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3 years ago

Please guyssssssssssss

Mathematics

1 answer:

marissa [1.9K]3 years ago

5 0

Answer:

X= 462000000

Step-by-step explanation:

I used a Calculator

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X² - 40x-129 resolve into factor

Lelu [443]

Answer:

(x - 43)(x + 3)

Step-by-step explanation:

consider the factors of the constant term (- 129) which sum to give the coefficient of the x- term (- 40)

the factors are - 43 and + 3 , since

- 43 × + 3 = - 129 and - 43 + 3 = - 40 , then

x² - 40x - 129 = (x - 43)(x + 3) ← in factored form

8 0

2 years ago

Jimmy has a piece of string that is 6 3/4 inches long, and Sam has a piece of string that is 7 1/2 inches long.

mafiozo [28]

Answer:

3/4 inches

Step-by-step explanation:

Subtract Jimmys string length from Sam's string length

7 1/2 - 6 3/4

Get a common denominator of 4

7 1/2*2/2 - 6 3/4

7 2/4 - 6 3/4

Borrow 1 ( or 4/4) from the 7

6 4/4 + 2/4 - 6 3/4

6 6/4 - 6 3/4

3/4

6 0

3 years ago

Read 2 more answers

Examine the system of equations.

Pachacha [2.7K]

Step-by-step explanation:

The lines have different slopes

.......,............

6 0

2 years ago

Order the following values from least to greatest.

maks197457 [2]

Answer:

-6 1/2

-1

these are in order from least to greatest

3 0

3 years ago

Read 2 more answers

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