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Rudik [331]
2 years ago
10

Question 6

Mathematics
2 answers:
IrinaVladis [17]2 years ago
8 0

Answer:

Positive, non-linear association

Step-by-step explanation:

Dmitry_Shevchenko [17]2 years ago
7 0

Answer:

positive, non-linear association

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Ms.Gallegos works 8 hours per day. She made $2,400 for working 10 days. How much money does Ms. Gallegos make per hour
ExtremeBDS [4]

Answer:

30 dollars per hour

5 0
2 years ago
Just answer question 3,4 and 5 only.. There will have a diagram because the question base on that. One thing, just show the work
irina [24]

Answer:

Step-by-step explanation:

Atachment in 1 minute.

7 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
How do you solve this?
Alchen [17]
The answer is (3x^2 - x +2 )= to AC
Area = 1/2 |BD| |AC|
(9x^3 - 3x^2 + 6x) = 1/2 (AC) 6x
3x ( 3x^2 - x +2 )= 1/2 (AC) 2(3x)
3^2 - x +2 = AC
3 0
3 years ago
Find the slope of the line graphed
Ray Of Light [21]

Answer:

The slope is \frac{3}{5} because if you do the rise over run method, it goes up 3 times and right 5 times, therefor being \frac{3}{5}.

5 0
3 years ago
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