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3 years ago

12

Please help me solve this!

1 answer:

romanna [79]3 years ago

3 0

So basically, the question is asking
out of all the girls, how many are going to a 4 year college so

we see that in the 4 year college row and girls collumn, the number is 67

and the total at the bottom of the girl's collumn is 129 so

67/129=0.519 which rounds up to 0.52

to convert decimal to percent, just multiply the number by 100 so
0.52 times 100=52% which is C

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What happens to the function f(x)=x4 when it becomes f′(x)=x4+5?

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2 years ago

A blank is a rectangular

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3 years ago

The surface area of a cylinder is increasing at a rate of 9 pi square meters per hour. The height of the cylinder is fixed at 3

Alekssandra [29.7K]

Answer:

$9\pi \text{ cubic meters per hour}$

Step-by-step explanation:

Since, the surface area of a cylinder,

$A= 2\pi r^2 + 2\pi rh$ ................(1)

Where,

r = radius,

h = height,

If $A= 36\pi\text{ square meters}, h = 3\text{ meters}$

$36\pi = 2\pi r^2 + 2\pi r(3)$

$18 = r^2 + 3r$

$\implies r^2 + 3r - 18=0$

$r^2 + 6r - 3r - 18 = 0$ ( by middle term splitting )

$r(r+6)-3(r+6)=0$

$(r-3)(r+6)=0$

By zero product property,

r = 3 or r = - 6 ( not possible )

Thus, radius, r = 3 meters,

Now, differentiating equation (1) with respect to t ( time ),

$\frac{dA}{dt}= 4\pi r\frac{dr}{dt} +2\pi(r\frac{dh}{dt} + h\frac{dr}{dt})$

∵ h = constant, ⇒ dh/dt = 0,

$\frac{dA}{dt} = 4\pi r \frac{dr}{dt} +2\pi h \frac{dr}{dt}$

We have, $\frac{dA}{dt}=9\pi\text{ square meters per hour}, r = h = 3\text{ meters}$

$9\pi = 4\pi (3) \frac{dr}{dt}+2\pi (3)\frac{dr}{dt}$

$9\pi = (12\pi + 6\pi )\frac{dr}{dt}$

$9\pi = 18\pi \frac{dr}{dt}$

$\implies \frac{dr}{dt} =\frac{1}{2}\text{ meter per hour}$

Now,

Volume of a cylinder,

$V=\pi r^2 h$

Differentiating w. r. t. t,

$\frac{dV}{dt}=\pi ( r^2 \frac{dh}{dt}+h(2r)\frac{dr}{dt})=\pi ((3)(6) (\frac{1}{2})) = 9\pi \text{ cubic meters per hour}$

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3 years ago

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3 years ago

chegg evaluate the following integral using trigonometric substitution. dx/square root(x^2-81), x > 9

vesna_86 [32]

Integration will be ln | ( x /9 ) + ( $\sqrt{(x/9)^{2} - 1 }$ ) | + c .

Given ∫ 1 dx / ( $x^{2}$ - 81 )

Put ,

x = 9 secθ

dx = 9 secθ tanθ dθ

and

$\sqrt{x^{2} - 9^{2} }$ = 9 tanθ

Substituting values in ∫ 1 dx / ( $x^{2}$ - 81 ) ,

∫ (9 secθ tanθ ) dθ / ( 9 tanθ )

∫ secθ dθ = ln | secθ + tanθ | + c

= ln | ( x /9 ) + ( $\sqrt{x^{2} - 9^{2} }$ / 9 ) | + c

= ln | ( x /9 ) + ( $\sqrt{(x/9)^{2} - 1 }$ ) | + c

Hence , the integration will be ln | ( x /9 ) + ( $\sqrt{(x/9)^{2} - 1 }$ ) | + c .

To learn more on integration follow link :

brainly.com/question/20156869

#SPJ4

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1 year ago