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hammer [34]
3 years ago
6

PLEASE HELP I INCLUDED REFERENCED EQUATIONS THROUGH LINKS TO IMGUR - 30 POINTS PLEASE HELP

Mathematics
1 answer:
tatuchka [14]3 years ago
7 0

Answer:

(a) 6x^{4} y^{4}

(b) x^{4} - 5x^{2} - 14x - 12

Step-by-step explanation:

(a)

(3xy^3)(2x^3y)

  • (3*2)(x*x^3)(y*y^3)
  • (6)(x^4)(y^4)

(b)

This is just the rainbow method or as i like to call it the multiplication attack. All the terms in the first bracket fire themselves on the ones in the second. Each term attacks all the other terms once. Once they fire (multiply) they stick and can be added up. To make it easier i like to imagine them as squads.

(x^2 +2x +3)[x^2 -2x-4]

This can be expressed now as

(x^2 +2x +3)[x^2] + (x^2 +2x +3)[-2x] + (x^2 +2x +3)[-4]

Now to expand

x^4 + 2x^3 + 3x^2 -2x^3 -4x^2 - 6x -4x^2 - 8x - 12

Group like terms (same x powers)

(x^4) + (2x^3-2x^3) + (3x^2-4x^2-4x^2) + (- 6x-8x) + (-12)

Simplify

x^{4} - 5x^{2} - 14x - 12

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The curves y = √x and y=(2-x) and the Cartesian axes form two distinct regions in the first quadrant. Find the volumes of rotati
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Answer:

Step-by-step explanation:

If you graph there would be two different regions. The first one would be

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If you rotate the first region around the "y" axis you get that

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And if you rotate the second region around the "y" axis you get that

{\displaystyle A_2 = 2\pi \int\limits_{1}^{2} x(2-x) dx = \frac{4\pi}{3} = 4.188 }

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If you revolve just the outer curve you get

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{\displaystyle A_1 =\pi \int\limits_{0}^{1} ( \sqrt{x})^2 dx = \frac{\pi}{2} = 1.5708 }

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Right ain’t it ?

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