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Crazy boy [7]
2 years ago
12

Solve the equation. 2.2= z - 1.1 z=?

Mathematics
1 answer:
fredd [130]2 years ago
7 0

Answer:

z = 3.3

Step-by-step explanation:

Given

2.2 = z - 1.1 ( add 1.1 to both sides )

3.3 = z

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Simplify the trigonometric expression sin(4x)+2 sin(2x) using Double-Angle
drek231 [11]

Answer:

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2 years ago
1. Four more than a number.<br> er bronto tngitoup et .I
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X + 4
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2 years ago
Adam spent $3.42 on orange juice. It costs .12 per ounce. How many did Adam buy?
SIZIF [17.4K]
Hi, 


 I would say that the answer is 28, but it still doesn't add up because when you multiply 12 by 28 you get $3.36 which isn't enough to meet his total, so I tried multiplying 12 by 29 and I got $3.48, which is six cents over his total. So I'm sorry if it is wrong I tried to help to my best ability.

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5 0
3 years ago
If there are 1000 grams in a kilogram in 454 grams in a pound how many pounds are there in a kilogram?
MaRussiya [10]

Answer:

2.2

Step-by-step explanation:

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6 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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