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2 years ago

How do you find degree measures? How would you solve this question?

Mathematics

1 answer:

MrRa [10]2 years ago

8 0

A. a=145 b=35 c=145 d=145 e=35 f=145 g=35

b. These are the pairs of vertical angles: a & c, b & 35, d & f, e & g.

Hope this was helpful.

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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

Kelsey’s bank charged her $17.50 for using her debit card at ATMs that are not owned by her bank 7 times in the last month. What

Katarina [22]

Answer:
i have no 1dea the answer. 18794398+1387o79=819

Step-by-step explanation:
1800000000-132987=90

4 0

2 years ago

Find the values that will make the equations true.

Alex17521 [72]

The value of a=7 will make the equation true

5 0

1 year ago

Solve the system of equations by subsitution.<br> b = 4a +2<br> 12a - 3b = -6

Vladimir [108]

Answer:

infinite solution

Step-by-step explanation:

hello

we substitute b = 4a + 2 in the second expression

12a -3*(4a+2) = -6

<=> 12a -12a -6 = -6

<=> -6 = -6

which is always true

so there is infinite solution

hope this helps

5 0

3 years ago

Write a sine function that has an amplitude of 5, a midline of 3 and a period of 1/8.

vodomira [7]

Answer:

y=5sin(16pix)+3

Step-by-step explanation:

Amp=5 means our curve is either y=5sin(bx+c)+d or y=-5sin(bx+c)+d.

y=sin(x) has period 2pi.

So y=sin(bx) has period 2pi/b.

We want 2pi/b=1/8.

Cross multiplying gives: 16pi=b

y=5sin(16pix+c)+d

d=3 since we want midline y=3.

y=5sin(16pix+c)+3

We can choose c=0 since we aren't required to have a certain phase shift.

y=5sin(16pix)+3

6 0

2 years ago