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ankoles [38]
2 years ago
10

Help with the linear equations

Mathematics
1 answer:
IgorC [24]2 years ago
7 0

Answer:

y \: intercept =  - 1 \frac{1}{2}

Step-by-step explanation:

The y intercept form for the equation of a line is

y = mx + c

You should note that <em>c</em> represents the y-intercept of the line (where the line touches the y-axis)

y = mx + c \\ \\ we \:were\: given \:the \:equation \:y = x - \frac{3}{2} \\ \\ therefore \: the\: value \:of \:c \:is\: - \frac{3}{2} \:or -1 \frac {1}{2}

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Answer:

1) F(x)=\frac{2x^{5} }{5}+x^{4}-\frac{x^{2} }{2}+C

2) f(x)=\frac{8x^5}{5}-\frac{3x^4}{2}+\frac{4x^3}{3}+D

Step-by-step explanation:

Applying the antiderivation rules:

F(x)= \frac{2x^{4+1} }{4+1}+4\frac{x^{3+1} }{3+1}-\frac{x^{1+1} }{1+1}+C

F(x)=\frac{2x^{5} }{5}+x^{4}-\frac{x^{2} }{2}+C

Checking by differentitation we have:

F'(x)=2*\frac{5x^{4} }{5}+4*\frac{4x^3}{4}+2*\frac{x}{2}\\F'(x)=2x^4+4x^3-x=f(x)

Which is demonstrated.

2) To find f we must antiderivate twice:

f'(x)=\int\limits {(32x^3-18x^2+8x)} \, dx\\f'(x)=8x^4-6x^3+4x^2+C\\f(x)=\int\limits( {8x^4-6x^3+4x^2)} \, dx\\ f(x)=8\frac{x^5}{5}-6\frac{x^5}{5}+4\frac{x^3}{3}+D\\ f(x)=\frac{8x^5}{5}-\frac{3x^4}{2}+\frac{4x^3}{3}+D

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