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Anestetic [448]
2 years ago
11

SOMEOne PLS EXPLAIN VERTICAL ASYMPTOTES TO ME IM DYING- i need to know how u get the asymptote(explanation) and answer T^TTTTTTT

TT

Mathematics
1 answer:
Kitty [74]2 years ago
4 0
A vertical asymptote is what you get when you try to divide by 0. To find where you get these, you need to look at the denominator and what values of x will make the denominator equal to 0.

In your denominator, you have (x+7)(x-5)(x-3).
What values of x makes (x+7)(x-5)(x-3)=0?
If x = -7, if x = 5, or if x = 3, then that entire expression will equal zero. (Same idea as when you solve equations by factoring.

Now the only place this can get trickier is if one of those factors — one of (x+7), (x-5), or (x-3) — also appears in the numerator. If that happens, then it’s more involved whether you have an asymptote or not. But that doesn’t happen in this example.

So the short version: Asymptotes happen when you try to divide by zero. Dividing by zero is not a good thing. So you just ask yourself, “What will make the denominator 0?”
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Answer:

B

Step-by-step explanation:

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3 years ago
x = c1 cos(t) + c2 sin(t) is a two-parameter family of solutions of the second-order DE x'' + x = 0. Find a solution of the seco
igomit [66]

Answer:

x=-cos(t)+2sin(t)

Step-by-step explanation:

The problem is very simple, since they give us the solution from the start. However I will show you how they came to that solution:

A differential equation of the form:

a_n y^n +a_n_-_1y^{n-1}+...+a_1y'+a_oy=0

Will have a characteristic equation of the form:

a_n r^n +a_n_-_1r^{n-1}+...+a_1r+a_o=0

Where solutions r_1,r_2...,r_n are the roots from which the general solution can be found.

For real roots the solution is given by:

y(t)=c_1e^{r_1t} +c_2e^{r_2t}

For real repeated roots the solution is given by:

y(t)=c_1e^{rt} +c_2te^{rt}

For complex roots the solution is given by:

y(t)=c_1e^{\lambda t} cos(\mu t)+c_2e^{\lambda t} sin(\mu t)

Where:

r_1_,_2=\lambda \pm \mu i

Let's find the solution for x''+x=0 using the previous information:

The characteristic equation is:

r^{2} +1=0

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r_1_,_2=0\pm \sqrt{-1} =\pm i

Therefore, the solution is:

x(t)=c_1cos(t)+c_2sin(t)

As you can see, is the same solution provided by the problem.

Moving on, let's find the derivative of x(t) in order to find the constants c_1 and c_2:

x'(t)=-c_1sin(t)+c_2cos(t)

Evaluating the initial conditions:

x(0)=-1\\\\-1=c_1cos(0)+c_2sin(0)\\\\-1=c_1

And

x'(0)=2\\\\2=-c_1sin(0)+c_2cos(0)\\\\2=c_2

Now we have found the value of the constants, the solution of the second-order IVP is:

x=-cos(t)+2sin(t)

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3 years ago
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