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professor190 [17]
3 years ago
8

Square root t+1 + square root t+4 = 5 solve

Mathematics
1 answer:
butalik [34]3 years ago
6 0
The answer to this is 1 because 4+1=5  and also 4 because of the same reason.
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If a sponsored link was delivered 100 times and 10 people clicked on it, then the number of impressions is 100, the number of cl
Rudik [331]

Answer:

Option A.

Step-by-step explanation:

It is given that a sponsored link was delivered 100 times and 10 people clicked on it.

Number of impressions = 100

Number of clicks = 10

The formula for click-through-rate (CTR) is

CTC=\dfrac{\text{Total measured clicks}}{\text{Total measured impression}}\times 100

CTC=\dfrac{10}{100}\times 100

CTC=10

The click-through-rate (CTR) would be 10%. Therefore, the correct option is A.

7 0
3 years ago
What is the volume of a cylinder with a radius of 9 inches and a height of 2 inches? Use 3.14 for pi. Round your answer to the n
Luden [163]
561.75 is your answer i hope i helped you get it right
7 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
BRAINLIEST TO CORRECT HURRY
Vlad [161]

Answer:

>

Step-by-step explanation:

Ummm...

0.20 is bigger than 0

4 0
3 years ago
HELP MEE PLEASEEEE I NEED THE CORRECT ANSWER
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It’s C oof oof oof oof
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3 years ago
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