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mixer [17]
2 years ago
9

Find the equation of the line that

Mathematics
1 answer:
Aleonysh [2.5K]2 years ago
4 0

Answer:

y = 3x/2-14

Step-by-step explanation:

We are given that the line is perpendicular to y = -2/3 and contains (4,-8).

<u>P</u><u>e</u><u>r</u><u>p</u><u>e</u><u>n</u><u>d</u><u>i</u><u>c</u><u>u</u><u>l</u><u>a</u><u>r</u><u> </u><u>D</u><u>e</u><u>f</u><u>.</u>

\displaystyle \large{m_1m_2 =  - 1}

Both slopes multiply each others equal to -1.

Finding another slope that is perpendicular to -2/3, substitite m1 = -2/3 in.

\displaystyle \large{ -  \frac{2}{3} m_2 =  - 1}

Multiply both sides by 3.

\displaystyle \large{ -  \frac{2}{3} m_2( 3) =  - 1(3)} \\  \displaystyle \large{ -  2 m_2=  - 3} \\  \displaystyle \large{  m_2=  \frac{3}{2} }

Therefore, another slope that is perpendicular to -2/3 is 3/2.

Then rewrite in slope-intercept form.

<u>S</u><u>l</u><u>o</u><u>p</u><u>e</u><u>-</u><u>I</u><u>n</u><u>t</u><u>e</u><u>r</u><u>c</u><u>e</u><u>p</u><u>t</u>

\displaystyle \large{y  = mx + b}

where m = slope and b = y-intercept; substitute m = 3/2 in.

\displaystyle \large{y  =  \frac{3}{2} x + b}

Since the line contains (4,-8), substitute x = 4 and y = -8 in and solve for b.

\displaystyle \large{ - 8 =  \frac{3}{2} (4) + b} \\  \displaystyle \large{ - 8 =  3(2)+ b} \\  \displaystyle \large{ - 8 =  6 +  b} \\  \displaystyle \large{ - 8 - 6 =  b} \\  \displaystyle \large{ - 14 =  b}

Therefore, b is -14; rewrite again in slope-intercept form.

Thus:-

\displaystyle \large{y  =  \frac{3}{2} x + b} \\  \displaystyle \large{y  =  \frac{3}{2} x  - 14}

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3 years ago
Find the indefinite integral using the substitution provided.
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Answer:  7\text{Ln}\left(e^{2x}+10\right)+C

This is the same as writing 7*Ln( e^(2x) + 10) + C

=======================================================

Explanation:

Start with the equation u = e^{2x}+10

Apply the derivative and multiply both sides by 7 like so

u = e^{2x}+10\\\\\frac{du}{dx} = 2e^{2x}\\\\7\frac{du}{dx} = 7*2e^{2x}\\\\7\frac{du}{dx} = 14e^{2x}\\\\7du = 14e^{2x}dx\\\\

The "multiply both sides by 7" operation was done to turn the 2e^(2x) into 14e^(2x)

This way we can do the following substitutions:

\displaystyle \int \frac{14e^{2x}}{e^{2x}+10}dx\\\\\\\displaystyle \int \frac{1}{e^{2x}+10}14e^{2x}dx\\\\\\\displaystyle \int \frac{1}{u}7du\\\\\\\displaystyle 7\int \frac{1}{u}du\\\\\\

Integrating leads to

\displaystyle 7\int \frac{1}{u}du\\\\\\7\text{Ln}\left(u\right)+C\\\\\\7\text{Ln}\left(e^{2x}+10\right)+C\\\\\\

Be sure to replace 'u' with e^(2x)+10 since it's likely your teacher wants a function in terms of x. Also, do not forget to have the plus C at the end. This is a common mistake many students forget to do.

To verify the answer, you can apply the derivative to it and you should get back to the original integrand of \frac{14e^{2x}}{e^{2x}+10}

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subtract 15 from each side and you get 10k = 100

divide each each by 10 and you get the answer of k = 10

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Answer:

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