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2 years ago

How do i solve this.

Mathematics

1 answer:

nignag [31]2 years ago

4 0

Let us set up some variables:

L: Length
W: Width

Now let us set up some equations based on the facts:

length of rectangle is 4 times its width --> L = 4W
perimeter of rectangle is 80 --> 2L + 2W = 80

Equations:

L = 4W -- equation 1

2L + 2W = 80 -- equation 2

Plug the value of L from equation 1 into equation 2

2(4W) + 2W = 80

8W + 2W = 80

10W = 80

W = 8 -- equation 3

Plug the value of W from equation 3 into equation 1:

L = 4W = 4 * 8 = 32

Therefore the length is 32 and the width is 8.

Hope that helps!

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Marissa and Callie both have babysitting jobs. Marissa charges $19 plus $1.50 per hour. Callie charges $15 plus $2.75 per hour.

aliya0001 [1]

Answer:

4 hours

Step-by-step explanation:

Write an equation

19 +1.5x < 15+2.75x

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4+1.5x<2.75x

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3 0

3 years ago

Read 2 more answers

A garden box has a perimeter of 18½ feet. If the length is 5 feet, what is the area of the garden box?

Travka [436]

Answer:

$area=\frac{85}{4}feet^{2}$

Step-by-step explanation:

1. Approach

Since it is given that the garden box is a rectangle, then the opposite sides are congruent. One can use this to their advantage, by setting up an equation that enables them to solve for the width of the rectangle. After doing so, one will multiply the width by the given length and solve for the area.

2. Solve for the width

It is given that the garden box is a rectangle. As per its definition, opposite sides in a rectangle are congruent. The problem gives the length and the perimeter of the rectangle, therefore, one can set up an equation and solve for the width.

$perimeter=18\frac{1}{2}\\\\length=5$

$2(length+width)=perimeter$

Substitute,

$2(5+width)=18\frac{1}{2}$

Conver the mixed number to an improper fraction. This can be done by multiplying the "number" part of the mixed number by the denominator of the fraction. Then add the result to the numerator.

$2(5+width)=\frac{37}{2}$

Inverse operations,

$2(5+width)=\frac{37}{2}\\/2\\\\5+width=\frac{37}{4}\\-5\\\\width=\frac{17}{4}$

3. Solve for the area

Now that one has solved for the width of the box, one must solve for the area. This can be done by multiplying the length by the width. Since the width is a fraction, one must remember, that when multiplying an integer by a fraction, one will multiply the integer by the numerator (the top of the fraction), and then simplify by reducing the fraction, if possible. Reducing the fraction is when one divides both the numerator and the denominator by the GCF (Greatest Common Factor).

$length=5\\\\width = \frac{17}{4}$