one of the acute angles is two time as large as the other acute angle. Find the measure of the two acute angles.

8.An angle has measure 44 degrees more than the measure of a supplement to it. What is the measure of the angle?

9.A line contains points (4, -3) and (7, -4). What is the slope of a line perpendicular to this line?

10.Two time the measure of the supplement of an angle is seven time the measure of the complement of the angle. Find the measure of the angle.

Algebra 1 Practice

Introduction for extra algebra 1 practice:

For extra practice we need to work out more problems in algebra 1. Only through this way we get extra practice in Algebra 1. Algebra 1 is one of the subdivisions of algebra which is mainly used to find the unknown variables with the known variables. Algebra 1 consists of algebraic expressions, conditions and polynomials. An algebraic expression represents a scale, a number gets added or subtracted or multiplied or divided on both the sides of the scale. The numbers are defined as constants. Algebra 1 has the part in complex numbers, matrices, vectors, real numbers etc


Example Problems for extra algebra 1 practice:

Example 1:

Evaluate the equation     7(-3x - 2) - (-x - 6) = -4(5x + 5) + 12

Solution:

Given the equation

7(-3x - 2) - (-x - 6) = -4(5x + 5) + 12

Multiply factors.
-21x - 14 + x + 6 = -20x - 20 +12

Grouping the terms.

-20x - 8 = -20x - 8

Add 20x + 8 to both sides, the above equation becomes

0 = 0

All real values are solution to this equation.

Example 2:

Simplify the expression    3(a -4) + 6b - 3(a -b -2) + 8

Solution:

Given the algebraic expression

3(a -4) + 6b - 3(a -b -2) + 8

Multiply factors.

= 3a - 12 + 6b -3a + 3b + 6 + 8

Grouping the above terms.

= 9b + 2

Example 3:

If x <4 -="" 4="" 5="" br="" nbsp="" simplify="" x="">
Solution:

Given the expression

|x - 4| - 5|-4|

If x <; 4 then x - 4 < 4

and if

x - 4 < 4 the |x - 4| = -(x - 4).

Substitute |x - 4| by -(x - 4) and |-4| by 4

|x - 4| - 5|-4|

= -(x - 4) -5(4)

= -x -16

Having problem with Discriminant of a Quadratic Equation keep reading my upcoming posts, i will try to help you.

Extra algebra 1 practice problems:


1) Evaluate the equation     6(-8x - 3) - (-5x - 5) = -8(2x + 4) + 9

Answer: x = 10/59

2) Simplify the expression    5(a -8) + 11b - 5(a -b +6) + 4

Answer: 16b + 64

Polynomials Practice

Introduction:

In arithmetic, polynomials practice are an expression of finite duration and it practice constructed from variables and constants, with only the operations of adding up, working out, development, and also non-negative, whole-number exponents.

For instance, x^2 − 4x + 7 is a on of polynomials, but x^2 − 4/x + 7x3/2 is not, because its next term involves division by the variable x and since its third expression contains a proponent and that is not a complete number.

Overview:

Polynomials are one of the either zero, or it can be practice as the sum of one or more without zero terms. The numeral of language is in restricted. These conditions consist of the some steady which may be multiplied a restricted number of variables.

The exponent on a variable of the idiom is called the degree of that variable in that term, the degree of the term is the calculation of the degrees of the variables in that phrase, and its degree of a polynomial is the largest degree of any one the term. Since x = x1, is the degree of a variable without a written exponent and is one. A term with no variables is called as stable term. The degree of an invariable term is 0.

For example:   - y is a term. The coefficient is –5, then variable of x and y, the degree of x is two, and the degree of y is one. The degree in which the entire term is the sum of the degrees of each variable in it, so in this example the degree is 2 + 1 = 3.

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Polynomial equations:

A polynomials equation is a one of the practice equation for which polynomials set equal to other polynomials. 3x^2+4x+y=0 is a polynomials equation. In case of a polynomials  practice equation the variable is considered as unknown, and one seeks to find the possible values for which both members of the equation evaluate to the same value (in general more than one solution may exist).

A polynomial equation is to be contrasted with the polynomial identity like (x+y) (x–y) =x^2–y^2, where both members represent the same polynomial in different forms, and as a consequence any evaluation of both members will give a valid equality.

