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2 years ago

Please help!! ALgebra 2 I will be h=giving out extra points the the brainiest answers

Mathematics

1 answer:

Arte-miy333 [17]2 years ago

4 0

Answer:

Step by Step Explanation

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PLEASE ANSWER ASAP!!!

Vesna [10]

Answer:

C=x (x+3)

Step-by-step explanation:

x cannot divide x+3 definitely so the denominators must be multiplied to get the least common denominator.

4 0

2 years ago

7y+65.7=2 true or false(don't answer that I just don't want this taken down(unless you do actually have the answer) is anyone in

Sophie [7]

It's false and same lol

8 0

2 years ago

Find the least common denominator for these two rational expressions 7/n^2 -8/7n

lianna [129]

Answer:

7n²

Step-by-step explanation:

7/n² - 8/7n

Least common multiple will be the least common denominator

L.C.M = 7n²

Dividing the L.C.M by denominators and multiplying with numerators;

((7×7) - (n×8))/7n²

= (49 - 8n)/7n²

7n² is the least common denominator

8 0

2 years ago

PLS HELP!! ALGEBRA!!

sineoko [7]

Horizontal distance = 48'

difference in elevation = 18"-6" = 12" = 1'

Slope of the sewer line = rise/run = 1/48 = 0.0208 = 2.08%

6 0

2 years ago

Integrate this two questions

tankabanditka [31]

Simplify the integrands by polynomial division.

$\dfrac{t^2}{1 - 3t} = -\dfrac19 \left(3t + 1 - \dfrac1{1 - 3t}\right)$

$\dfrac t{1 + 4t} = \dfrac14 \left(1 - \dfrac1{1 + 4t}\right)$

Now computing the integrals is trivial.

$\displaystyle \int \frac{t^2}{1 - 3t} \, dt = -\frac19 \int \left(3t + 1 - \frac1{1-3t}\right) \, dt \\\\ = -\frac19 \left(\frac32 t^2 + t + \frac13 \ln|1 - 3t|\right) + C \\\\ = \boxed{-\frac{t^2}6 - \frac t9 - \frac{\ln|1-3t|}{27} + C}$

where we use the power rule,

$\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C ~~~~ (n\neq-1)$

and a substitution to integrate the last term,

$\displaystyle \int \frac{dt}{1-3t} = -\frac13 \int \frac{du}u \\\\ = -\frac13 \ln|u| + C \\\\ = -\frac13 \ln|1-3t| + C ~~~ (u=1-3t \text{ and } du = -3\,dt)$

$\displaystyle \int \frac t{1+4t} \, dt = \frac14 \int \left(1 - \frac1{1+4t}\right) \, dt \\\\ = \frac14 \left(t - \frac14 \ln|1 + 4t|\right) + C \\\\ = \boxed{\frac t4 - \frac{\ln|1+4t|}{16} + C}$

using the same approach as above.

5 0

1 year ago