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Verizon [17]
3 years ago
8

Plz work this out ASAP plz

Mathematics
1 answer:
andre [41]3 years ago
3 0

Answer:

0.143059

Step-by-step explanation:

\dfrac{\sqrt{19 + 6^2}}{7.2^2} =

= \dfrac{\sqrt{55}}{51.84}

= 0.143059

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The answer would be 516.66666667 is your answer have a very nice day
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Find all real values of x such that f(x) = g(x) <br>for f (x) = x^2 + 1 g(x) = 3x - x^2​
gogolik [260]
cant figure it out yet i’ll
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What is the measure of an interior angle of a regular 9 sided polygon?
Talja [164]

Answer:

140

Step-by-step explanation:

3 0
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
I need help seeing what I did wrong
Makovka662 [10]

Answer:  The two roots are x = 3/2 and x = -2

=========================================================

Explanation:

You have the right idea so far. But the two numbers should be 3 and -4 since

  • 3*(-4) = -12
  • 3+(-4) = -1

The -1 being the coefficient of the x term.

This means you need to change the -3x and 4x to 3x and -4x respectively. The other inner boxes are correct.

---------

Refer to the diagram below to see one way to fill out the box method, and that helps determine the factorization.

If we place a 2x to the left of -2x^2, then we need an -x up top because 2x*(-x) = -2x^2

Then based on that outer 2x, we need a -2 up top over the -4x. That way we get 2x*(-2) = -4x

So we have the factor -x-2 along the top

The last thing missing is the -3 to the left of 3x. Note how -3*(-x) = 3x in the left corner and -3*(-2) = 6 in the lower right corner.

We have the factor 2x-3 along the left side.

---------

The two factors are (2x-3) and (-x-2) which leads to the factorization (x+3)(-x+2)

The last thing to do is set each factor equal to 0 and solve for x

  • 2x-3 = 0 solves to x = 3/2 = 1.5
  • -x-2 = 0 solves to x = -2

The two roots are x = 3/2 and x = -2

8 0
3 years ago
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