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3 years ago

Which function is undefined when Theta = StartFraction pi Over 2 EndFraction radians?

Mathematics

2 answers:

Varvara68 [4.7K]3 years ago

6 0

Is this some type of code?

I can't read code, i mean -- huh, i am not a robot.

lina2011 [118]3 years ago

6 0

Theta = (pi/2) and theta = (3pi/2)

Please correct me if I’m wrong???

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Solve for x and y simultaneously <br><br>Pls help me

LenKa [72]

Step-by-step explanation:

Are the roots equal or unequal

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3 years ago

The equation for the circle below is x2 + y2 = 16. What is the length of the<br> circle's radius?

ikadub [295]

Answer:

Step-by-step explanation:

Because 16 is the diameter and to find the radius you need to divide by 2

8 0

3 years ago

Solve for x: 4x + one half(2x + 4) > 12

zalisa [80]

Answer:

x > 2

Step-by-step explanation:

Given

4x + $\frac{1}{2}$ (2x + 4) > 12 ← distribute and simplify left side

4x + x + 2 > 12

5x + 2 > 12 ( subtract 2 from both sides )

5x > 10 ( divide both sides by 5 )

x > 2

3 0

3 years ago

Find the equations of the tangents to the curve x = 9t2 + 9, y = 6t3 + 6 that pass through the point (18, 12).

Oxana [17]

Check the attached file for the answer.

6 0

4 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