Practice Vinculum

Introduction to Practice Vinculum:

Mathematical symbol which is used for grouping the mathematical expression is known as vinculum. A horizontal bar below the numerator and over the denominator is said to be vinculum. Vinculum is in the form of fraction, radical or in the parenthesis. For example, the mathematical expression `x + y - z` is expressed as `x + (y - z)` in the parenthesis form. Students practice the problems of vinculum with simple solutions to solve. Let us see about practice vinculum in this article. Having problem with Simplifying Rational Expressions Calculator keep reading my upcoming posts, i will try to help you.

General Format for Practice Vinculum

The format for vinculum is

In the form of fraction `(a + b)/ (a - b)`
In the form of radical is `sqrt(a - b)`
In the form of parentheses is `((a + b))/ ((a - b)) `
Worked Examples to Practice Vinculum

Example 1 to Practice Vinculum:

Solve the arithmetic expression `(27 + 37)/ (13 - 5)` .

Solution:

Step 1:

Given arithmetic expression is `(27 + 37)/ (13 - 5)` .

Step 2:

Solve the numerator and denominator of the expression, we get,

Adding the numerator, we get,

`27 + 37 = 64`

Subtracting the denominator, we get,

`13 - 5 = 8`

Step 3:

Dividing the fraction by placing the numerator and denominator, we get,

`64/8 = 8`

Step 4:

Dividing the fraction we get 8.

Hence, the solution for solving the vinculum is 8.

Example 2 to Practice Vinculum:

Solve the arithmetic expression `(36 + 6)/ (12 - 10)` .

Solution:

Step 1:

Given arithmetic expression is `(18 + 6)/ (12 - 9)` .

Step 2:

Solve the numerator and denominator of the expression, we get,

Adding the numerator, we get,

`18 + 6 = 24`

Subtracting the denominator, we get,

`12 - 9 = 3`

Step 3:

Dividing the fraction by placing the numerator and denominator, we get,

`24/3 = 8`

Step 4:

Dividing the fraction we get 8.

Hence, the solution for solving the vinculum is 8.

Example 3 to Practice Vinculum:

Solve the arithmetic expression `sqrt(27 + 22)` .

Solution:

Step 1:

Given arithmetic expression is `sqrt(27 + 22)` .

Step 2:

Solve the above expression, we get,

Adding the expression, we get,

`sqrt(49)`

Step 3:

Taking square root we get,

`sqrt(49) = 7`

Step 4:

Hence, the solution for solving the vinculum is 7. Please express your views of this topic free math problems for kids by commenting on blog.

Practice Problems to Practice Vinculum

Problem 1:

Solve the arithmetic expression `(23 + 4)/ (16 - 7)` .

Problem 2:

Solve the arithmetic expression `sqrt(109 + 12)` .

Solutions:

1. 3

2. 11

3rd Grade Maths

Introduction to 3rd grade maths:

3rd grade mathematics covers operations like addition, subtraction, multiplying and division the numbers. They also learn the word problem of addition, subtraction, multiplying and division,four digit numbers and their place values, successor and predecessor etc. We will see some examples from 3rd grade mathematics.

Tutoring on Addition for 3rd Grade Math Homework

Like two digit numbers we will start to add the numbers from left side. First, we will add ones, then tens, then hundred and at last thousand.

Example: 2 4 5 6

+ 2 5 6 3

5 0 1 9

Home work for practice:

(a)    4915 + 3456 =

(b)   2456 + 6754 =

(c)    1345 + 4326 =

(d)    1298 + 4264 =

(e)     3427 + 6743 =

Answer: (a) 8371 (b) 9210 (c) 5671 (d) 5562 (e) 10170

Word problems:

(a) In a city there are 140 shops of toys, 200 shops of grossary items and 300 shops of fancy dresses. Find out the total number of shops?

(b)   A shopkeeper sold 700 greeting cards in one year and 650 cards in the next year.     How many cards did he sell in two years?

Solution: (a) Number of toy shops in a city = 140

Number of grossary shops        = 200

Number of fancy dresses shops = 300

The total number of shops are = 140 + 200 + 300

= 640 shops Ans.

Solution (b) Number of greeting cards sold in one year = 700

Number of greeting cards sold in next year = 650

The total number of cards sold in two years = 700 + 650 = 1350

1350 cards Ans.

