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3 years ago

Which two integers are square root 35 between

Mathematics

1 answer:

omeli [17]3 years ago

3 0

It's between 5 and 6.
5 squared is 25 and 6 squared is 36 so the square root is between them.

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The room can hold a maximum of 400 people. If there are already 134 people

Olegator [25]

Answer:

x+ 134 is less than or equal to 400 is the answer

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2 years ago

You have the same bowl, with 5 orange, 6 blue, 3 green, 4 red and 7 yellow candies. You take out an orange candy from the bowl.

love history [14]

Answer:

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Step-by-step explanation:

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3 years ago

Sam is using the expression 8mn in class. What is the value of the expression when m = 4 a n d n = 12 ?

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Answer:

8 x 4 x 12 = 384

Step-by-step explanation:

8 x 4 x 12 = 384

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3 years ago

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A pizza place offers 4 different cheeses and 10 different toppings. In how many ways can a pizza be made with 1 cheese and 4 top

mylen [45]

Answer:

Step-by-step explanation:

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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