# List all the factor pairs for 48 make a tabletop to help

Answer: 1  and 48 are a factor pair of 48 since 1 x 48= 48

2 and 24 are a factor pair of 48 since 2 x 24= 48

3 and 16 are a factor pair of 48 since 3 x 16= 48

4 and 12 are a factor pair of 48 since 4 x 12= 48

6 and 8 are a factor pair of 48 since 6 x 8= 48

8 and 6 are a factor pair of 48 since 8 x 6= 48

12 and 4 are a factor pair of 48 since 12 x 4= 48

16 and 3 are a factor pair of 48 since 16 x 3= 48

24 and 2 are a factor pair of 48 since 24 x 2= 48

48 and 1 are a factor pair of 48 since 48 x 1= 48

## Related Questions

AA
SSS
SAS
Not Similar

vertically opposite angles and alternate pair

so by aa similarity they are similar

I just need number 1

b is the awnser

Step-by-step explanation:

B

Step-by-step explanation:

There are 96 squares in total. Yard is 3 feet, so you would divide the total by 3.

Find the equation of the line through the points (-3,-3) and (2,-1) using point-slope form. Then rewrite theequation in slope-intercept form.

Step-by-step explanation:

The point-slope equation is y-y₁=m(x-x₁). Since we don't know our slope, we can use the formula to find the slope. All we have to do is use the coordinate we were given and plug it into the formula.

Now that we have the slope, we can fill out the point-slope equation.

y-(-3)=2/5(x-(-3))

y+6=2/5(x+3)

This is the point-slope form.

Now, we can distribute and solve to get slope-intercept form.

y+6=2/5x+6/5

y=2/5x-24/5

The equation of the line through the points (-3,-3) and (2,-1) can be found using point-slope form. It is y = (2/5)x - 9/5 in slope-intercept form.

### Explanation:

To find the equation of a line using the point-slope form, we need to determine the slope of the line and use one of the given points to write the equation. Firstly, let's find the slope of the line using the formula: m = (y2 - y1) / (x2 - x1). Plugging in the coordinates (-3, -3) and (2, -1) into the formula gives us m = (-1 - (-3)) / (2 - (-3)) = 2/5. Now, we can choose one of the points (for example, (-3, -3)) and use the point-slope form equation: y - y1 = m(x - x1). Substituting the values, we get y - (-3) = (2/5)(x - (-3)). Simplifying the equation yields y + 3 = (2/5)(x + 3), which is the equation of the line in point-slope form.

To rewrite the equation in slope-intercept form y = mx + b, we need to isolate the y variable. Distributing the (2/5) to (x + 3) in the point-slope form equation gives us y + 3 = (2/5)x + 6/5. Subtracting 3 from both sides gives us y = (2/5)x + 6/5 - 3. Simplifying further, the equation becomes y = (2/5)x - 9/5. Therefore, the equation of the line through (-3, -3) and (2, -1) in slope-intercept form is y = (2/5)x - 9/5.

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Sarah buys 8 packages, each with 20 balloons, and uses 1/4 of them to decorate the hall. Write and evaluate an expression to find how many balloons Sarah has left.​

The fraction is defined as a number in the form of a/b where a nd b are integers and b is not equal to zero. Sarah has left with 120 balloons.

### What is a Fraction?

A fraction is used to represent a portion or part of anything. It consists of two parts a numerator and a denominator. For example 1/4 is a fraction where numerator is 1 and denominator is 4.

The total number of packages = 8

Number of balloons in each package = 20.

The portion of balloons use to decorate hall = 1/4.

Suppose the remaining balloons are x.

Thus the expression for remaining balloons can be written as,

x = 20 × 8 - 1/4×(20 × 8)

x = 160 - 40

= 120.

Hence, Sarah has left with 120 balloons and the expression for left balloons is x = 20 × 8 - 1/4×(20 × 8).

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(8 × 20) ÷ 4

40

Step-by-step explanation:

8 × 20 = 160

160/4 = 40

Find the discount on a leather recliner with a regular price of \$260 if the recliner is 30% off. What is the sale price of the recliner?The discount on the leather recliner is

260x 30% = 78
260-78=\$182
The sale price of the recliner is \$182
The discount on the leather recliner is \$78
It is 182, hope this helps!

among a group of students 50 played cricket 50 played hockey and 40 played volleyball. 15 played both cricket and hockey 20 played both hockey and volleyball 15 played cricket and volley ball and 10 played all three. if every student played at least 1 game find the no of students and how many students played only cricket, only hockey and only volley ball

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Explanation:

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

Cricket only= 30

Volleyball only = 15

Hockey only = 25

Number of students that play cricket= n(C)

Number of students that play hockey= n(H)

Number of students that play volleyball = n(V)

From the question, we have that;

n(C) = 50, n(H) = 50, n(V) = 40

Number of students that play cricket and hockey= n(C∩H)

Number of students that play hockey and volleyball= n(H∩V)

Number of students that play cricket and volleyball = n(C∩V)

Number of students that play all three games= n(C∩H∩V)

From the question; we have,

n(C∩H) = 15

n(H∩V) = 20

n(C∩V) = 15

n(C∩H∩V) = 10

Therefore, number of students that play at least one game

n(CᴜHᴜV) = n(C) + n(H) + n(V) – n(C∩H) – n(H∩V) – n(C∩V) + n(C∩H∩V)

= 50 + 50 + 40 – 15 – 20 – 15 + 10

Thus, total number of students n(U)= 100.

Note;n(U)= the universal set

Let a = number of people who played cricket and volleyball only.

Let b = number of people who played cricket and hockey only.

Let c = number of people who played hockey and volleyball only.

Let d = number of people who played all three games.

This implies that,

d = n (CnHnV) = 10

n(CnV) = a + d = 15

n(CnH) = b + d = 15

n(HnV) = c + d = 20

Hence,

a = 15 – 10 = 5

b = 15 – 10 = 5

c = 20 – 10 = 10

Therefore;

For number of students that play cricket only;

n(C) – [a + b + d] = 50 – (5 + 5 + 10) = 30

For number of students that play hockey only

n(H) – [b + c + d] = 50 – ( 5 + 10 + 10) = 25

For number of students that play volleyball only

n(V) – [a + c + d] = 40 – (10 + 5 + 10) = 15

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