All categories

3 years ago

The answer/ why it’s right

Mathematics

2 answers:

brilliants [131]3 years ago

8 0

Answer:

The answer is option 2.

Step-by-step explanation:

It is given that the formula of the volume of rectangular prism is, V = length×width×height. Then you have to substitute the values into the formula to find the volume :

length = 3x

width = 7x

height = 15x

V = (3x) × (7x) × (15x)

= 315x³

pantera1 [17]3 years ago

4 0

<h2>AREA OF RECTANGULAR PRISM=AREA OF cuboid =l×b×h</h2><h2>=15x×7x×3x</h2><h2>=315x³</h2>

FOLLOW ME FOR CLEARING YOUR NEXT DOUBT

You might be interested in

Why do you switch the inequality sign when multiplying or dividing by a negative?

Andreas93 [3]

If we multiply or divide both sides of an inequality by a negative number, you have to flip the inequality. Hope this helps :)

5 0

3 years ago

What is 5x + + 25x + 16 +

bazaltina [42]

2(15x + 8) I hope this helps you

3 0

3 years ago

Convert 0.45 to a percent. Remember to add the percent sign.<br> 0.45%<br> 4.5%<br> 45%<br> 50%

notka56 [123]

45% is the answer to this question

3 0

3 years ago

How to i solve for v for this following equation -19v = -17v – 14

EastWind [94]

The answer will be v=7

8 0

3 years ago

Read 2 more answers

Use the given information to find the exact value of the trigonometric function

eimsori [14]

$\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}$

Half Angle Formula

$\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}$ $\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}$ $\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}$ $\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\ \\ \end{gathered}$

Answer:

$\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}$

Checking:

$\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{ (3rd quadrant)} \end{gathered}$

Also,

$\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}$

The answer is none of the choices

7 0

1 year ago