Tutoring on Subtraction for 3rd Grade Math Homework

To subtract the numbers, always subtract lowest number from the highest number.

Example: Subtract 214 5 from 5478

5 4 7 8

- 2 1 4 5

3 3 3 3

Home work practice:

(a)    5675 – 3241 =

(b)   3678 – 1625 =

(c)    6666 – 2323 =

(d)   9567 – 2235 =

(e)    7 883 – 2378 =



Answer: (a) 2434 (b) 2053 (c) 4343 (d) 7332 (e) 5505

Word problems :

(a) The cost of one bicycle is 2000 and other bicycle is 4500. What is the difference between their cost?

(b) What should be added to 2340 to get 5000?

Answer: (a) Cost of one bicycle is = 2000

Cost of other bicycle is = 4500

Difference between their costs = 4500 – 2000

= 2500 Ans.

(c)    First we subtract 2340 from 5000

5000 – 2340 = 2660

2340 – 2660 = 5000

Number should be added 2660 Ans.

Solve Math Manipulative

Introduction to solve math manipulative

The math manipulative is the organization of agrees learn the little offspring understand the mathematics concepts. Next we observe the objective support to the math manipulative. A manage of use math manipulative is that consent to the student mathematical thoughts any symbols to substance. In this content we are going to discuss about solve math manipulative. The following are the examples involved in solve math manipulative.

Sample Problem for Solve Math Manipulative:

Solve math manipulative problem 1:

Work out the area of the circle? The radius of a circle is 5 inches.

Solution:

The area of the circle formula is A = `pi` r2

r = 5

A = 3.14 * 5 * 5

= 3.14 * 25

= 78.5

The area of the 4 inches circle is A = 78.5

Solve math manipulative problem 2:

Work out the area of the circle? The diameter of a circle is 6 cm.

Solution:

The area of the circle formula is A =`pi` r2

To find the radius formula is r = diameter / 2

r = 6 / 2

r = 3

A = 3.14 * 3 * 3

= 3.14 * 9

= 28.26

The area of the 4 inches circle is A = 28.26

Solve math manipulative problem 3:

Find the x value: X+8=10

Solution:

X+8=10

Subtract both sides on -8

x+8-8=10-8

x= 2

Answer:-x = 2

Solve math manipulative problem 4:

Find the x value: X-14=11

Solution:

X-14=11

Both sides adding by 14

X-14+14=11+14

X = 25

Answer:-x= 25

Solve math manipulative problem 5:

8a2 * 5a2 +7 b3*5b3

Solution:

8a2 * 5a2 +7 b3*5b3

Multiply the a and b terms

40a2+2+35b3+3

40a4+35b6

Answer:- 40a4+35b6

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Practice Problem for Solve Math Manipulative:

Find the value of a: a+8=15

Answer:-a= 7

Find the value of a: 5a-55=30

Answer:-A=17

Find the value of a: (5a-65)/2=8

Answer:- A = 21

9a^2 +7a^2 +5b3-b3

Answer: - 1 6a^2+4b^3

8a^2 * 5a^3 +6b^3*7b^3

Answer: - 40a^5+42b^6

Equality Symbol

Introduction to equality symbol:

Equality expression allows all the operation in mathematics. The equality should be denoted by a symbol ‘=’.

Example: p = q. It represents that the terms at both sides are equal which is represented by a symbol ‘=’. We can do all the arithmetic operations at this equality expression without change the meaning of equality. This equality is used for solving the equations.

Properties of Equality Symbol:

The properties of equality symbol are,

Addition property of equality symbol
Subtraction property of equality
Multiplication property of equality
Division property of equality
Addition property of equality:

It states that the addition of a number at equality cannot change the meaning of equality symbol.

Example: p + q = r + q

Subtraction property of equality:

The subtraction of a number at equality cannot change the meaning of the equality symbol

Example: p – q = r – q

Multiplication property of equality symbol:

The multiplication of a number at equality cannot change the meaning of equality symbol.

Example: p . q = r . q

Division property of equality:

The division of a number at equality cannot change the equality symbol.

Example: p/q = r/q

Example Problems to Equality Symbol:

Example: 1

Solve: 5 + b = 25

Solution:

Given 5 + b = 25

Subtract a number 5 at both sides for getting unknown value b.

