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3 years ago

What is the slope-intercept equation of the line going through (-2,5) and (1,-1)?

Mathematics

1 answer:

jarptica [38.1K]3 years ago

3 0

Answer:

y = -2x+1

Step-by-step explanation:

The slope is found by

m = (y2-y1)/(x2-x1)

= (-1-5)/(1--2)

= -6/(1+2)

= -6/3

= -2

Then we can use point slope form to make an equation

y-y1 = m(x-x1)

y-5 = -2(x--2)

y-5 = -2(x+2)

Distribute

y-5 = -2x -4

Add 5 to each side

y-5+5 = -2x-4+5

y = -2x+1

This is in point slope form

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If you're using the app, try seeing this answer through your browser: brainly.com/question/2822772

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Evaluate the indefinite integral:

$\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx}\\\\\\
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Make the following substitution:

$\mathsf{sin\,x=u\quad\Rightarrow\quad cos\,x\,dx=du}$

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$=\mathsf{\displaystyle\int\! \frac{1}{u^2}\,du}\\\\\\
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Integrate it by applying the power rule:

$\mathsf{=\dfrac{u^{-2+1}}{-2+1}+C}\\\\\\
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Now, substitute back for u = sin x, so the result is given in terms of x:

$\mathsf{=-\,\dfrac{1}{sin\,x}+C}\\\\\\
\mathsf{=-\,csc\,x+C}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\displaystyle\int\! \frac{cos\,x}{sin^2\,x}\,dx=-\,csc\,x+C} \end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>indefinite integral substitution trigonometric trig function sine cosine cosecant sin cos csc differential integral calculus</em>

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