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sergiy2304 [10]
1 year ago
8

The figures below are similar. The labeled sides are corresponding.4 yd1 ydP1 = ?P2 = 4 ydWhat is the perimeter of the larger sq

uare?P1 = yardsSubmit

Mathematics
1 answer:
juin [17]1 year ago
5 0

In this case, we'll have to carry out several steps to find the solution.

Step 01:

data:

smaller square

side = 1 yd

larger square:

side = 4 yd

Step 02:

geometry:

perimeter:

larger square perimeter:

perimeter = s + s + s + s = 4s

perimeter = 4(4yd) = 16 yd

The answer is:

perimeter = 16 yd

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

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\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.

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\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

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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.

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Wheres the diagram or else the q cant be done


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