Prime Factorization

Prime factorization is one of the most important concepts in mathematics. Before learning about prime factorization, let’s have a look at prime numbers and related concepts in this post. Prime number is a number that is exactly divisible by only 1 and itself. For example: Madhumita bought 5 action figures toys /b> for her children. Here, the number 5 in action figures toys is a prime number because it is divisible by only 1 and 5. 2, 3, 5, 7 etc. are examples of prime numbers. Now let’s move on to the concept of prime factorization hereafter.

Prime factorization is the process of finding the prime number multiplying which we can get a certain number. For example Sheena bought 8 brightest flashlight toys for her baby. She bought these brightest flashlight toys from online kids’ shop. Find the prime factors of 8: 8 = 2 * 2* 2. Here, 2 is the prime factor of 8. A number that doesn’t have any prime factors except itself is a prime number. An exact divisor of a number is called a factor. For example: Meenu bought 30 crib mobile toys for orphan kids. She bought these crib mobile toys in wholesale. Now, 30 is exactly divisible by 10 and therefore, 10 is a factor of 30. And 5 * 3 * 2 = 30, therefore, 5, 3 and 2 are the prime factors of 30.

Examples on Prime Factorization:
1. Maria bought 60 apples from the big market of the city. Find the prime factors of 60:
60 = 2 * 2* 3* 5
Here, 2 * 2 * 3 are the prime factors of 60
2. Rupa bought 18 chocolates from the corner shop to distribute among the kids. Find the prime factors of 18:
18 = 2 * 3 * 3
Here, 2 * 3* 3 are the prime factors of 18.
These are the basics about prime factors and prime factorization.

How to Use Algebra

Algebra:

Algebra word problems are very useful to answer real-life problems. Remember the famous words of Albert Einstein

Definition:

That branch of math which treats of the relations and properties of quantity by means of letters and other symbols.

A branch of mathematics in which signs, usually letters of the alphabet, represent numbers or members of a specified set and are used to represent quantities and to express general relationships that hold for all members of the set.


How to use Algebra:


An algebraic expression is every combination of variables and constants with mathematical operations (addition, subtraction, multiplication, division, roots and powers). Writing algebraic expressions is a method of determining and combining constants, variables, and math convention using the order of operations. The next shows how to put in writing algebraic expressions, with an example in parentheses.

Step 1:

Write down what the algebraic expression is to show. (A toy line has toys of 3 different sizes and both with a different price. Small costs $5, medium costs $10, and large costs $15. Write as an algebraic expression.)

Step 2:

Determine the constant(s). (5, 10, 15 for price of each size toy.)

Step 3:

Determine the variable(s). (x, y, z for the quantity of each size toy.)

Step 4:

Determine the operations that need to be done. (Multiply both prices by the quantity and find the sum total of all 3 sizes.)

Step 5:

Mingle the constants, variables, and operations using the conventions of the order of operations. This is the algebraic expression. (5x + 10y + 15z)

I have recently faced lot of problem while learning Division of Rational Numbers, But thank to online resources of math which helped me to learn myself easily on net.

Use Algebra expressions with detailed solutions:


Problem 1:

Answer the expression 5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

Solution:

5(-3x - 2) - (x - 3) = -4(4x + 5) + 13

-15x - 10 - x + 3 = -16x - 20 +13

-16x - 7 = -16x – 7

0 = 0

Problem 2:

Reduce the algebraic expression 2(a -3) + 4b - 2(a -b -3) + 5

Solution:

= 2(a -3) + 4b - 2(a -b -3) + 5

= 2a - 6 + 4b -2a + 2b + 6 + 5

= 6b + 5

Problem 3:

Reduce |x - 2| - 4|-6|

Solution:

= |x - 2| - 4|-6|

Alternate |x - 2| by -(x - 2) and |-6| by 6.

