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MrRissso [65]
3 years ago
11

please ANSWER at least ONE question CORRECTLY FOR 8 POINTS & ANSWER at least THREE CORRECTLY to be MARKED BRAINLIEST! (you d

on’t have to do all of them, but it would be greatly appreciated if you did.)

Mathematics
1 answer:
4vir4ik [10]3 years ago
8 0

Answer:

a. yes

b. no

c. no

d. no

e. yes

Step-by-step explanation:

Here, we want to answer yes or no for each of the numbers

a) Yes

If we compare with the general form

y = mx + b

m

is slope and b is y-intercept

so, 2 is slope and 3 is y intercept; so 3 in y-intercept is (0,3)

b. No

This is the point-slope form we are trying to

get

Evaluating this, we have that

y-3 = 2x - 6

y = 2x-6 + 3

y = 2x-3

c) NO

Just insert the values and check here

-1 = 2(5) + 3

it does not work

d) No

Substitute

-4 = 2(-5) + 3

e) Yes

This is the point-slope form

y + 3 = 2x + 6

y = 2x + 6-3 = 2x + 3

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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
3 years ago
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