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sdas [7]
3 years ago
5

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

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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]

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9% chance

Step-by-step explanation:

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Sam is using the expression 8mn in class. What is the value of the expression when m = 4 a n d n = 12 ?
goldfiish [28.3K]

Answer:

8 x 4 x 12 = 384

Step-by-step explanation:

8 x 4 x 12 = 384

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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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8 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
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