= |x - 2| - 4|-6| = -(x - 2) -4(6)

= -x -22

Problem 4:

Estimate f (2) - f (1), f(x) = 6x + 1

Solution:

f (2) - f (1) = (6*2 + 1) - (6*1 + 1)

= 6

Problem 5:

Answer the equation |-2x + 2| -3 = -3

Solution:

|-2x + 2| -3 = -3

|-2x + 2| = 0

x = 1

Finding Intercepts

Introduction for finding intercepts:

In coordinate geometry, the y-intercept is the y-value of the point where the graph of a function or relation intercepts the y-axis of the coordinate system. If the curve in question is given as y = f(x), the y-intercept is found by calculating f(0). Functions which are undefined at xy-intercept. Some 2-dimensional mathematical relationships such as circles, ellipses, and hyperbolas can have more than one y-intercept. Because functions associate x values to no more than one y value as part of their definition, they can have at most one y-intercept.                                                  Source – Wikipedia.



The general rules for finding intercepts:


Rule for finding X intercepts:

To find an x-intercept, put y = 0 in the equation.

Your x-intercept will be written as a point (x, 0).

Rule for finding Y intercepts:

To find a y-intercept, x = 0 in the equation. Your y-intercept will be written as a point (0, y).

There are two special types of lines are to be considered these are horizontal lines and vertical lines.

1. A horizontal line is in the form y = k; that is, the y-value is always a constant value.

2. A vertical line is in the form x = h; that is, the x-value is always a constant value.


Examples for finding X and Y intercepts:

Examples for finding X and Y intercepts are given below:

1) find the x Intercept in an equation y = 5x-1.

Solution:

Let the given equation is y = 5x-1.

We have to put y = 0 in the above equation

Y = 5x-1

0 = 5x-1

Add 1 on the both sides

0+1 = 5x-1+1

1 = 5x

1/5 = x

Therefore the x-intercept is (1/5, 0)

Example 2

Find the x intercept for x=2y+3

To find the x - intercept:

Solution:

We have to put y=0 in the above equation

X = 2y+3

X = 2(0) + 3

X = 0 + 3

X = 3

Therefore the x-intercept is (3, 0)

3. To find the y - intercept for y= 5x-1

Solution:

We have to put x = 0 in the above equation

Y = 5x-1

Y = 5(0)-1

Y = 0-1

Y = -1

Therefore the y-intercept is (0,-1).

4)find the y – intercept for x = 2y + 3

Solution:

We have to put x = 0 in the above equation

X = 2y+3

0 = 2y+3

-3 = 2y+3-3

-3 = 2y

Y = -3/2

Therefore the y-intercept is (0,-3/2).

Free Algebra Practice

Introduction to free algebra practice:

Algebra comes from the Arabic word al-gebra. It represents as the “reunion of the broken words”. In mathematics, Letters like a, b, c, x are been used to represent the numbers, values, complex numbers, integers and these unite with the general operations to form the algebraic expressions. If two algebraic expressions are equated we get the algebraic equations. In free algebra practice we shall discuss about the examples and practice problems.

Solved examples - free algebra practice:

Example problem: 1

Solve the two consecutive numbers have a sum of 81. find the  numbers?

Solution:

Let x = The first Consecutive Number

Since the numbers are consecutive, meaning one number comes right after the other, the second number must be one more than the first. So, x + 1 equals the second number.

Let x + 1 is a 2nd consecutive

The problem says that the sum of the two numbers is 81. This can be shown in the equation like the following:

x + (x + 1) = 81

The equation which you just wrote can be solved as follows:

Initial Equation

x + (x + 1) = 81

After combine like terms

2x + 1 = 81

After subtracting 1 from each side

2x = 80

After dividing each side by 2

x = 40

Answer : x = 40


Solving linear equations - free algebra practice:

Solve the following linear equations: x + 2y + 3z = 14, 3x + y + 2z = 11, 2x + 3y + z = 11.

Solution:

x + 2y + 3z = 14 (1)

3x + y + 2z = 11 (2)

2x + 3y + z = 11 (3)

Consider the equations (1) and (3)

(1) ? x + 2y + 3z = 14

(3) × 3 ? 6x + 9y + 3z = 33 (subtracting)

–5x – 7y = –19

5x + 7y = 19 (4)

Consider the equations (2) and (3)

(2) ? 3x + y + 2z = 11

(3) × 2 ? 4x + 6y + 2z = 22 (subtracting)

–x – 5y = –11

x + 5y = 11 (5)

Consider the equations (4) and (5)

(4) ? 5x + 7y = 19

(5) × 5 ? 5x + 25y = 55 (subtracting)

–18y = –36

? y = 2

Substitute y = 2 in (5) we get

x + 5(2) = 11

x + 10 = 11

x + 10 - 10 = 11 - 10

? x = 1

Substitute x = 1, y = 2 in (3) we get

2(1) + 3(2) + z = 11

2 + 6 + z = 11

8 - 8 + z = 11 - 8

? z = 3

The solution is x = 1, y = 2, z = 3.

