The height of the box is 6 inches. the length of the box is 14 inches and the width of the box is 4 inches.

Mathematics

1 answer:

Amanda [17]3 years ago

5 0

The answer would be C because to find surface area you have to do this: How many faces does the shape have? 6. What are the faces? Front/Back Left/Right Top/Bottom. So then you would find out what Front/Bottom is so you look at the front shape what's the base of it? 14 inches. What's the height of it? 6 inches. 14×6 is 84 now you add 84 and 84 because front ANDROID back so it's 168. Save that number now you do the same to the other sides and add up all the answers at the end to find the surface area. DONT FORGET THIS IS NOT FOR TRIANGULAR PRISMS!

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How do I differentiate this?

disa [49]

Answer:

$\frac{8}{\left(-x+2\right)^2}$ & $x = 1, x = 3$

Step-by-step explanation:

I'm sure you are familiar with the product rule,

If y = u*v => dy/dx = u * dv/dx + v * dy/dx <----- product rule

In this case:

$u=3x+2,\:v=\left(2-x\right)^{-1},\\=>\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

Now remember the sum rule:

$\frac{d}{dx}\left(3x+2\right) = \frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(2\right),\\\frac{d}{dx}\left(3x\right) = 3,\\\frac{d}{dx}\left(2\right) = 0\\\frac{d}{dx}\left(3x+2\right) = 3$

For this second bit we apply the chain rule:

$\frac{d}{dx}\left(\left(2-x\right)^{-1}\right) = -\frac{1}{\left(2-x\right)^2}\frac{d}{dx}\left(2-x\right),\\\frac{d}{dx}\left(2-x\right) = -1,\\\\=> -\frac{1}{\left(2-x\right)^2}\left(-1\right)\\=> \frac{1}{\left(2-x\right)^2}$

If we substitute these values back into the expression... $\frac{d}{dx}\left(3x+2\right)\left(2-x\right)^{-1}+\frac{d}{dx}\left(\left(2-x\right)^{-1}\right)\left(3x+2\right)$

...we get the following:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)$

The rest is just pure simplification:

$3\left(2-x\right)^{-1}+\frac{1}{\left(2-x\right)^2}\left(3x+2\right)\\= \frac{3}{-x+2}+\frac{3x+2}{\left(-x+2\right)^2}\\= \frac{3\left(-x+2\right)}{\left(-x+2\right)^2}+\frac{3x+2}{\left(-x+2\right)^2}\\\\= \frac{3\left(-x+2\right)+3x+2}{\left(-x+2\right)^2}\\\\= \frac{8}{\left(-x+2\right)^2}$

Now let's equate this to equal 8 for the second bit and solve for x:

$\frac{8}{\left(-x+2\right)^2}=8,\\\frac{8}{\left(-x+2\right)^2}\left(-x+2\right)^2=8\left(-x+2\right)^2,\\8=8\left(-x+2\right)^2,\\\left(-x+2\right)^2=1,\\x = 1, x = 3$

4 0

3 years ago

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Luden [163]

You have to do whats in the parentheses from left to right then subtract the answer you get by 20

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= -576.

You have to use pemdas for your check off
P=parentheses
E=exponet
M=multiplication
D=division
A=add
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HOPE IM CORRECT

4 0

2 years ago

Read 2 more answers

Hey guys!!! I forgot how to do this :/ lol<br> whats 10+4÷2?

Travka [436]

Answer:

Step-by-step explanation:

8 0

2 years ago

Please answer this correctly without making mistakes

satela [25.4K]

Answer: