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3 years ago

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What is the equation of the line that passes through the point (-6,-5)(−6,−5) and has a slope of \frac{3}{2}

1 answer:

kvv77 [185]3 years ago

6 0

Answer:

y = 3/2x + 4

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Is this function linear or nonlinear?

Alekssandra [29.7K]

Answer:

yes

Step-by-step explanation:

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3 years ago

If fg + gh = fh,which of the following statements must be true?

Oksanka [162]

The only safe conclusion is that point G lies on line FH or that point G lies somewhere between line FH. We cannot conclude that point G is the midpoint of line FH eventhough by virtue of definition of midpoint, the given equation is a proof equation. If G were to be midpoint, segment FG must be equal to segment GH in line FH.

5 0

3 years ago

Plz awnser because Aleks is dumb

Black_prince [1.1K]

Since you know it's a square, you also know that all the sides are 3 2/3. So to find area you multiply, so 3 2/3 x 3 2/3. An easy way to do this is convert it to a mixed number (11/3) and then multiply:
11/3 x 11/3 = 121/9

8 0

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:

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

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

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

notice how all the constants were combined into one big constant C.

7 0

3 years ago

What is the solution to the system of equations?

MariettaO [177]

Answer:

x=-4 y=-1

Step-by-step explanation:

Let's solve your system by substitution.

−3x−8y=20;y=5x+19

Rewrite equations:

y=5x+19;−3x−8y=20

Step: Solve y=5x+19for y:

y=5x+19

Step: Substitute5x+19foryin−3x−8y=20:

−3x−8y=20

−3x−8(5x+19)=20

−43x−152=20(Simplify both sides of the equation)

−43x−152+152=20+152(Add 152 to both sides)

−43x=172

−43x

−43

=

172

−43

(Divide both sides by -43)

x=−4

Step: Substitute−4forxiny=5x+19:

y=5x+19

y=(5)(−4)+19

y=−1(Simplify both sides of the equation)

Answer:

x=−4 and y=−1

7 0

3 years ago