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Problems for practice - free algebra practice:

Problem 1:

The quadratic polynomial is divided by (x–1), (x+1) and (x–2) leaves the remainders 2, 4, 4 respectively, find the quadratic polynomial.

solution:The required quadratic polynomial is q(x) = x2 – x + 2.

Problem: 2

Solve the value of a and b if ax3 + bx2 + 7x + 9 and x3 + ax2 – 2x + b – 4 when divided by x + 2 leave remainders –13 and –16 respectively.

Solution: a –1, b = –4

Learning Practice of Statistics

Introduction to learning practice of statistics :

Statistics may be defined as the science of collection, presentation, analysis and interpretation of numerical data. Information is collected, presented and organized in the form of tables, graphs etc analyzed and then inferences are drawn from them. Statistical method is a technique used to obtain, analyze, summarise, compare and present the numerical data.Learning Statistics in the singular is a science which investigate the statistical methods and deals with their application. I like to share this Degrees of Freedom Statistics with you all through my article.


Definition for some statistical terms for learning practice of statistics

Data:

The collection of a particular type of information in the form of numerical figures is called, a set of data. This set of data obtained in the original form is called a set of raw data. Each numbered figure in the set of data is called an observation.

Ex: 35, 28, 26, 30, 32, 35, 26, 31, 36, 28

Array:

It is very difficult to draw any inference from this raw set of data. So we arrange it in ascending or descending order of size. The above set of data arranged in ascending order is:

Ex: 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 30 ,31.

Mean:

The arithmetic mean in statistics is the same as 'average' in arithmetic.

Mean of Ungrouped or Raw data:

The mean of a set of data is found out by dividing the sum of all the observations by the total number of observations in the date. We denote the mean by x' (read “x bar”).

Mean = sum of observations / number of observations

Median:

The median of a set of numbers is the middle number when all the numbers are arranged in order of size. ie. In descending or ascending order.

Mode:

The mode of a set of numbers is the number which occurs most frequently in the set. If no number occur more than once, the set of data Is said to have no mode. If different numbers occur the same number of times, the set of data has more than one mode. Having problem with how to long divide decimals keep reading my upcoming posts, i will try to help you.

Let us see some practice problems on these.


learning practice of statistics - Example problems


1) find the mean and median of the following number 39, 37, 38, 28, 30, 35, 36.

Solution:

Mean = 39+37+38+28+30+35+36 / 7

= 34.7

Median = 39, 38, 37, 36, 35, 30, 28

= 36

2) find the mode of the following number 52, 58, 58, 58, 65, 73, 73, 73.

Solution:

two modes: 58 and 73.

Maths Tricky Questions

Introduction to maths tricky questions

Math is a study of quantity. Here is an article that provides some tricky questions in math that will help you in cracking various placement tests as well as other class tests.Lets see some of the tricky questions on math and their solutions. I like to share this Solve Math Problems Free with you all through my article.

Solutions to maths tricky questions

1. The average age of a class is 15.8 years. The average age of the boys in the class is 16.4 years while that of the girls is 15.4 years. What is the ratio of boys to girls in the class?

Solution: Let the ratio be k:1

Then, k * 16.4 + 1 * 15.4 = (k+1) * (15.8)

16.4 k + 15.4         = 15.8k + 15.8

16.4k - 15.8k        = 15.8 - 15.4

0.6k        = 0.4

k   = 0.4 / 0.6

or   k = `(2)/(3)`

So, required ratio = `(2)/(3)`  : 1 = 2 : 3  Answer

2. A sum of Rs. 1000 is lent to be returned in 11 monthly instalments of Rs. 100 each, interest being simple. What is the rate of interest?