5 + b – 5 = 25 – 5

b = 20

To check:

5 + 20 = 25

25 = 25

Answer: b= 25

Example: 2

Solve: n - 22 = 69

Solution:

Given n - 22 = 69

Add a number 22 at both sides for getting unknown value n.

n – 22 + 22 = 69 - 22

n = 47

Example: 3

Solve: 7 p = 147

Solution:

Given 7 p = 147

Divide a common number 7 at both sides for getting p

`7/7`  p = `147/7`

p =21

Practice Problems to Equality Symbol:

Problem: 1

Solve 56 + m = 165

Answer: 109

Problem; 2

Solve 90 n = 180

Answer: 2

The Equation Y Kx is Called

Introduction to the equation y  kx is called:
If  x and y are the two variables, y is directly proportional to x and there is a constant k exists. (Constant k is always not equal to zero). This is called as direct proportionality. It is denoted as y ∝ x.

If we want to remove the proportionality symbol, then we use the constant k. It also can be written as y = k x.

where,  k =` y / x` .   k is called a proportionality constant (or) constant of proportionality. Now, we are going to see some of the problems on the equation y  kx is called an direct proportionality. Having problem with Addition of Polynomials keep reading my upcoming posts, i will try to help you.

Example Problem Related to the Equation Called as Y Kx:

Example problem 1:

Let us assume the variable y is directly proportional to x. Given y = 60 and x = 20. Write an equation which is called as a direct proportionality that relates x and y.

Solution:

Given y = 60 and x = 20

Direct proportionality, y ∝ x

It also can be written as y = k x

Substituting the given x and y values in the above equation,

60 = k * 20

Divide by 20 on both sides of the equation

`60 / 20 = (20k) / 20`

k = 3

Substitute the value of k in the equation y = kx.

So, the equation is y = 3x. I have recently faced lot of problem while learning greatest integer function, But thank to online resources of math which helped me to learn myself easily on net.

Additional Problem Related to the Equation Called as Y Kx:

Example problem 2:

Let us assume x and y are the variables. If y varies directly as x and the value of y = 66 when the value of x = 11.

1)      Find the value of proportionality constant

2)      Calculate the value of y when x = 22

3)      Calculate the value of x when y = 33

Solution:

1) Given y = 66 and x = 11

Direct proportionality, y ∝ x

It also can be written as y = k x

Substituting the given x and y values in the above equation,

66 = k * 11

Divide by 11 on both sides of the equation

`66 / 11 = (11k) / 11`

k = 6

So, the proportionality constant k =6.

2) Substitute the value of x = 22 and the proportionality constant k = 6 in the equation y = k x

y = 6* 22

y = 132

So, the value of y = 132.

3) Substitute the value of y = 33 and the proportionality constant k = 6 in the equation y = k x

33 = 6 x

Divide by 6 on both sides of the equation

`33 / 6 = (6x) / 6`

x = 5.5

So, the value of x = 5.5.

Practice Problems Related to the Equation Y Kx:

1) Let us assume the variable y is directly proportional to x. Given y = 26 and x = 13. Write an equation which is called as a direct proportionality that relates x and y. (Answer:  y =  2x).

2) Let us assume the variable y is directly proportional to x. Given y = 50 and x = 10. Write an equation which is called as a direct proportionality that relates x and y. (Answer: y = 5x).

Repeated Factors

Introduction to repeated factors:

In math, a word factor is a number which can divide a particular number. The factors may be composite numbers or prime numbers. The numbers 0 and 1 are not considered as factors. Repeated factors are the factors which are repeated. Let us see the details of repeated factors with some examples.

Explanation to Repeated Factors:

Consider a number 32.

Factors of a number 32 is 2 x 2 x 2 x 2 x 2 = 2^5

Here 2 is a factor that is repeated. So, we can say that 2 is a repeated factor for a number 32. These repeated factors may be composite numbers or prime numbers.

Example Problems to Repeated Factors:

Example: 1

Which of the following is the repeated factor of a number 125?

a) 11

b) 10

c) 25

d) 5

Solution:

Given number: 125

Factors of a number 125 = 5 x 5 x 5

= 53

Answer: d

Example: 2

Which of the following is the repeated factor of a number 81?

a) 2

b) 3

c) 9

d) 0

Solution:

Given number: 81

Factors of a number 81 is 3, 9

3 x 3 x 3 x 3 = 81

9 x 9 = 81

Repeated factor is 3

Answer: b

Example: 3

Which of the following is the repeated factor of a number 512?

a) 11

b) 512

c) 4

d) 2

Solution:

Given number: 512

512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2

= 29

Repeated factor is 2.

