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Natalka [10]
3 years ago
9

A 42 inch wire is bent into the shape of a rectangle whose width is twice its

Mathematics
2 answers:
kirza4 [7]3 years ago
6 0

Answer: Choice A

2(2L + L)  = 42

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

Explanation:

The length is L, which is some placeholder for a positive number.

The width is twice as much as this, so it's 2L

The perimeter is found by adding up the four sides (two of which are L, the other two are 2L)

So we have L+L+2L+2L = 2(2L+L) representing the perimeter.

Or we could use this formula

P = 2(L+W)

where L and W are the length and width respectively.

Either way, we end up with 2(2L + L) = 42

Alborosie3 years ago
5 0

Answer:

42 = 2( l+2l)

Step-by-step explanation:

width = 2 length

Perimeter = 2 (l+w)

42 = 2( l+2l)

Divide each side by 2

42/2 = 2 (3l) /2

21 = 3l

Divide by 3

21/3 = 3l/3

7 = l

The length is 7

width is 2*l = 2*7 = 14

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The approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

<h3>What is depreciation?</h3>

Depreciation is to decrease in the value of a product in a period of time. This can be given as,

FV=P\left(1-\dfrac{r}{100}\right)^n

Here, (<em>P</em>) is the price of the product, (<em>r</em>) is the rate of annual depreciation and (<em>n</em>) is the number of years.

Two different cars each depreciate to 60% of their respective original values. The first car depreciates at an annual rate of 10%.

Suppose the original price of the first car is x dollars. Thus, the depreciation price of the car is 0.6x. Let the number of year is n_1. Thus, by the above formula for the first car,

0.6x=x\left(1-\dfrac{10}{100}\right)^{n_1}\\0.6=(1-0.1)^{n_1}\\0.6=(0.9)^{n_1}

Take log both the sides as,

\log 0.6=\log (0.9)^{n_1}\\\log 0.6={n_1}\log (0.9)\\n_1=\dfrac{\log 0.6}{\log 0.9}\\n_1\approx4.85

Now, the second car depreciates at an annual rate of 15%. Suppose the original price of the second car is y dollars.

Thus, the depreciation price of the car is 0.6y. Let the number of year is n_2. Thus, by the above formula for the second car,

0.6y=y\left(1-\dfrac{15}{100}\right)^{n_2}\\0.6=(1-0.15)^{n_2}\\0.6=(0.85)^{n_2}

Take log both the sides as,

\log 0.6=\log (0.85)^{n_2}\\\log 0.6={n_2}\log (0.85)\\n_2=\dfrac{\log 0.6}{\log 0.85}\\n_2\approx3.14

The difference in the ages of the two cars is,

d=4.85-3.14\\d=1.71\rm years

Thus, the approximate difference in the ages of the two cars, which  depreciate to 60% of their respective original values, is 1.7 years.

Learn more about the depreciation here;

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Answer:

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Step-by-step explanation:

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