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Sergio [31]
3 years ago
9

Find an equation for the line that passes through the points (5.-5) and (-4,-2)

Mathematics
1 answer:
ANEK [815]3 years ago
4 0

Answer:

\large\boxed{y=-\dfrac{1}{3}x-\dfrac{10}{3}\to x+3y=-10}

Step-by-step explanation:

The slope-intercept form of an equation of a line:

y=mx+b

<em>m</em><em> - slope</em>

<em>b</em><em> - y-intercept</em>

The formula of a slope:

m=\dfrac{y_2-y_1}{x_2-x_1}

We have the points (5, -5) and (-4, -2).

Substiute:

m=\dfrac{-2-(-5)}{-4-5}=\dfrac{-2+5}{-9}=\dfrac{3}{-9}=-\dfrac{1}{3}

Put the value of the slope and the coordinates of the point (5, -5) to the equation of a line:

-5=-\dfrac{1}{3}(5)+b

-5=-\dfrac{5}{3}+b               <em>add 5/3 to both sides</em>

-\dfrac{15}{3}+\dfrac{5}{3}=b\\\\-\dfrac{10}{3}=b\to b=-\dfrac{10}{3}

Finally we have the equation of a line in the slope-intercept form:

y=-\dfrac{1}{3}x-\dfrac{10}{3}

Convert to the standard form <em>(Ax + By = C)</em>:

y=-\dfrac{1}{3}x-\dfrac{10}{3}          <em>multiply both sides by 3</em>

3y=-x-10              <em>add x to both sides</em>

x+3y=-10

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
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Answer:

f'(x) = 4x^2 - 3 for x \le -3

Step-by-step explanation:

See attachment for proper question

Given

f(x) = -\frac{1}{2}\sqrt{x + 3}

For

x \ge -3

Required

Determine the inverse function

f(x) = -\frac{1}{2}\sqrt{x + 3}

Replace f(x) with y

y = -\frac{1}{2}\sqrt{x + 3}

Swap the positions of x and y

x = -\frac{1}{2}\sqrt{y + 3}

Multiply both sides by -2

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Square both sides

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Make y the subject

y = 4x^2 - 3

The inverse has been solved. So, we need to replace y with f'(x)

f'(x) = 4x^2 - 3

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x \le -3

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f'(x) = 4x^2 - 3 for x \le -3

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Is square root 1 minus sine squared theta = cos Θ true? If so, in which quadrants does angle Θ terminate?
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Answer:

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Step-by-step explanation:

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