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2 years ago

Let the graph of g be a translation 3 units up, followed by a vertical stretch by a factor of 2 and a reflection in the x-

Mathematics

1 answer:

Keith_Richards [23]2 years ago

5 0

Answer:

The answer is $g(x)= -6x(x-1)+3$

Step-by-step explanation:

we know that f(x) is written as $3x^{2}-3x$

to show that the graph is being translated up, vertically stretching, and reflecting across the x axis, we need to factor the function.

$f(x)= 3x^{2}-3x \\ 3x(x-1)\\$

now we can incorporate the changes to this graph.

multiply 3x by 2 and -1. (2 for a stretch and -1 for a reflection).

$-6x(x-1)$

tack on the +3 to show a translation of 3 units up.

$-6x(x-1)+3$

therefore, $g(x)= -6x(x-1)+3$

hope this helps!

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The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi

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The height of cone is decreasing at a rate of 0.085131 inch per second.

Step-by-step explanation:

We are given the following information in the question:

The radius of a cone is decreasing at a constant rate.

$\displaystyle\frac{dr}{dt} = -7\text{ inch per second}$

The volume is decreasing at a constant rate.

$\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}$

Instant radius = 99 inch

Instant Volume = 525 cubic inches

We have to find the rate of change of height with respect to time.

Volume of cone =

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

Instant volume =

$525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}$

Differentiating with respect to t,

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

Putting all the values, we get,

$\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131$

Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.

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