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Helga [31]
3 years ago
5

Point C is on the segment AB which has endpoints A(-1, 0) and B(3, 8). Point C is three times as far from point A as it is from

point B. What are the coordinates of point C?
Mathematics
1 answer:
Viefleur [7K]3 years ago
8 0

Answer:

  C = (2, 6)

Step-by-step explanation:

The coordinates of point C can be found as the weighted average of the endpoint coordinates. The weights are the reverse of the relative segment lengths.

For AC : CB = 3 : 1, we have ...

  C = (A +3B)/(1+3) = ((-1, 0) +3(3, 8))/4 = (-1+9, 0+24)/4

  C = (2, 6)

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5 0
3 years ago
F(x) = 3x + x3
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Answer:

Please check the explanation.

Step-by-step explanation:

Given

f(x) = 3x + x³

Taking differentiate

\frac{d}{dx}\left(3x+x^3\right)

\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'

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solving

\frac{d}{dx}\left(3x\right)

\mathrm{Take\:the\:constant\:out}:\quad \left(a\cdot f\right)'=a\cdot f\:'

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\mathrm{Apply\:the\:common\:derivative}:\quad \frac{d}{dx}\left(x\right)=1

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=3

now solving

\frac{d}{dx}\left(x^3\right)

\mathrm{Apply\:the\:Power\:Rule}:\quad \frac{d}{dx}\left(x^a\right)=a\cdot x^{a-1}

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Thus, the expression becomes

\frac{d}{dx}\left(3x+x^3\right)=\frac{d}{dx}\left(3x\right)+\frac{d}{dx}\left(x^3\right)

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f'(x) = 3 + 3x²

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substituting the value  f'(x) = 15 in f'(x) = 3 + 3x²

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3x² = 15-3

3x² = 12

Divide both sides by 3

x² = 4

\mathrm{For\:}x^2=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt{f\left(a\right)},\:\:-\sqrt{f\left(a\right)}

x=\sqrt{4},\:x=-\sqrt{4}

x=2,\:x=-2

Thus, the value of x​ will be:

x=2,\:x=-2

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