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2 years ago

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Javier has a model of a cone with a radius of 3 inches and a height of 10 inches. Javier increased his model’s volume by doublin

g its radius and height. What is the volume of the larger cone in cubic inches?

2 answers:

yan [13]2 years ago

8 0

Answer:

240 pi in. ^2

Step-by-step explanation:

Just took the test. Correct me if I'm wrong!

Thank you

Tisa B Sora

statuscvo [17]2 years ago

3 0

Answer:

$240 \pi\ in^{3}$

Step-by-step explanation:

we know that

If two figures are similar, then the ratio of its volumes is equal to the scale factor elevated to the cube

Let

z------> the scale factor

x-----> the volume of the larger cone

y-----> the volume of the original cone

$z^{3}=\frac{x}{y}$

In this problem we have

$z=2$ -----> scale factor

substitute

$2^{3}=\frac{x}{y}$

$8=\frac{x}{y}$

$x=8y$

That means-----> The volume of the larger cone is 8 times the volume of the original cone

Find the volume of the original cone

$V=\frac{1}{3}\pi r^{2} h$

we have

$r=3\ in$

$h=10\ in$

substitute

$V=\frac{1}{3}\pi (3^{2})(10)=30 \pi\ in^{3}$

therefore

The volume of the larger cone is equal to

$8*30 \pi\ in^{3}=240 \pi\ in^{3}$

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Aleonysh [2.5K]

Answer:

The partial fraction decomposition is $\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{50}{x + 1}+\frac{-29}{\left(x + 1\right)^{2}}+\frac{-54}{x + 2}$ .

Step-by-step explanation:

Partial-fraction decomposition is the process of starting with the simplified answer and taking it back apart, of "decomposing" the final expression into its initial polynomial fractions.

To find the partial fraction decomposition of $\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}$ :

First, the form of the partial fraction decomposition is

$\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{A}{x + 1}+\frac{B}{\left(x + 1\right)^{2}}+\frac{C}{x + 2}$

Write the right-hand side as a single fraction:

$\frac{- 4 x^{2} + 13 x - 12}{\left(x + 1\right)^{2} \left(x + 2\right)}=\frac{\left(x + 1\right)^{2} C + \left(x + 1\right) \left(x + 2\right) A + \left(x + 2\right) B}{\left(x + 1\right)^{2} \left(x + 2\right)}$

The denominators are equal, so we require the equality of the numerators:

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Expand the right-hand side:

$- 4 x^{2} + 13 x - 12=x^{2} A + x^{2} C + 3 x A + x B + 2 x C + 2 A + 2 B + C$

The coefficients near the like terms should be equal, so the following system is obtained:

$\begin{cases} A + C = -4\\3 A + B + 2 C = 13\\2 A + 2 B + C = -12 \end{cases}$

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