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3 years ago

Find the slope intercept form of the liner function using the points (-6,6) and (-2,4) Show all work

Mathematics

1 answer:

olga2289 [7]3 years ago

4 0

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

the equation of a line in slope- intercept form is

y = mx + c ( m is the slope and c the y-intercept )

to calculate m use the gradient formula

m = ( y₂ - y₁ ) / (x₂ - x₁ )

with (x₁, y₁ ) = (- 6, 6 ) and (x₂, y₂ ) = (- 2, 4 )

m = $\frac{4-6}{-2+6}$ = $\frac{-2}{4}$ = - $\frac{1}{2}$

y = - $\frac{1}{2}$ x + c is the partial equation

to find c substitute either of the 2 points into the partial equation

using (- 2, 4 ), then

4 = 1 + c ⇒ c = 4 - 1 = 3

y = - $\frac{1}{2}$ x + 3 ← in slope-intercept form

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3 years ago

First make a substitution and then use integration by parts to evaluate the integral. (Use C for the constant of integration.) x

e-lub [12.9K]

Answer:

$(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

Step-by-step explanation:

Ok, so we start by setting the integral up. The integral we need to solve is:

$\int x ln(5+x)dx$

so according to the instructions of the problem, we need to start by using some substitution. The substitution will be done as follows:

U=5+x

du=dx

x=U-5

so when substituting the integral will look like this:

$\int (U-5) ln(U)dU$

now we can go ahead and integrate by parts, remember the integration by parts formula looks like this:

$\int (pq')=pq-\int qp'$

so we must define p, q, p' and q':

p=ln U

$p'=\frac{1}{U}dU$

$q=\frac{U^{2}}{2}-5U$

q'=U-5

and now we plug these into the formula:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int \frac{\frac{U^{2}}{2}-5U}{U}dU$

Which simplifies to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\int (\frac{U}{2}-5)dU$

Which solves to:

$\int (U-5)lnUdU=(\frac{U^{2}}{2}-5U)lnU-\frac{U^{2}}{4}+5U+C$

so we can substitute U back, so we get:

$\int xln(x+5)dU=(\frac{(x+5)^{2}}{2}-5(x+5))ln(x+5)-\frac{(x+5)^{2}}{4}+5(x+5)+C$

and now we can simplify:

$\int xln(x+5)dU=(\frac{x^{2}}{2}+5x+\frac{25}{2}-25-5x)ln(5+x)-\frac{x^{2}+10x+25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}-\frac{5x}{2}-\frac{25}{4}+25+5x+C$

$\int xln(x+5)dU=(\frac{x^{2}-25}{2})ln(5+x)-\frac{x^{2}}{4}+\frac{5x}{2}+C$

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7 0

3 years ago

Y=4x-6 and passes through point 8,12

umka2103 [35]

Answer:

Slope of required line if both lines are parallel: y=4x-20

Slope of required line if both lines are perpendicular: y=-1/4x+14

Step-by-step explanation:

We need to find equation of line while we are given another line having equation y=4x-6 and passes through point (8,12)

<u><em>Note: Since it is not given if the required line is parallel or perpendicular to the given line. I will solve for both cases.</em></u>

If Both lines are parallel the equation of new line will be:

We need to find slope and y-intercept to write equation of new line.

When lines are parallel there slopes are same

So, slope of given line y=4x-6 is 4 (Compare with general equation y=mx+b, m is slope so m=4)

Slope of new line is: m=4

Now finding y-intercept

Using slope m=4 and point (8,12) we can find y-intercept

$y=mx+b\\12=4(8)+b\\12=32+b\\b=12-32\\b=-20$

y-intercept of new line is: b=-20

Equation of required line having slope m=4 and y-intercept b=-20 is

$y=mx+b\\y=4x-20$

If Both lines are perpendicular the equation of new line will be:

We need to find slope and y-intercept to write equation of new line.

When lines are perpendicular there slopes are opposite of each other

So, slope of given line y=4x-6 is 4 (Compare with general equation y=mx+b, m is slope so m=4)

Slope of new line is: m=-1/4

Now finding y-intercept

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Equation of required line having slope m=-1/4 and y-intercept b=14 is

$y=mx+b\\y=-\frac{1}{4} x+14$

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3 years ago

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lubasha [3.4K]

Answer:

x=400 m is the correct answer

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3 years ago