???? Please help!!!!!!!

Mathematics

1 answer:

trasher [3.6K]3 years ago

6 0

Answer:

It would probably be C

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11. If f(x)=4x-9, what is the equation for f^-1(x)?

FrozenT [24]

F(x) = 4x - 9

let f(x) = y, this implies that x = f⁻¹(y)

y = 4x - 9 Let us solve for x.

4x - 9 = y

4x = y + 9

x = (y + 9)/4

Recall that x = f⁻¹(y),

x = (y + 9)/4

f⁻¹(y) = (y + 9)/4

That means that for f⁻¹(x)

f⁻¹(x) = (x + 9)/4

Hope this explains it.

3 0

3 years ago

Hi! I need help with the attached question in calc. Thank you:)

Elena L [17]

Answer:

3π square units.

Step-by-step explanation:

We can use the disk method.

Since we are revolving around AB, we have a vertical axis of revolution.

So, our representative rectangle will be horizontal.

R₁ is bounded by y = 9x.

So, x = y/9.

Our radius since our axis is AB will be 1 - x or 1 - y/9.

And we are integrating from y = 0 to y = 9.

By the disk method (for a vertical axis of revolution):

$\displaystyle V=\pi \int_a^b [R(y)]^2\, dy$

So:

$\displaystyle V=\pi\int_0^9\Big(1-\frac{y}{9}\Big)^2\, dy$

Simplify:

$\displaystyle V=\pi\int_0^9(1-\frac{2y}{9}+\frac{y^2}{81})\, dy$

Integrate:

$\displaystyle V=\pi\Big[y-\frac{1}{9}y^2+\frac{1}{243}y^3\Big|_0^9\Big]$

Evaluate (I ignored the 0):

$\displaystyle V=\pi[9-\frac{1}{9}(9)^2+\frac{1}{243}(9^3)]=3\pi$

The volume of the solid is 3π square units.

Note:

You can do this without calculus. Notice that R₁ revolved around AB is simply a right cone with radius 1 and height 9. Then by the volume for a cone formula:

$\displaystyle V=\frac{1}{3}\pi(1)^2(9)=3\pi$