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3 years ago

Write the equation of a line in slope intercept form that contains the points (3,-5) and (-3,-3)

Mathematics

1 answer:

ASHA 777 [7]3 years ago

4 0

Answer:

y = -1/3x - 4

Step-by-step explanation:

slope is change in y over the change in x

-3 + 5 = 2

-3 - 3 = -6

slope is -2/6 or -1/3

y+3 = -1/3 (x + 3)

y + 3 = -1/3x -1

y = -1/3x - 4

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3 years ago

Evaluate the integral. (remember to use absolute values where appropriate. Use c for the constant of integration.) 5 cot5(θ) sin

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$I=5\int \frac{cos^{4}\theta }{sin\theta }\times cos\theta d\theta \\\\I=5\int \left ( 1-sin^{2}\theta \right )^{2}\times \frac{cos\theta }{sin\theta }d\theta \\put\ \sin\theta =t\\\\dt=cos\theta d\theta \\\\I=5\int\frac{t^{4}+1-2t^{2}}{t}dt\ \ \ \ \ \ \ \ \ \ \because (a-b)^2=a^2+b^2-2ab\\\\I=5\left ( \int t^{3}dt + \int \frac{1}{t} -2\int t \right )dt$

by using the integration formula

we get,

$\\I=5\left ( \frac{t^{4}}{4} +logt -t^{2}\right )\\\\I=\frac{5}{4}t^{4}+5\log t-5t^{2}+c$

now put the value of t=\sin\theta in the above equation

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$\int 5\cot^5\theta \sin^4\theta d\theta=\frac{5}{4}sin^{4}\theta+5\log \sin\theta - 5sin^{2} \theta+c$

hence proved

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