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11Alexandr11 [23.1K]
2 years ago
11

HELP ON FIRST 3 PLS

Mathematics
1 answer:
rodikova [14]2 years ago
4 0

Answer:

THE answer are

  • A' ( -5,-2 )
  • G' (-8,1 )
  • E' (4 -9 )

HOPE IT HELP YOU.

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Answer:

Step-by-step explanation:

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Write the log equation as an exponential equation. You do not need to solve for x.
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the answer is - 1/4
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H(t)=-16(t-1)^2+16 <br> What is the maximum height the ball reaches and how long does it take?
Aleksandr-060686 [28]

Given t = time and h = height

h(t) =  - 16(t - 1)^{2}  + 16

From the equation.

y = a(x - h)^{2}  + k

h is the axis od symmetry

(h, k) is the vertex.

In the equation, h is 1 and k is 16.

So the maximum value is 16 at t = 1.

The ball reaches maximum at 16 and it takes t = 1 to reach the maximum point.

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2 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:

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How to write the linear function passing through the points (3,-5) and (-8,-5)
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Check the pictrue below.

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