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Natali [406]
4 years ago
8

Find the sum and express it in simplest form. (-4a + c + 3) + (-3a + 5c)

Mathematics
1 answer:
Pie4 years ago
7 0
(-4a+c+3) + (-3a + 5c)

-4a-3a+c+5c+3 <--- Combine Like Terms


Answer: -7a+6c+3
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An experiment consists of three steps. There are four possible results on the first step, three possible results on the second s
jeyben [28]

Answer:

24 steps

Step-by-step explanation:

Given

Step\ 1 = 4

Step\ 2 = 3

Step\ 3 = 2

Required

The total number of experimental outcomes (n)

This calculated by multiplying the results in each step.

n = Step\ 1 * Step\  2 * Step\ 3

n = 4*3*2

n = 24

6 0
3 years ago
Use the method of variation of parameters to find a particular solution of the given differential equation. Then check your answ
olga_2 [115]

Answer:

Therefore the complete primitive is

y=c_1 e^{2y}+c_2e^{3t}+e^{t}

Therefore the general solution is

y=c_1e^{2t}+c_2e^{3t}+e^t

Step-by-step explanation:

Given Differential equation is

y''-5y'+6y=2e^t

<h3>Method of variation of parameters:</h3>

Let y=e^{mt} be a trial solution.

y'= me^{mt}

and y''= m^2e^{mt}

Then the auxiliary equation is

m^2e^{mt}-5me^{mt}+6e^{mt}=0

\Rightarrow m^2-5m+6=0

\Rightarrow m^2  -3m -2m +6=0

\Rightarrow m(m  -3) -2(m -3)=0

\Rightarrow  (m-3)(m-2)=0

\Rightarrow  m=2,3

∴The complementary function is C_1e^{2t}+C_2e^{3t}

To find P.I

First we show that e^{2t} and e^{3t} are linearly independent solution.

Let y_1=e^{2t}  and y_2= e^{3t}

The Wronskian of y_1 and y_2 is \left|\begin{array}{cc}y_1&y_2\\y'_1&y'_2\end{array}\right|

                                                =\left|\begin{array}{cc}e^{2t}&e^{3t}\\2e^{2t}&3e^{3t}\end{array}\right|

                                                 =e^{2t}.3e^{3t}-e^{2t}.2e^{3t}

                                                  =e^{5t} ≠ 0

∴y_1 and y_2 are linearly independent.

Let the particular solution is

y_p=v_1(t)e^{2t}+v_2(t)e^{3t}

Then,

Dy_p= 2v_1(t)e^{2t}+v'_1(t)e^{2t}+3v_2(t)e^{3t}+v'_2(t)e^{3t}

Choose v_1(t) and v_2(t) such that

v'_1(t)e^{2t}+v'_2(t)e^{3t}=0 .......(1)

So that

Dy_p= 2v_1(t)e^{2t}+3v_2(t)e^{3t}

D^2y_p= 4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}

Now

4v_1(t)e^{2t}+9v_2(t)e^{3t}+ 2v'_1(t)e^{2t}+3v'_2(t)e^{3t}-5[2v_1(t)e^{2t}+3v_2(t)e^{3t}] +6[v_1e^{2t}+v_2e^{3t}]=2e^t

\Rightarrow  2v'_1(t)e^{2t}+3v'_2(t)e^{3t}=2e^t .......(2)

Solving (1) and (2) we get

v'_2=2 e^{-2t}    and  v'_1(t)=-2e^{-t}

Hence

v_1(t)=\int (-2e^{-t}) dt=2e^{-t}

and  v_2=\int 2e^{-2t}dt =-e^{-2t}

Therefore y_p=(2e^{-t}) e^{2t}-e^{-2t}.e^{3t}

                     =2e^t-e^t

                    =e^t

Therefore the complete primitive is

y=c_1 e^{2y}+c_2e^{3t}+ e^{t}

<h3>Undermined coefficients:</h3>

∴The complementary function is C_1e^{2t}+C_2e^{3t}

The particular solution is y_p=Ae^t

Then,

Dy_p= Ae^t and D^2y_p=Ae^t

\therefore Ae^t-5Ae^t+6Ae^t=2e^t

\Rightarrow 2Ae^t=2e^t

\Rightarrow A=1

\therefore y_p=e^t

Therefore the general solution is

y=c_1e^{2t}+c_2e^{3t}+e^t

4 0
3 years ago
A rocket is launched at the rate of 11 feet per second from a point on the ground 15 feet from an observer. To 2 decimal places
iragen [17]
Let the angle of elevation is x and the height of the rocket from the ground is y

tanx = y/15

by differentiating both sides with respect to T

sec²x·dx/dt = (dy/dt)/15

at y = 30 , the hypotenuse of the triangle = 15√5

sec²x=(15√5/15)²=5

5 dx/dt = 11/15

dx/dt = 11/75 rad/sec
3 0
4 years ago
Read 2 more answers
If DF = 51 and EF = 32, then DE = ?
Verizon [17]

Answer:

Step-by-step explanation:

51 + 32 = 83 for DE

7 0
4 years ago
The measures of two angles of a triangle are 36° and 28°. What is the measure of the third angle?
galina1969 [7]

Answer: 116 degrees.

Step-by-step explanation:

Interior angles of a triangle add to 180.

Knowing the first two angles are 36 and 28, you can subtract them from 180 to find the answer:

180 - 36 - 28 = 116

6 0
3 years ago
Read 2 more answers
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