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S_A_V [24]
2 years ago
8

Find the surface area of the cylinder in terms of x.

Mathematics
2 answers:
Alekssandra [29.7K]2 years ago
7 0
Surface area in units of PI = 2 x r x h +2 x r^2

first picture:

r = 7, h = 18

2 x 7 x 18 + 2 x 7^2

252 + 98

350 PI cm^2

2nd picture:  

2 x 9 x 20  +2 x 9^2

360 + 162

522 PI in^2


svlad2 [7]2 years ago
4 0
Surface area = 2πr² + 2πrh  

Surface are = 2π (7)² + 2π(7)(18) = 98π + 252π = 350π cm²


Surface area = 2πr² + 2πrh  

Surface are = 2π (9)² + 2π(9)(20) = 162π + 360π = 522π cm²
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If the speed limit on a interstate is 45 and someone is going 85 how much more are they going than the speed limit?
-Dominant- [34]

Answer:

40 miles over the speed limit

Step-by-step explanation:

You want the difference between the two numbers so you subtract 45 from 85 which gives you 40

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PLE}ASE HELPP!!!!!!!!!!!!!
Dimas [21]
70 degrees! This is the answer. Notice that the triangle ACF is isosceles, therefore, 35, 35 degrees, and 110 the vertex at F, then AB is 180 deg, so 70 left!

 
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2 years ago
How do you simplify (4 – 5) – (13 – 18 + 2) ?
denpristay [2]

Answer:

2

Step-by-step explanation:

4-5= -1

13-18+2= (-5+2)= -3

-1-(-3)=2

Don't forget your order of operations

P (parenthesis)

E (exponents)

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Find F"(x) if f(x) = cot (x)
hammer [34]

f(x)=\cot x\implies f'(x)=-\csc^2x\implies\boxed{f''(x)=2\csc^2x\cot x}

If you don't know the first derivative of \cot, but you do for \sin and \cos, you can derive the former via the quotient rule:

\cot x=\dfrac{\cos x}{\sin x}

\implies(\cot x)'=\dfrac{\sin x(-\sin x)-\cos x(\cos x)}{\sin^2x}=-\dfrac1{\sin^2x}=-\csc^2x

or if you know the derivative of \tan:

\cot x=\dfrac1{\tan x}

\implies(\cot x)'=-(\tan x)^{-2}\sec^2x=-\dfrac{\sec^2x}{\tan^2x}=-\dfrac{\frac1{\cos^2x}}{\frac{\sin^2x}{\cos^2x}}=-\dfrac1{\sin^2x}=-\csc^2x

As for the second derivative, you can use the power/chain rules:

(-\csc^2x)'=-2\csc x(\csc x)'=-2\csc x(-\csc x\cot x)=2\csc^2x\cot x

or if you don't know the derivative of \csc,

\csc x=\dfrac1{\sin x}

\implies(-\csc^2x)'=\left(-(\sin x)^{-2}\right)'=2(\sin x)^{-3}(\sin x)'=\dfrac{2\cos x}{\sin^3x}

which is the same as the previous result since

\csc^2x\cot x=\dfrac1{\sin^2x}\dfrac{\cos x}{\sin x}=\dfrac{\cos x}{\sin^3x}

4 0
3 years ago
What is the answer for. 2x+16y-36y-9+1-x+3x
poizon [28]

After solving 2x+16y-36y-9+1-x+3x we get 4x-20y-8

Step-by-step explanation:

We need to solve 2x+16y-36y-9+1-x+3x

Combining the like terms:

2x+16y-36y-9+1-x+3x\\=2x-x+3x+16y-36y-9+1\\=4x-20y-8

So, After solving 2x+16y-36y-9+1-x+3x we get 4x-20y-8

Keywords: Polynomials

Learn more about Polynomials at:

  • brainly.com/question/1563227
  • brainly.com/question/11207748
  • brainly.com/question/4390083

#learnwithBrainly

3 0
3 years ago
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