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statuscvo [17]
2 years ago
12

If x > 2, find lx-2l Help pleaseee

Mathematics
1 answer:
Lyrx [107]2 years ago
8 0
<h3>Answer:  x-2</h3>

Explanation:

If x > 2, then x-2 > 0 after subtracting 2 from both sides.

Since x-2 is always positive when x > 2, this means the absolute value bars around the x-2 aren't needed. The results of |x-2| and x-2 are perfectly identical.

For example, if we tried something like x = 5, then

  • x-2 = 5-2 = 3
  • |x-2| = |5-2| = |3| = 3

Both outcomes are 3. I'll let you try other x inputs.

So because |x-2| and x-2 are identical, this means |x-2| = x-2 for all x > 2.

In short, we just erase the absolute value bars.

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Answer:

Step-by-step explanation:

You can't ever let this go negative. At least not at the grade you are in.

It can be 0.

So the domain must start at x = 7

sqrt(5*7 - 35) = sqrt(0) = 0

x can have any value (including 7) between 7 and infinity. If you choose a number less than 7 (like 6) the square root will go negative and that's not to be done.

So the interval is

7 ≤ x < ∞

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2 years ago
A flower bed has the shape of a rectangle 21 feet long and 9 feet wide. What is its area in square yards?
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3 years ago
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3 years ago
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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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