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3 years ago

Write an equation perpendicular to the line y = 3x + 5 through the point (6, -5). State your answer as y=mx+b.

Mathematics

1 answer:

liubo4ka [24]3 years ago

3 0

Answer:

y=(-1/3) x - 3

Step-by-step explanation:

A line that is perpendicular to it has a slope that is an opposite reciprocal, which in this case is -1/3. Now you use point slope form which is y-y1=m(x-x1) that would be y+5=-1/3(x-6). Distribute the -1/3 and get y+5= -1/3 x+ 2. Subtract 5 from both sides and get y=-1/3 x - 3

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Step-by-step explanation:

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3 years ago

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olga_2 [115]

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3 years ago

Solve 73 make sure to also define the limits in the parts a and b

Aleks04 [339]

73.

$f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}$

$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3$ $\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3$

Since we can't divide by zero, we need to find when:

$x^4-2x^2+144=0$

But before, we can factor the numerator and the denominator:

$\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}$

Now, we can conclude that the vertical asymptotes are located at:

$\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}$

so, for x = -3:

$\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty$ $\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty$

For x = 4:

$\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$ $\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty$

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11 months ago

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Anna71 [15]

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3 years ago

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What is the answer to 19+x/2=4?

salantis [7]

19+x/2=4
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subtract 19 on each side
x=8-19
subtract
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