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2 years ago

Find the vertical, horizontal, and slant asymptotes, if any for y=x^3/ (x-2)^4

1 answer:

7 0

This has both a horizontal and a vertical asymptote. There are no slant (oblique) asymptotes cuz the degree of the numerator is not higher than that of the denominator. If the degree of the numerator is less than the degree of the denominator, which is our case here, then the horizontal asymptote is 0. But we also have a vertical asymptote, which occurs where the denominator = 0. We all know that we break every rule known to mankind if we try to divide by 0, so there also a vertical asymptote at x = 2.

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A circular table has a diameter of 20 inches.what is the area of the table

Effectus [21]

Hello!

To find the area of a circle you do $\pi r ^{2}$

the "r" stands for radius which is half the diameter

So we do $\pi * 10^{2}$

Which gives us the answer of 314.16

Hope this helps!

3 0

2 years ago

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Jana can spend, at most, $200 on school clothes.

Soloha48 [4]

Answer:

x <= 200

Step-by-step explanation:

Your answer choices are unclear so I cannot specify a letter. The answer is

x<=200.

4 0

2 years ago

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Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .

Solving for F(s) on the last equation, $F(s)=\frac{s}{s^2+25}$ , then the Laplace transform we were searching is $-F'(s)=\frac{s^2-25}{(s^2+25)^2}$

3 0

2 years ago

Can u help me solve this pls

lisov135 [29]

Answer:

First option: (Substitute)

2x+3 (x+1.5) = 12

3 0

2 years ago

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What is the sum of the geometric series in which a1 = −5, r = 3, and an = −405?

xxMikexx [17]

$a_1=-5;\ r=3;\ a_n=-405\\\\a_n=a_1r^{n-1}\\\\subtitute\\\\-5\cdot3^{n-1}=-405\ \ \ |divide\ both\ sides\ by\ (-5)\\\\3^{n-1}=81\\\\3^{n-1}=3^4\iff n-1=4\to n=5\\\\S_n=\dfrac{a_1(1-r^n)}{1-r}\\\\S_5=\dfrac{-5(1-3^5)}{1-3}=\dfrac{-5(1-243)}{-2}=\dfrac{5\cdot(-242)}{2}=5\cdot(-121)=-605\\\\\\Answer:\boxed{S_5=-605}$

4 0

3 years ago