G(x)=4+3(x-1), evaluate for g(6)

The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubi

storchak [24]

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

$V = \frac{1}{3}\pi r^{2}h$

Differentiate the equation with respect to time t, such that

$\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)$

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)$

To differentiate the product,

Let r² = u, so that

$\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)$

Then, using product rule

$\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}]$

Since $u = r^{2}$

Then, $\frac{du}{dr} = 2r$

Using the Chain's rule

$\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}$

∴ $\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})]$

Then,

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

Now,

From the question

$\frac{dr}{dt} = 7 m/min$

$\frac{dV}{dt} = 236 m^{3}/min$

At the instant when $r = 99 m$

and $V = 180 m^{3}$

We will determine the value of h, using

$V = \frac{1}{3}\pi r^{2}h$

$180 = \frac{1}{3}\pi (99)^{2}h$

$180 \times 3 = 9801\pi h$

$h =\frac{540}{9801\pi }$

$h =\frac{20}{363\pi }$

Now, Putting the parameters into the equation

$\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}]$

$236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)]$

$236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386]$

$708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}$

$708 = 30790.75 \frac{dh}{dt} + 76.36$

$708 - 76.36 = 30790.75\frac{dh}{dt}$

$631.64 = 30790.75\frac{dh}{dt}$

$\frac{dh}{dt}= \frac{631.64}{30790.75}$

$\frac{dh}{dt} = 0.021 m/min$

Hence, the rate of change of the height is 0.021 meters per minute.

3 0

2 years ago

How do I solve (7-6i)(-8+3i)?

Alekssandra [29.7K]

I = {2.666666667, 1.166666667

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2 years ago

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MDEG

Margaret [11]

The answer would be 84

5 0

2 years ago

Help!! Create a set of coordinates to model a relation that is a linear function.

grigory [225]

Answer:

You can group a ratio or a multiple of x or y to prove a linear function.

To set coordinates randomly pick a title ie) rise in price for matches over 40 years.

$14 yr 10 $20 yr 11 etc. $25 year 12 etc.

We show yr 0 = 0 yr 1 = 8 and if 8 is the price we have a ratio start of 1:8 upon year 1. we then pinpoint the data what year was $16 and we know that yr 10 = $14 so yr 11 = $16.

Once we can write a format which isn't asked we can prove the relationship target of the graph would be x8

As the x y relationship coordinates can be shown here.

= 1 , 8

2 ,16

3 ,24

4, 32

and then change number of years to decades. To make a linear equation work we could change the rate upon the decade that shows a more stable rate of change to be of significance and easier to read.

Step-by-step explanation:

A linear function is a type of function of x and y proves a single line.

When a given ratio or rate of increase occurs ie) xy = 1/8 or 8/1 we can set the 1-4 decades spaced out on a graph and go up by decades since 1980 = decade 1, decade 2 decade 3 decade 4

for x value and for y we have price the actual data of change.

Therefore y = price change from $8 - $32 in last 40 years to appeal to advertisers who want to be ethical and fair for customers who pay more than $32 a game, they look for linear graphs that can show least amount cost of a ticket and average price ticket and compare success stories in advertising to crowds to further testing graphs before advertising so that companies can test advertising before sponsorship which is one way of investment, that can help ease costs of selection of tickets and go full circle for the financier of such games.They need linear graphs to compare to other business as each linear graphs can show better stability. So it is a good example to show costs and prices as prices demonstrate exactly how companies grow compared to their competitors.

7 0

2 years ago

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A square is drawn below what is the value of x

Andru [333]

Answer:

You donf have the picture shown, how is anybody supposed to work on it

3 0

2 years ago

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