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3 years ago

Please help fast if you can

Mathematics

2 answers:

Artemon [7]3 years ago

7 0

115. because 180-65=115.

Sonbull [250]3 years ago

5 0

That is A 115 hopefully this help

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Why is 6 1/2 equal to the square root of 6

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Because It Can Be Simplified Many Ways

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3 years ago

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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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8 months ago

Inverse of f(x)=3/2x +6

kvasek [131]

Answer:

f(x) = (x - 6)/(3/2)

Step-by-step explanation:

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3 years ago

Mark is pitching a tent he ties a rope from a nail to the top of the tent what is the length of the rope between the nail and th

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I believe that it is 7

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3 years ago

How do u find the slope of (6,-10), (-15,15)

Sergeu [11.5K]

$m = \frac{y_{2} - y_{1}}{x_{2}- x_{1}} = \frac{15 - (-10)}{-15 - 6} = \frac{15 + 10}{-21} = \frac{25}{-21} = -1\frac{4}{21}$

   y - y₁ = m(x - x₁)
y - (-10) = -1⁴/₂₁(x - 6)
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2 years ago