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Dovator [93]
3 years ago
7

How do I find surface area in a rectangular prisms

Mathematics
2 answers:
-Dominant- [34]3 years ago
7 0

Answer:to find any geomtric shape draw like a solid box, if you can faltten it out and add diagrams,and labels

but volume is diffrent

Step-by-step explanation:

if helpful can you mark me brainest or give me a thanks

taurus [48]3 years ago
4 0
Here is an example or you can just fold it out.

You might be interested in
Q(x)= 1/2x - 3; q(x) =-4. Find the value of x
Lelu [443]

Answer:

q(x) =-5

Step-by-step explanation:

I am solving this using Function Notation.

We are given:

q(x)= \frac{1}{2}x - 3

While we are to find:

q(x) =-4

In Function Notation, replace all the "x" you see with the value you are given.

So, in this case,

q(x)= \frac{1}{2}x - 3\\\\=\frac{1}{2}x - 3

Would become

q(x)= \frac{1}{2}x - 3\\\\=\frac{1}{2}(-4) - 3

Now we can solve.

\frac{1}{2}(-4) - 3

Use PEMDAS; Parenthesis first;

\frac{1}{2}(-\frac{4}{1})

Cross divide: 1 cancels 1, -4 ÷ 2 = -2

Hence, we are left with

(-2) - 3

-5

Therefore, the value of x is -5

8 0
2 years ago
I have 20 movies 85% are action how many are action
Alona [7]
20*0.85
17
17 are action movies
8 0
3 years ago
Read 2 more answers
Tell me the answer and how u did it!
Elena L [17]
The answer is c. I just did the math it is hard to explain how I did it though. Hopefully this helps you! :)
7 0
3 years ago
Read 2 more answers
One question simple please help
igor_vitrenko [27]

Answer:

y = \frac{1}{3}x - 9

Step-by-step explanation:

Standard equation of a line is y = mx + b, where m is the slope.

Given line y = - 3x + 78, slope, m₁ = -3

<em><u>To find the line perpendicular to the given line.</u></em>

The lines are perpendicular to each other if the product of their slopes = - 1

That is,

         m_1 \times m_2 = -1

So the slope of new line is

                                                   -3 \times m_2 = -1\\\\m_2 = \frac{-1}{-3} = \frac{1}{3}

Therefore , equation \ of \ new \ line \ \\\\(y - y_1) = m_2(x - x_1) , where \ (x_1 , y_1) = (9, -6 )\\\\(y - (-6))= \frac{1}{3}(x -9)\\\\y + 6 = \frac{1}{3} x - 3\\\\y = \frac{1}{3}x -3-6\\\\y = \frac{1}{3}x -9\\\\

7 0
3 years ago
Y = (1/4)x^2 - (1/2)lnx..over the interval (1, 7e) ...what is the arc length ?
xenn [34]
So, f[x] = 1/4x^2 - 1/2Ln(x) 
<span>thus f'[x] = 1/4*2x - 1/2*(1/x) = x/2 - 1/2x </span>
<span>thus f'[x]^2 = (x^2)/4 - 2*(x/2)*(1/2x) + 1/(4x^2) = (x^2)/4 - 1/2 + 1/(4x^2) </span>
<span>thus f'[x]^2 + 1 = (x^2)/4 + 1/2 + 1/(4x^2) = (x/2 + 1/2x)^2 </span>
<span>thus Sqrt[...] = (x/2 + 1/2x) </span>
6 0
3 years ago
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