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Free_Kalibri [48]
3 years ago
12

For all pairs of real numbers M and V where M = 3V + 6, V = ?

Mathematics
1 answer:
Ede4ka [16]3 years ago
6 0
M = 3V + 6 \ \ | (-6)\\ \\M-6=3V+6-6 \\ \\M-6 =3V \ \ /:3 \\ \\V=\frac{M }{3}-\frac{6}{3} \\ \\V=\frac{M}{3 }- 2


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13. A jeweller bought an engagement ring for €1200 and
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Ava's car used 6 gallons of gas to drive 132 miles. At what rate does her car use gas in gallons per mile? Express your answer i
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if the equation 2x^2+ bx+ 5=0 has no real solutions, which of the following must be true A. b^2 < 10 B. b^2> 10 C. b^240​
melisa1 [442]

Answer:

C. b²< 40

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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

So, the roots are given by:

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)

3 0
3 years ago
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