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2 years ago

Please help thanks and show work

Mathematics

1 answer:

My name is Ann [436]2 years ago

6 0

First find the area of the triangle at the end.

Area of a triangle is 1/2 x base x height.

The triangle is a right triangle with a base of 4 and height of 3

area = 1/2 x 4 x 3 = 6 square cm

Now to find the volume multiply the area of the triangle by the length of the prism.

volume = 6 square cm x 11 cm = 66 cubic cm.

answer: 66 cm^3

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3 years ago

What is -2b + 4> -6 is ?

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3 years ago

The table models the population, P(n), of a colony of bacteria over n days.

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2 years ago

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

zvonat [6]

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;

brainly.com/question/25297296

4 0

2 years ago