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svetlana [45]
4 years ago
5

Gina's mother has plans to carpet a floor with a width of 35 feet and a length of 50 feet. A 1.5-foot border of carpet around th

e room will be a different color than the rest of the carpet, as shown below. What is the area of the floor that is surrounded by the border?

Mathematics
1 answer:
neonofarm [45]4 years ago
7 0
<h3>Answer:  1504 square feet</h3>

=========================================

Explanation:

For the larger rectangle, the bottom length is 50 ft. Subtract off two copies of 1.5 to get 50-1.5-1.5 = 47 ft, which is the bottom length of the smaller rectangle. We subtract off two copies of 1.5 as there are two sides (left and right) of a border that is 1.5, which is not part of the smaller rectangle.

The same is done for the vertical component. We start with 35 as the larger vertical side and it goes to 35-1.5-1.5 = 32

The smaller rectangle has dimensions 47 ft by 32 ft. The area is 47*32 = 1504 square feet. We can write "square feet" as "ft^2" without the quotes. Another abbreviation is "sq ft".

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

Side note: the area of just the border is equal to the difference of the areas of the larger and smaller rectangle.

The larger rectangle has area 50*35 = 1750 sq ft, so the area of just the border only is 1750 - 1504 = 246 sq ft

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If 11250 = m(600) + b and 7650 = m(400) + b, how do I find the values of b and m? I forgot how to compare when I have two differ
Lelu [443]

Answer:

b=450 and m=18

Step-by-step explanation:

This exercise is an example of <em>linear system equations</em>. A <em>system of linear equations</em> is a set of (linear) equations that have more than one unknown. The unknowns appear in several of the equations, but not necessarily in all of them. What these equations do is relate the unknowns to each other.

The easy way to solve this problem is by using the<em> reduction method</em>.The <em>reduction method</em> consists of operating between the equations, such as adding or subtracting both equations, so that one of the unknowns disappears. Thus, we obtain an equation with a single unknown.

Ordering the equations as a system equations.

\left \{ {{600m+b=11250} \atop {400m+b=7650}} \right.

Using the reduction method, let's substract the second equation with the first equation, in order to clear m.

600m + b = 11250

<u>-400m -  b = -7650</u>

200m       = 3600    -------->  m = 3600/200 -----> m = 18

We can substitute the value of m in any of the two equations to obtain the unknow b.

Let's substitute the value of m in the second equation.

400(18) + b = 7650

7200 + b = 7650 -----> b = 7650 - 7200 -------> b = 450

We can check this values in both equations.

600m + b = 11250 -----> 600(18) + 450 = 11250

400m + b = 7650 ----->  400(18) + 450 = 7650

Satisfying the result of both equations.

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3 years ago
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zimovet [89]
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