What is the volume of this triangular prism?

Kellen wants to rent a banquet room in a restaurant for her cousin’s baby shower. The restaurant charges $350 for the banquet ro

irina [24]

Answer:

35

Step-by-step explanation:

To solve this problem, we can make an equation using the values provided to us. The equation is shown below:

$350 + 32.50x = 1,500$

The first step is to subtract 350 from 1,500.

$32.50x=1,500-350$

$32.50x=1,150$

The last step is to divide 1,150 by 32.50 to get <em>x</em> by itself.

$x=\frac{1,150}{32.50}$

$x=35.38$

Kellen can have 35 people at the shower.

5 0

3 years ago

8 A right rectangular prism is completely filled with small cubes. Each small cube

aleksandrvk [35]

A-20

i hope this helps!!

5 0

3 years ago

Please help me! I will give brainliest and points!

hoa [83]

Answer:

5^5

Step-by-step explanation:

He has 5 choices day 1

5 choices day 2

5 choices day 3

5 choices day 4

5 choices day 5

5*5*5*5*5 = 5 ^5

8 0

2 years ago

Read 2 more answers

What is 2/9b=1/3b-7 the answer

Travka [436]

If you would like to know solve the equation 2/9 * b = 1/3 * b - 7, you can calculate this using the following steps:

2/9 * b = 1/3 * b - 7
2/9 * b - 1/3 * b = -7
2/9 * b - 3/9 * b = -7
-1/9 * b = -7
1/9 * b = 7 /*9
b = 7 * 9
b = 63

The correct result would be 63.

5 0

3 years ago

Please please help!!

shepuryov [24]

Answer:

$\large\boxed{x=0\ and\ x=\pi}$

Step-by-step explanation:

$\tan^2x\sec^2x+2\sec^2x-\tan^2x=2\\\\\text{Use}\ \tan x=\dfrac{\sin x}{\cos x},\ \sec x=\dfrac{1}{\cos x}:\\\\\left(\dfrac{\sin x}{\cos x}\right)^2\left(\dfrac{1}{\cos x}\right)^2+2\left(\dfrac{1}{\cos x}\right)^2-\left(\dfrac{\sin x}{\cos x}\right)^2=2\\\\\left(\dfrac{\sin^2x}{\cos^2x}\right)\left(\dfrac{1}{\cos^2x}\right)+\dfrac{2}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}=2$

$\dfrac{\sin^2x}{(\cos^2x)^2}+\dfrac{2-\sin^2x}{\cos^2x}=2\\\\\text{Use}\ \sin^2x+\cos^2x=1\to\sin^2x=1-\cos^2x\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-(1-\cos^2x)}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{2-1+\cos^2x}{\cos^2x}=2\\\\\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{1+\cos^2x}{\cos^2x}=2$

$\dfrac{1-\cos^2x}{(\cos^2x)^2}+\dfrac{(1+\cos^2x)(\cos^2x)}{(\cos^2x)^2}=2\qquad\text{Use the distributive property}\\\\\dfrac{1-\cos^2x+\cos^2x+\cos^4x}{\cos^4x}=2\\\\\dfrac{1+\cos^4x}{\cos^4x}=2\qquad\text{multiply both sides by}\ \cos^4x\neq0\\\\1+\cos^4x=2\cos^4x\qquad\text{subtract}\ \cos^4x\ \text{from both sides}\\\\1=\cos^4x\iff \cos x=\pm\sqrt1\to\cos x=\pm1\\\\ x=k\pi\ for\ k\in\mathbb{Z}\\\\\text{On the interval}\ 0\leq x$

4 0

3 years ago