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Mamont248 [21]
3 years ago
5

Write the equation of the line perpendicular to

Mathematics
1 answer:
r-ruslan [8.4K]3 years ago
5 0
Solving for y, we add 5y to both sides and subtract 4, getting 9x-4=5y. Dividing both sides by 5, we get 9x/5-4/5=y. Since the slope is 9/5 (since 9/5*x=9x/5), we multiply it by -1 and find the reciprocal of it to get -5/9 as the perpendicular slope, so -5x/9+b=y. Plugging 1 in for x and -6 in for y, we get -5*1/9+b=-6 and by adding 5/9 to both sides we get -5-4/9=b , and since in y=mx+b y and x are variables, we end up with y=-5x/9+(-5-4/9) for slope intercept form.

To get it into standard form, we need it in ay+cx=b with a, b, and c being constants. Adding 5x/9 to both sides, we end up with y+5x/9=(-5-4/9) for standard form


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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.

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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

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\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

We acquire the exact same answer.

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