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4 years ago

8

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

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