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Galina-37 [17]
3 years ago
5

Find f(x) - g(x) *

Mathematics
1 answer:
Studentka2010 [4]3 years ago
6 0

Answer:

3x^3+13x^2-3x+7

Step-by-step explanation:

I dont know if your question is complete.

f(x) - g(x)

(4x^3+6x^2-3x+9)-(x^3-7x^2+2)\\4x^3+6x^2-3x+9-x^3+7x^2-2\\3x^3+13x^2-3x+7

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Read 2 more answers
Sami has 89 video games. He keeps 14 for himself and distributes the remaining video games to each
Natasha2012 [34]

Answer:

89=14-5x

Step by step

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5 0
4 years ago
Find parametric equations for the path of a particle that moves along the circle x2 + (y − 1)2 = 16 in the manner described. (En
ArbitrLikvidat [17]

Answer:

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t, c) x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right), y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right).

Step-by-step explanation:

The equation of the circle is:

x^{2} + (y-1)^{2} = 16

After some algebraic and trigonometric handling:

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = 1

\frac{x^{2}}{16} + \frac{(y-1)^{2}}{16} = \cos^{2} t + \sin^{2} t

Where:

\frac{x}{4} = \cos t

\frac{y-1}{4} = \sin t

Finally,

x = 4\cdot \cos t

y = 1 + 4\cdot \sin t

a) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

b) x = 4\cdot \cos t, y = 1 + 4\cdot \sin t.

c) x = 4\cdot \cos t'', y = 1 + 4\cdot \sin t''

Where:

4\cdot \cos t' = 0

1 + 4\cdot \sin t' = 5

The solution is t' = \frac{\pi}{2}

The parametric equations are:

x = 4\cdot \cos \left(t+\frac{\pi}{2}  \right)

y = 1 + 4\cdot \sin \left(t + \frac{\pi}{2} \right)

7 0
3 years ago
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