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3 years ago

Solve the equation: 15x11+(29x42)=?

Mathematics

2 answers:

Semenov [28]3 years ago

4 0

Look at the attached picture

Hope it will help you...

Good luck on your assignment

~pragya

Katyanochek1 [597]3 years ago

3 0

Answer:

1,383

Step-by-step explanation:

15x11+(29x42)

165 + (1,218)

= 1,383

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(3x − 4)(2x)2 + 2x − 1).

zzz [600]

(3x−4)(2x)(2)+2x−1

Distribute:

=12x2+−16x+2x+−1

Combine Like Terms:

=12x2+−16x+2x+−1=(12x2)+(−16x+2x)+(−1)=12x2+−14x+−1

Answer:=12x2−14x−1

hope this helps! was there meant to be a parenthesis by the 2 + 2x - 1??

4 0

3 years ago

Y’all please help me with this I don’t understand

Yanka [14]

Answer: I believe it’s 8

Step-by-step explanation:

It’s 13 minus 5

3 0

3 years ago

Read 2 more answers

Answer the question on the attachment

Nonamiya [84]

Answer:

The answer to your question is below

Step-by-step explanation:

Part A.

-3x + 7y = 13

Solve the equation for "y"

7y = 3x + 13

y = $\frac{3}{7} x + 13$

Part B.

Two lines are parallel if their slopes are the same

Two lines are perpendicular if their slope are opposite reciprocals

Slope of the first line = $\frac{3}{7}$

Slope of the second line = $\frac{3}{7}$

Conclusion: the lines are parallel because their slopes are the same.

5 0

3 years ago

Which equation has no solution?

dybincka [34]

So, we start with this.
$5+2\left(3+2x\right)=x+3\left(x+1\right)$
Simplify.
$4x+11=4x+3$
Subtract 11 from both sides.
$4x+11-11=4x+3-11$
Simplify.
$4x=4x-8$
Subtract 4x from both sides.
$4x-4x=4x-8-4x$
Simplify
$0=-8$
This equation has no solution, the answer would be B!
Have a nice day! :)

7 0

3 years ago

Read 2 more answers

Integrate xe^(-x^2).

Gekata [30.6K]

You can easily integrate it using a simple substitution:

$\large\begin{array}{l} \mathsf{\displaystyle\int\!x\,e^{-x^2}\,dx}\\\\ =\mathsf{\displaystyle\int\!\left(-\frac{1}{2}\right)\cdot (-2)x\,e^{-x^2}\,dx}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\int\!\,e^{-x^2}\cdot (-2x)\,dx\qquad\quad(i)}\\\\ \end{array}$

$\large\begin{array}{l} \textsf{Let}\\\\ \mathsf{-x^2=u,}\quad\textsf{then}\quad\mathsf{-2x\,dx=du}\\\\\\ \textsf{so (i) becomes}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\int\!e^u\,du}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\cdot e^u+C}\\\\ =\mathsf{\displaystyle-\,\frac{1}{2}\cdot e^{-2x}+C}\\\\\\ \boxed{\begin{array}{c} \mathsf{\displaystyle\int\!x\,e^{-x^2}\,dx=-\,\frac{1}{2}\,e^{-x^2}+C} \end{array}}\qquad\checkmark \end{array}$

If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2159730

$\large\textsf{I hope it helps. :-)}$

Tags: integrate substitution indefinite integral exponential composite calculus

7 0

4 years ago