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Umnica [9.8K]
3 years ago
8

There were 6 pizza pies at the summer barbeque at Abby's House. She shares the pizza with her 7 friends. If all 8 people get the

same amount of pizza, what 1. portion of pizza can each person have? Use pictures, diagrams, or math operations to show your work.
Mathematics
1 answer:
vodka [1.7K]3 years ago
8 0

6 pies, 8 people

6 pies divided by 8 people

so 6/8

that is 0.75 or 3/4 or 75%

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A rectangle has a length 6 more than it's width if the width is decreased by 2 and the length decreased by 4 the resulting has a
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Answer:

Length of original rectangle: 11 units.

\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}

\text{Perimeter of new rectangle}=20

Step-by-step explanation:

Let x represent width of the original rectangle.  

We have been given that a rectangle has a length 6 more than it's width. S the length of the original rectangle would be x+6.

We have been given that when the width is decreased by 2 and the length decreased by 4 the resulting has an area of 21 square units.

The width of new rectangle would be x-2.

The length of new rectangle would be x+6-4=x+2.

The area of new rectangle would be (x+2)(x-2).

Now we will equate area of new rectangle with 21 and solve for x as:

(x+2)(x-2)=21

Applying difference of squares, we will get:

x^2-2^2=21

x^2-4=21

x^2-4+4=21+4

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Since width cannot be negative, so we will take positive square root of both sides.

\sqrt{x^2}=\sqrt{25}

x=5

Therefore, the width of original rectangle is 5 units.

Length of the original rectangle would be x+6\Rightarrow x+5=11.

Therefore, the length of original rectangle is 11 units.

\text{Area of original rectangle}=5\times 11

\text{Area of original rectangle}=55    

Therefore, area of the original rectangle is 55 square units.

Now we will find ratio of the original rectangle area to the new rectangle area as:

\frac{\text{Area of original rectangle}}{\text{Area of new rectangle}}=\frac{55}{21}

We know that perimeter of rectangle is two times the sum of length and width.

\text{Perimeter of new rectangle}=2((x+2)+(x-2))

\text{Perimeter of new rectangle}=2((5+2)+(5-2))

\text{Perimeter of new rectangle}=2(7+3)

\text{Perimeter of new rectangle}=2(10)

\text{Perimeter of new rectangle}=20

Therefore, the perimeter of the new rectangle is 20 units.

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