Ask question

Ask question

All categories

Ad libitum [116K]

3 years ago

11

Find the surface area of a rectangular prism that is 16 inches long, 12 inches wide, and 5 inches high.

1 answer:

OverLord2011 [107]3 years ago

4 0

We have an equation for the surface area of a rectangular prism:
A= 2 (w*l+ l*h+ h*w)

Plug l= 16 in, w= 12 in, and h= 5 in into the above equation, we have:
A= 2 (12 in*16 in+ 16 in* 5 in+ 5 in* 12 in)= 664 in^(2).

The final answer is 664 in^(2).

Hope this is helpful~

You might be interested in

The diagonals of a rectangle are always

guapka [62]

Answer:

The diagonals of a rectangle are always congruent

7 0

2 years ago

Read 2 more answers

What is the length of line segment KJ?

Karolina [17]

Please consider the attached file.

We can see that triangle JKM is a right triangle, with right angle at M. Segment KM is 6 units and segment MJ is 3 units. We can also see that KJ is hypotenuse of right triangle.

We will use Pythagoras theorem to solve for KJ as:

$KJ^2=KM^2+MJ^2$

$KJ^2=6^2+3^2$

$KJ^2=36+9$

$KJ^2=45$

Now we will take positive square root on both sides:

$\sqrt{KJ^2}=\sqrt{45}$

$KJ=\sqrt{9\cdot 5}$

$KJ=3\sqrt{5}$

Therefore, the length of line segment KJ is $3\sqrt{5}$ and option D is the correct choice.

5 0

3 years ago

Hi everone can you help i will be sending more qusetions each one is 10 points thank you

kkurt [141]

Answer:

144 cubic inches

Step-by-step explanation:

volume = l x w x h. This = 8 x 6 x 3.

Hope this helps :)

8 0

2 years ago

Read 2 more answers

Can you guys help me with this question

VLD [36.1K]

Answer:

2+2=4 quick mafs

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Solve sin 0 + 1 = cos20 on the interval 0 ≤ 0 < 2pi. Show work please!

yan [13]

Answer:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}\frac{2\pi}{3}\frac{4\pi}{3}$

Step-by-step explanation:

You need 2 things in order to solve this equation: a trig identity sheet and a unit circle.

You will find when you look on your trig identity sheet that

$cos(2\theta)=1-2sin^2(\theta)$

so we will make that replacement, getting everything in terms of sin:

$sin(\theta)+1=1-2sin^2(\theta)$

Now we will get everything on one side of the equals sign, set it equal to 0, and solve it:

$2sin^2(\theta)+sin(\theta)=0$

We can factor out the sin(theta), since it's common in both terms:

$sin(\theta)(2sin(\theta)+1)=0$

Because of the Zero Product Property, either

$sin(\theta)=0$ or

$2sin(\theta)+1=0$

Look at the unit circle and find which values of theta have a sin ratio of 0 in the interval from 0 to 2pi. They are:

$\theta=\frac{\pi}{2},\frac{3\pi}{2}$

The next equation needs to first be solved for sin(theta):

$2sin(\theta)+1=0$ so

$2sin(\theta)=-1$ and

$sin(\theta)=-\frac{1}{2}$

Go back to your unit circle and find the values of theta where the sin is -1/2 in the interval. They are:

$\theta=\frac{2\pi}{3},\frac{4\pi}{3}$

7 0

3 years ago