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3 years ago

What is the slope of a line that is perpendicular to the line whose equation is 8y−5x=118y−5x=11?

2 answers:

5 0

<span>8y−5x=11
8y = 5x + 11
y = 5/8x + 11/8 has slope = 5/8
</span><span>a line that is perpendicular to the line, slope is opposite and reciprocal so slope = - 8/5

answer
slope = -8/5</span>

Harrizon [31]3 years ago

4 0

Answer:

The slope of a line that is perpendicular to the line whose equation is $8y-5x=11$ is $m=-\frac{8}{5}$

Step-by-step explanation:

Given : Equation $8y-5x=11$

To find : What is the slope of a line that is perpendicular to the line whose equation is given?

Solution :

First we find the slope of the given line,

The general slope form of line is $y=mx+b$ where m is the slope of the line and b is the y-intercept.

Re-write the given equation into general form,

$8y-5x=11$

Take 5x to another side,

$8y=5x+11$

Divide both side by 8,

$y=\frac{5}{8}x+\frac{11}{8}$

On comparing with general form,

The slope of the line is $m=\frac{5}{8}$

We know,

When two line are perpendicular one slope is negative reciprocal of another.

If the slope of line is $m=\frac{5}{8}$

Then the slope of perpendicular line on this line is $m=-\frac{8}{5}$

Therefore, The slope of a line that is perpendicular to the line whose equation is $8y-5x=11$ is $m=-\frac{8}{5}$

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Answer:

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Step-by-step explanation:

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b) y' - 5y = 4e^7x

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c) dy/dx + 2y = 2/(1+e^4x)

$\frac{dy}{dx}+2y=\frac{2}{1+e^{4x}}\\p(x)=2\\Q(x)=\frac{2}{1+e^{4x}}\\\mu(x)=\int P(x)dx=\int 2dx=2x\\y=e^{-\mu(x)}\int {e^{\mu(x)}Q(x)dx}\\y=e^{-2x}\int {e^{2x}\frac{2}{1+e^{4x}}}dx\\y=e^{-2x}arctan(e^{2x})+ce^{-2x}$

d) 1/2 di/dt + i = 4cos(3t)

$\frac{1}{2}\frac{di}{dt}+i=4cos(3t)\\\frac{di}{dt}+2i=8cos(3t)\\p(t)=2\\Q(t)=8cos(3t)\\\mu(t)=\int P(t)dt=\int 2dt=2t\\i=e^{-\mu(t)}\int {e^{\mu(t)}Q(t)dt}\\i=e^{-2t}\int {e^{2t}8cos(3t}dt\\i=e^{-2t}\left(\frac{8\left(3e^{2t}\sin \left(3t\right)+2e^{2t}\cos \left(3t\right)\right)}{13}+C\right)$

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drek231 [11]

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3 years ago

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Aleks04 [339]

Hi there!

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3 years ago