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Sati [7]
3 years ago
7

Graph the equation by plotting three

Mathematics
1 answer:
Jlenok [28]3 years ago
6 0

Answer:

-3y = 2x - 7

y = -2/3x - 7/3

Step-by-step explanation:

y cannot be negative nor a whole number.

To get y by itself you have to divide the -3 over to the other side.

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Which of the following is a polynomial?
Artyom0805 [142]

Answer:

C

Step-by-step explanation:

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3 years ago
Consider the polynomial p(x) = x^3 + 4x^2 + 6x − 36.
Nady [450]

Answer:

a. attached graph; zero real: 2

b. p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)

c. the solutions are 2, -3-3i and -3+3i

Step-by-step explanation:

p(x) = x³ + 4x² + 6x - 36

a. Through the graph, we can see that 2 is a real zero of the polynomial p. We can also use the Rational Roots Test.

p(2) = 2³ + 4.2² + 6.2 - 36 = 8 + 16 + 12 - 36 = 0

b. Now, we can use Briott-Ruffini to find the other roots and write p as a product of linear factors.

2 |  1     4     6    -36

     1      6    18     0

x² + 6x + 18 = 0

Δ = 6² - 4.1.18 = 36 - 72 = -36 = 36i²

√Δ = 6i

x = -6±6i/2 = 2(-3±3i)/2

x' = -3-3i

x" = -3+3i

p(x) = (x - 2)(x + 3 + 3i)(x + 3 - 3i)

c. the solutions are 2, -3-3i and -3+3i

4 0
3 years ago
A ball of radius 15 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid. Hint: The u
OverLord2011 [107]

Answer:

The volume of the ball with the drilled hole is:

\displaystyle\frac{8000\pi\sqrt{2}}{3}

Step-by-step explanation:

See attached a sketch of the region that is revolved about the y-axis to produce the upper half of the ball. Notice the function y is the equation of a circle centered at the origin with radius 15:

x^2+y^2=15^2\to y=\sqrt{225-x^2}

Then we set the integral for the volume by using shell method:

\displaystyle\int_5^{15}2\pi x\sqrt{225-x^2}dx

That can be solved by substitution:

u=225-x^2\to du=-2xdx

The limits of integration also change:

For x=5: u=225-5^2=200

For x=15: u=225-15^2=0

So the integral becomes:

\displaystyle -\int_{200}^{0}\pi \sqrt{u}du

If we flip the limits we also get rid of the minus in front, and writing the root as an exponent we get:

\displaystyle \int_{0}^{200}\pi u^{1/2}du

Then applying the basic rule we get:

\displaystyle\frac{2\pi}{3}u^{3/2}\Bigg|_0^{200}=\frac{2\pi(200\sqrt{200})}{3}=\frac{400\pi(10)\sqrt{2}}{3}=\frac{4000\pi\sqrt{2}}{3}

Since that is just half of the solid, we multiply by 2 to get the complete volume:

\displaystyle\frac{2\cdot4000\pi\sqrt{2}}{3}

=\displaystyle\frac{8000\pi\sqrt{2}}{3}

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