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Artist 52 [7]
2 years ago
9

What is the total surface area of rectangular prism?

Mathematics
2 answers:
Elena L [17]2 years ago
7 0

Answer:

So surface area is calculated by
2lw + 2lh + 2hw = SA
L is length
W is width
H is Height

SA is surface area.

So plug it in, (reminder this equation only works for cubes, rectangular prisms)

2(3)(6) + 2(3)(7) + 2(7)(6) = ?

So now follow order of operations to solve.
36 + 42 + 84 = 162

So 162m^2 is the surface area.

algol132 years ago
4 0

Answer:

162 [m²].

Step-by-step explanation:

required surface is:

A=(3*7+6*7+3*6)*2=162 [m²].

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Answer:

4.5

Step-by-step explanation:

I think it will work if you Divide 54 over 12 = 4.5

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Kelsey painted one-fourth of her bedroom in three-sevenths of an hour. At this rate, how long would it take her to paint the ent
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May I please get help. ​
Karo-lina-s [1.5K]

Answer:

y = -\frac{2}{3} +6

Step-by-step explanation:

To be parallel to the original line, the slope has to be the same (-\frac{2}{3}). For the line to go through the point (6,2) the y-intercept has to be 6.

The graph shows the original line (black) and new line that goes through the point (yellow).

Hope it helps!

3 0
2 years ago
If sinx = p and cosx = 4, work out the following forms :<br><br><br>​
Kay [80]

Answer:

$\frac{p^2 - 16} {4p^2 + 16} $

Step-by-step explanation:

I will work with radians.

$\frac {\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)} {[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]}$

First, I will deal with the numerator

$\cos^2 \left(\frac{\pi}{2}-x \right)+\sin(-x)-\sin^2 \left(\frac{\pi}{2}-x \right)+\cos \left(\frac{\pi}{2}-x \right)$

Consider the following trigonometric identities:

$\boxed{\cos\left(\frac{\pi}{2}-x \right)=\sin(x)}$

$\boxed{\sin\left(\frac{\pi}{2}-x \right)=\cos(x)}$

\boxed{\sin(-x)=-\sin(x)}

\boxed{\cos(-x)=\cos(x)}

Therefore, the numerator will be

$\sin^2(x)-\sin(x)-\cos^2(x)+\sin(x) \implies \sin^2(x)- \cos^2(x)$

Once

\sin(x)=p

\cos(x)=4

$\sin^2(x)-\cos^2(x) \implies p^2-4^2 \implies \boxed{p^2-16}$

Now let's deal with the numerator

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)]

Using the sum and difference identities:

\boxed{\sin(a \pm b)=\sin(a) \cos(b) \pm \cos(a)\sin(b)}

\boxed{\cos(a \pm b)=\cos(a) \cos(b) \mp \sin(a)\sin(b)}

\sin(\pi -x) = \sin(x)

\sin(2\pi +x)=\sin(x)

\cos(2\pi-x)=\cos(x)

Therefore,

[\sin(\pi -x)+\cos(-x)] \cdot [\sin(2\pi +x)\cos(2\pi-x)] \implies [\sin(x)+\cos(x)] \cdot [\sin(x)\cos(x)]

\implies [p+4] \cdot [p \cdot 4]=4p^2+16p

The final expression will be

$\frac{p^2 - 16} {4p^2 + 16} $

8 0
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Answer:

Is there an image to visualize the field?

Step-by-step explanation:

I would love to help.

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