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s2008m [1.1K]
2 years ago
5

Wayne and Mack have

Mathematics
2 answers:
Murljashka [212]2 years ago
8 0

Answer:

18

Step-by-step explanation:

\sf Twice = *2\\9 * 2 = 18

Darya [45]2 years ago
6 0

Answer:

18

Step-by-step explanation:

mack has 18cats because she has 2×9 cats as Wayne 2×9 is 18

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Find the bearing of the ship. Round to the nearest tenth of a degree if necessary.
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  117°

Step-by-step explanation:

We assume you want to report your bearing as an angle measured CW from North. For conventional angles greater than 90°, it will be ...

  450° -angle = 450° -333° = 117°

The bearing of the ship is 117°.

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If the dimensions of a rectangle are A plus 3b and A minus 3b what is the area of the rectangle
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we notice this is a special case

(a+b)(a-b)=a^2-b^2

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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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