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2 years ago

What is the equation of a line parallel to -x+3y=-6 that passes through 9,-1

Mathematics

1 answer:

Natali [406]2 years ago

5 0

Answer:

${ \tt{ - x + 3y = - 6}} \\ { \tt{3y = x - 6}} \\ { \tt{y = \frac{1}{3}x - 2 }}$

• For parallel line, their gradients are the same:

→ m = ⅓

• From general equation of a line;

$→ \: { \tt{y = mx + c}} \\$

• For the point (9, -1):

$→ \: { \tt{ - 1 = (9 \times \frac{1}{3} ) + c}} \\ \\→ \: { \tt{ - 1 = 3 + c}} \\ \\ → \: { \tt{c = - 4}}$

therefore:

$→ \: { \boxed{ \tt{y = \frac{1}{3}x - 4 }}}$

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Answer:

(a) $f'(x)=-\frac{2}{x^3}$

(b) $y=-0.25x+0.75$

Step-by-step explanation:

The given function is

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(b)

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The coordinates of point of tangency are (2,0.25).

The slope of tangent at x=2 is

$m=(\frac{dy}{dx})_{x=2}=f'(x)_{x=2}$

Substitute x=2 in equation 2.

$f'(2)=\frac{-2}{(2)^3}=\frac{-2}{8}=\frac{-1}{4}=-0.25$

The slope of the tangent line at x=2 is -0.25.

The slope of tangent is -0.25 and the tangent passes through the point (2,0.25).

Using point slope form the equation of tangent is

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$y=-0.25x+0.5+0.25$

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3 years ago

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Answer:

Step-by-step explanation:

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