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3 years ago

Write an equation in point-slope form for the line that passes through the given points (1,-3),(4,-15)

1 answer:

5 0

ANSWER: y= -4x+1
slope-intercept form is y=mx+b

to find m, which is the slope, we used
m=(y2-y1)/(x2-x1) (rise over run)
=(-15-(-3)/4-1)
= -12/3
= -4

to find b, choose one of the points and use the slope -4 and plug it in:
y=m(x)+b
-3= -4(1)+b
-3= -4+b
1=b

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3 years ago

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KIM [24]

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3 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

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2 years ago

Find the surface area of this square pyramid

coldgirl [10]

Answer:

the answer should be 96

Step-by-step explanation:

hope this helps :)

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3 years ago

Plz factorize this equation: 144-9c^2

nydimaria [60]

-9(c+4)(c-4)
this is the answer to your question

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3 years ago