Solution:    Rs. 100 + SI on Rs. 1000 for 11 months

= Rs. 1000 + SI on Rs. 100 for (1 + 2 + 3 + 4 + ..... + 10) months

= Rs. 1000 SI on Rs. 100 for `(110)/(2)`  months

= Rs 1000 + SI on Rs. 100 for 55 months

SI on Rs. 100 for 55 months = Rs. 100

So, Rate =`(100 * 100 * 12)/(100 * 55)`  %

= 21.82 % Answer

3. Ramlal is four times as old as his son. Four years later the sum of their ages will be 43 years. What is the present age of the son?

Solution: Let present age of son = x years

Present age of Ramlal = 4x years

Four years later:

Son's age = x + 4

Ramlal's age = 4x + 4

According to the given condition:

x + 4 + 4x + 4  = 43

5x + 8 = 43

5x= 43 - 8

5x= 35

x= `(35)/(5)`  = 7

So, present age of son = 7 years Answer

4. A man can row at 5 km/hr in still water and the velocity of the current is 1 km/hr. It takes him 1 hr to row to a place and back. How far is the place?

Solution: Distance = 1 x `(5^2 - 1^2)/(2 * 5)` 52 - 12 / 2 x 5

= `(24)/(10)`   = 2.4 km  Answer

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Maths Tricky Questions (Continued)


Some more maths tricky questions with the solutions :-

5. Bucket P has thrice the capacity has bucket Q. It takes 60 turns for bucket P to fill the empty drum. How many turns it will take for both the buckets P and Q, having each turn together to fill the empty drum?

Solution: Let capacity of P be x litre

Then, capacity of Q = x/3 litre

capacity of drum = 60x litres

Required number of turns = `(60x)/(x + x/3)`

= `(3 * 60x)/(3x+x)`

= `(180x)/(4x)`  = 45 turns Answer

6. Rajan invests 65% of his intended investment in machinery and 20% on raw material in his business. If he has a balance of $6000 what was the intended investment?

Solution: Let intended investment be x

Investment spent on machinery = 65%

Investment spent on raw materials = 20%

So, total percentage of investment spent = 65% + 20% = 85%

Percentage of investment left = (100% = 85%) = 15%

According to the statement:

15% of x = 6000

`(15x)/(100)`   = 6000

15x = 6000 x 100

x = `(600000)/(5)`  = 40000

So, intended investment = $40000 Answer

7. By selling an article at $310 a merchant loses 7% on his outlay. Find the percentage of profit or loss when he sells the article at $350 ?

Solution: When article is sold at $310, percentage lost = 7%

So, remaining = 93%

Now, 93 : x   =  310 : 350

93 x 350 = 310 * x

x    = 105 %

i.e. Profit = (105 - 5) % = 5% Answer

8. Two trains each 300 m in length are running on the parallel linesin opposite directions with the speed of 70 km/hr and 50 km/hr respectively. In how much time will they cross each other completely?

Solution: t = `(300)/((70+50) * 5/18)`

= `(300)/((120) * 5/18)`

= 9 sec Answer

9. One year ago the ratio between father's and son's age was 4:1. The ratio of their ages after 4 years will be 3:1. Find the ratio of their ages after 15 years.

Solution: Let present age of father = x

present age of son = y

According to the statement:

x-1 : y -1  = 4:1

So, 4(y - 1) = x - 1

4y - 4  = x - 1

4y - x  = - 1 + 4

-x + 4y = 3 -----> (1)

Also, x + 4 : y + 4 = 3:1

So, 3 (y + 4) = x + 4

3y + 12  = x + 4

3y - x      = 4 - 12

-x + 3y     = -8  -------> (2)

Subtracting (2) from (1), we get,

4y - x - (-x + 3y) = 3 - (-8)

4y - x + x - 3y     = 11

y = 11

Substituting the value of y in equation (1)

-x + 4y = 3

-x + 44 = 3

-x = 3 - 44

- x = -41   or x = 41

So, present age of father = x = 41 years

present age of son = y    = 11 years

Note: If you require any more help in Maths Tricky Questions, our group of expert math tutors will help you in doing do.

rct math practice

Introduction to RCT math practice

Rct math practice includes aspects of Algebra, ratio, proportion, percent, rational numbers, probability, permutation, Statistics, geometry and Consumer Mathematics. In this article we shall discuss about some important rct math practice problem s. let us we discuss about algebra problems, permutation problem and combinations problems.


