Given f(x) = x + 2, describe how setting k = 4 affects the slope and y-intercept of the graph of g compared to the graph of f.

A globe has a diameter of 12 inches. It fits inside a cube-shaped box that has a side length of 12 inches.

AlekseyPX

The volume, rounded to the nearest hundredth of a cubic inch, of the space inside the box that is not taken up by the globe is 823.68 inches³

<h3>Volume of a sphere</h3>

The globe is a sphere. Therefore, the volume is as follows:

volume = 4 / 3 πr³

where

r = 12 /2 = 6 inches

volume = 4 / 3 × π × 6³

volume = 864 / 3 π

volume = 288 × 3.14 = 904.32inches³

<h3>Volume of a cube</h3>

The box is a cube. Therefore,

volume = L³

where

L = length

Therefore,

volume = 12³

volume = 1728 inches³

Therefore, the volume of the space inside the box that is not taken up by the globe is as follows:

volume not taken up = 1728 - 904.32 = 823.68 inches³

learn more on volume here: brainly.com/question/15385003

4 0

2 years ago

Factor completely.<br> 1^3-8x^2+16

Ratling [72]

Answer:

-8x^2+17

Step-by-step explanation:

Evaluate the exponent.

1^3-8x^2+16

1-8x^2+16

Add the numbers.

17-8x^2.

Rearrange terms (final answer).

-8x^2+17.

3 0

3 years ago

Write the given expression as an algebraic expression in x. tan(2 cos^-1(x))

miv72 [106K]

$\bf tan\left[ 2cos^{-1}(x) \right]\implies tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]
\\\\\\
\textit{if we say }cos^{-1}\left( \frac{x}{1} \right)=\theta\textit{ that means }tan\left[ 2cos^{-1}\left( \frac{x}{1} \right) \right]\iff tan(2\theta)\\\\
-----------------------------\\\\
cos^{-1}\left( \frac{x}{1} \right)=\theta\implies cos(\theta)=\cfrac{x}{1}\cfrac{\leftarrow adjacent=a}{\leftarrow hypotenuse=c}\\\\
-----------------------------\\\\
$

$\bf \textit{again, using the pythagorean theorem to get the opposite side}
\\\\\\
c^2=a^2+b^2\implies \pm\sqrt{c^2-a^2}=b\implies \pm\sqrt{1^2-x^2}=b
\\\\\\
\pm\sqrt{1-x^2}=b\\\\
-----------------------------\\\\
tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\implies tan(2\theta)=\cfrac{2\cdot \frac{\pm\sqrt{1-x^2}}{x}}{1-\left( \frac{\pm\sqrt{1-x^2}}{x} \right)^2}
\\\\\\$

$\bf tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{1-\frac{1-x}{x}}\implies 
tan(2\theta)=\cfrac{\frac{\pm2\sqrt{1-x^2}}{x}}{\frac{x-1+x}{x}}
\\\\\\
tan(2\theta)=\cfrac{\pm2\sqrt{1-x^2}}{x}\cdot \cfrac{x}{2x-1}
\implies 
tan(2\theta)=\cfrac{\pm 2\sqrt{1-x^2}}{2x-1}$

6 0

4 years ago

Two-fifths of a number is decreased by three. Find the value of the expression if the number is twenty-five.

Sindrei [870]

Let's take the number as x.

two-fifths of a number = 2x / 5
decreased by 3 = (2x/5) - 3

(2x/5) - 3
(2*25/5) - 3
(2*5) - 3
10 - 3
7

8 0

3 years ago

Read 2 more answers

(a^3b^12c^2)*(a^5c^2)/(b^5c^4)^0=

drek231 [11]

I dont understand what yopu typed out.

8 0

3 years ago