All categories

2 years ago

How to solve 2+6|2x-6| > 74

Mathematics

1 answer:

lesya692 [45]2 years ago

8 0

Answer:

2 + 6 + 2x + 6 > 74

x > 9

You might be interested in

Using the defects method, which of these relationships represents the Law of Cosines if the measure of the included angle betwee

horrorfan [7]

Asked and answered elsewhere.
brainly.com/question/1410592

_____
Google doesn't recognize the terms "defects method" except in association with various postings of this same problem.

4 0

3 years ago

#24 Help needed :P thanks

alukav5142 [94]

Have you got your answer yet?

7 0

3 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

Helppp pleaseeee I’ll give brainliste

a_sh-v [17]

Answer:

F and B

Step-by-step explanation:

4 0

2 years ago

I need help plzz!! Just box number 4, TY

nydimaria [60]

12.36% I think sorry if its wrong ;-;

3 0

2 years ago