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2 years ago

A right cylinder has a diagonal length of 37 and a total surface area of 492.

Mathematics

1 answer:

Aneli [31]2 years ago

8 0

Answer:

Step-by-step explanation:

A right cylinder has a diagonal length of 37 and a total surface area of 492.

What is the height of the cylinder?

A. 35

B. 42

C. 25

D. 17

E. 32

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An electric current, I, in amps, is given by I=cos(wt)+√8sin(wt), where w≠0 is a constant. What are the maximum and minimum valu

exis [7]

Take the derivative with respect to t
$- w \sin(wt) + \sqrt{8} w cos(wt)$
the maximum and minimum values occur when the tangent line is zero so we set the derivative to zero
$0 = -w \sin(wt) + \sqrt{8} w cos(wt)$
divide by w
$0 =- \sin(wt) + \sqrt{8} cos(wt)$
we add sin(wt) to both sides

$\sin(wt)= \sqrt{8} cos(wt)$
divide both sides by cos(wt)
$\frac{sin(wt)}{cos(wt)}= \sqrt{8} \\ \\ arctan(tan(wt))=arctan( \sqrt{8} ) \\ \\ wt=arctan(2 \sqrt{2)}$ OR $\\$ $wt=arctan( { \frac{1}{ \sqrt{2} } )$
(wt)=2(n*pi-arctan(2^0.5))
(wt)=2(n*pi+arctan(2^-0.5))
where n is an integer
the absolute max and min will be

$I=cos(2n \pi -2arctan( \sqrt{2} ))$
since 2npi is just the period of cos
$cos(2arctan( \sqrt{2} ))= \frac{-1}{3} 
$
substituting our second soultion we get
$I=cos(2n \pi +2arctan( \frac{1}{ \sqrt{2} } ))$
since 2npi is the period
$I=cos(2arctan( \frac{1}{ \sqrt{2}} ))= \frac{1}{3}$
so the maximum value = $\frac{1}{3}$
minimum value = $- \frac{1}{3}$

4 0

3 years ago

there are 239 people in the first six rows of the movie theater if the same number of people are in the next six rows about how

bija089 [108]

Answer:

478 people

Step-by-step explanation:

All you need to do for this problem is simply multiply 239 by 2. If the first 6 rows are 239 people, and so are the next 6, that's all you need to do.

239*2=478

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2 years ago

Help?? I seriously don’t understand?

Anvisha [2.4K]

Answer:

Step-by-step explanation:

3(x-2)

Replace x as 12

3(12-2)

3(10)

3 times 10

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2 years ago

Need quick answers has to show work

Zielflug [23.3K]

Answer:

Step-by-step explanation:

For some read I can’t see the pictures

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3 years ago

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

adoni [48]

Answer: The required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Step-by-step explanation:

Since we have given that

$y=\ln[x(2x+3)^2]$

Differentiating log function w.r.t. x, we get that

$\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [x'(2x+3)^2+(2x+3)^2'x]\\\\\dfrac{dy}{dx}=\dfrac{1}{[x(2x+3)^2]}\times [(2x+3)^2+2x(2x+3)]\\\\\dfrac{dy}{dx}=\dfrac{4x^2+9+12x+4x^2+6x}{x(2x+3)^2}\\\\\dfrac{dy}{dx}=\dfrac{8x^2+18x+9}{x(2x+3)^2}$

Hence, the required derivative is $\dfrac{8x^2+18x+9}{x(2x+3)^2}$

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3 years ago