RCT math practice example problems

Example 1:

How many three-digit numbers can be formed by using the digits 1, 2, 3, 4, 5.

We have to determine the total number of three digit numbers formed by using the digits 1, 2, 3, 4, 5.

Clearly, the repetition of digits is allowed.

A three digit number has three places viz. units, ten’s and hundred’s. Unit’s place can be filled by any of the digits 1, 2, 3, 4, 5. So unit’s place can be filled in 5 ways.

Similarly, each one of the ten’s and hundred’s place can be filled in 5 ways.

Total number of required numbers

= 5 × 5 × 5 = 125

Example 2:

Solve for x:

6 x - 4 = 2 x – 10

Subtract 2x from both sides of the equation

6x – 2x – 4 = 2x – 2x – 10

4x – 4 = - 10

Add 4 to both sides of the equation

2x – 4 + 4 = - 12 + 4

2x = - 8

Divided by 2 both side of the equation

x = - 4

The answer x is – 4

Example 3:

If nC3 = nC6, find 12Cn

Solution:

nC4 = nC6   n = 3 + 6 = 9

Now 12Cn = 12C9

= 12C (12 ? 9) = 12C3

=12 × 11
1 × 2× 3

= 22

Example 4:

Find the number of different 4-letter words with or without meanings that can be formed from the letters of the word ‘NUMBER’

Solution:

There are 6 letters in the word ‘NUMBER’.

So, the number of 4-letter words

= the number of arrangements of 6 letters taken 4 at a time

= 6P4

= 360

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RCT math practice problems

Question 1:

In how many ways can 5 gentlemen and 5 ladies sit together at a round table, so that no two ladies may be together?

Answer:2880

Question 2:

How many different signals can be made by hoisting 6 differently colored flags one above the other, when any number of them may be hoisted at one time?

Answer:1956

Solving Joint Variation

Introduction to solving joint variation:

Joint variation is like a direct variation in this it has two or more variables. The variables which are in right side of the equal sign vary equally. The value of the dependent variable increases when the values of the independent variable increases and the value of the dependent variable decreases when the value of the independent variable decreases. Having problem with Type of Functions keep reading my upcoming posts, i will try to help you.

General Form – Solving Joint Variation:

The general form of the joint variation is y = kxz , where k is a constant and x and z are variables.

Example Problems – Solving Joint Variation:

Example 1 – Solving joint variation:

Identify whether the given expression a=kbc is joint variation with the values b=2, c=3 and b=4 and c=5 with k = 1.

Solution:

The given expression is `a= kbc` .

Now substitute the given values in the given expression

When `b` =2 and `c` = 3 then `a` = `1*2*3 ` `=>` `a` = `6` .

When `b` = 4 and `c` = 5 then `a` = `1*4*5` `=>` `a` = `20`

When the value of `b` and `c` jointly directly varies then the value of `a` also varies directly. Please express your views of this topic Least Common Multiple Definition by commenting on blog.

Example 2 – Solving joint variation:

Identify whether the given expression y=kxz is joint variation with the values x =5, z=6 and x=6 and z = 7 with k = 2.

Solution:

The given expression is `y=kxz.`

Now substitute the given values in the given expression

When `x` =5 and `z` = 6 then `y` = `2*5*6` `=>` ` y` = `60` .

When `x` = 6 and `z` = 7 then `y` = `2*6*7` `=>`` y` = `84`

When the value of x and z jointly directly varies then the value of y also varies directly.

Example 3 – Solving joint variation:

Find the value of constant for the equation a=kbc where a = 3, b=4 and c = 1.

Solution:

The given expression is `a = kbc`

Now the above expression can be written as `k` = `a/(bc)`

`k` = `a/(bc)`

`k ` = `3/(4*1)`

`k` = `3/4`

`k` = 0.75

The value of `k` is 0.75.

Example 4 – Solving joint variation:

Find the value of z for the equation y = kxz where k = 3, y=4 and x = 1.

Solution:

The given expression is `y = kxz`

Now the above expression can be written as `z` = `y/(kx)`

`z` = `y/(kx)`

`z` = `4/(3*1)`

`z ` = `4/3`

`z` = `1.33`