Answer: d

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Practice Problems to Repeated Factors:

Problem: 1

Which of the following is the repeated factor of a number 64?

a) 6

b) 0

c) 2

d) 64

Answer: c

Problem: 2

Which of the following is the repeated factor of a number 27?

a) 6

b) 3

c) 2

d) 27

Answer: b

Regression of Y on X

Introduction to regression of y on x:

Regression equations, also known as estimating equations, are algebraic expressions of the regression lines. Since there are two regression lines, there are two regression equations - the regression equation of X on Y is used to describe the variations in the values of X for given changes in Y and the regression equation of Y on X is used to describe the variation in the values of Y for given changes in X. I like to share this Logistic Regression Model with you all through my article.

Expression of Regression of Y on X

The regression equation of Y on X is expressed as follows:
Yc= a + bX
In this equation a and b are constants (fixed numerals values) which determines the position of the line completely. These constants are called the parameters  of the line. If the value of either or both of them is changed, another line is determined. The parameter 'a' determines the intercept, i.e., what will be the value of Y ( dependent variable) when X (independent variable) takes the value zero. The parameter 'b' determines the slope of the line i.e., the change in Y per unit change in X. The symbol Yc stands for the value of Y computed from the relationship for a given X. Please express your views of this topic need help with math homework for free by commenting on blog.

If the values of the constants 'a' and 'b' are obtained, the line is completely determined. But the question is how to obtain these values. The answer is provided by the methods of Least Squares which states that the line should be drawn through the plotted points in such a manner that the sum of the squares of the deviations of the actual Y values from the computed Y values is the least, or other words, in order to obtain a line which fits the points best `sum` (Y - Yc)2 should be minimum. Such a line is known as the line of 'best fit.'

A Straight Line Fitted by least Squares has the Following Characteristics(regression of Y on X)

It gives the best fit to the data in the sense that it makes the sum of the squared deviations from the line, `sum` (Y - Yc)2 ,  smaller than they would be from any other straight line. This property accounts for the name 'Least Square.'
The deviation above the line equal those below the line, on the average. This means that the total of the positive and the negative is zero , or `sum` (Y - Yc) = 0.
The straight line goes through the overall mean of the data `( barX barY )`.
When the data represent a sample from a larger population, the least squares line is a 'best' estimate of the population regression line.


With a little algebra and differential calculus it can be shown that the following two equations, if solved simultaneously, will yield values of the parameters a and b such that the least squares requirement is fulfilled :

`sumY = Na + bsum X`

`sumXY = a sumX + b sumX^2`

These equations are usually called the normal equations. In the equations `sumX, sumY, sumXY, sumX^2`  indicates totals which are computed from the observed pairs of values of two variables X and Y to which the least squares estimating line is to be  fitted and N is the number of observed pairs of values.

Regression Equation of X on Y

Regression Equation of X on Y is expressed as follows:
Xc = Na +bY

To determine the values of a and b the following two normal equations are to be solved simultaneously.

`sumX = Na + bsum Y`

`sumXY = a sumY + b sumY^2`

Illustration of Regression of Y on X

From the following data obtain the Regression Equation of Y on X
X               6               2               10                4               8
Y               9              11                5                8                7
Solution. :

OBTAINING REGRESSION EQUATION

`X`    `Y`    `XY`    X^2
 6     9     54        36
 2    11     22         4
10     5     50       100
 4     8     32        16
 8     7     56        64
`sumX = 30`    `sumY = 40`    `sumXY = 214`


Regression Equation of Y on X :
Yc = a + bX
To determine the value of a and b the following two normal equations are to be solved
`sumX = Na + bsumX`
`sumXY = asumX + bsumX^2`
Substituting the values

40 = 5 a + 30 b                                                        .................(i)
214 = 30a + 220b                                                     ..................(ii)
Multiplying equation (i) by 6 : 240 = 30a +180b                                       ..................(iii)
214 = 30a + 220b                                                     ..................(iv)

Deducting equation (iv) from (iii)
- 40b =  26
or b = -0.65

Substituting the value of b in equation (i)
40 = 5a + 30 (-0.65)
or   5a = 40 + 19.5 = 59.5
or   a  = 11.9

Putting the values of a and b in the equation, the regression of Y on X is
Y = 11.9 - 0.65X