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3 years ago

I don't know how to do three and four

Mathematics

2 answers:

xenn [34]3 years ago

5 0

Angle 3 is also 104: and angle 5 and 2 are 76: and angle 6 is 104: 3 and 8 are 104: and 7 and are 76

angle 1 and 6 are verticle angles
angles 7 and 8 are suplimentary
angles 4 and 5 are alternate exterior angles
andgle 2 and 4 are im not sure

dmitriy555 [2]3 years ago

5 0

If angle one equals 104 then
Angle2: 208
Angle3: 312
Then keep adding 104 degrees to the rest of the angles

Hope I helped

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2x+14y=-108 9x+2y= 2 Solve the following system of equations using any method

Orlov [11]

2x+14y=-108
63x+14y= 14 (multiply the second equation by 7 and subtract)
_____________
-61x=-122
x=-122/-61
x=2
2(2)+14y=-108
4+14y=-108
14y=-112
y=-112/14
y=-8

4 0

3 years ago

Does anyone wanna go on zooom

alina1380 [7]

Answer: no lol

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

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

Any face that is not a base

Ann [662]

Answer:

In Mathematics Geometry, lateral face is said be the side of a 3D-figure in that is not a base.

Please check the attached figure to visual the concept.

Step-by-step explanation:

In Mathematics Geometry, lateral face is said be the side of a 3D-figure in that is not a base.

The faces in in a prism or pyramid which are not bases are basically the lateral faces.

For example, the lateral faces are basically parallelograms in Triangular prism which are not the bases.

Please check the attached figure to visual the concept.

7 0

2 years ago

Onii-Chan help me whit 5-8

lina2011 [118]

The answer to 5-8 is 3

4 0

3 years ago