Ask question

Ask question

All categories

3 years ago

8

Which one is correct? Please hurry <3

2 answers:

PilotLPTM [1.2K]3 years ago

8 0

Answer:

PEMDAS so multiply or divide the coefficient

Step-by-step explanation:

Levart [38]3 years ago

4 0

First one (20 characters) hope this helped

You might be interested in

This is due in 8 minutes help pls :/

Oksana_A [137]

A*4/9b*5

(In case you don’t understand what that means) A to the power of 4 divided by 9 multiplied by b to the power of 5. This is the most you can simplify it to

8 0

3 years ago

Read 2 more answers

Please Help!!!! Please!!!!!!

RoseWind [281]

The answer it's C. I already did all the multiplications for you

6 0

3 years ago

F(x) = Square root of quantity x plus seven. ; g(x) = 8x - 11 Find f(g(x)). (1 point)

Ahat [919]

Answer:

2 sqrt(2x-1)

Step-by-step explanation:

f(x) = sqrt(x+7)

g(x) = 8x-11

f(g(x))=

Place g(x) in for x in the function f(x)

f(g(x)) = sqrt( 8x-11 +7)

= sqrt( 8x -4)

Factor out 4

= sqrt( 4(2x-1)

= 2 sqrt(2x-1)

6 0

3 years ago

Read 2 more answers

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 bell rings every 2 hours, a second bell rings every 3 hours, and a third bell rings every 4 hours. if all 3 bells rind at 9:00

MrRa [10]

Answer:

9:00 P.M.

Step-by-step explanation:

All 3 bells ring at 9AM

1st bell: every 2 hours:

9AM, 11AM, 1PM, 3PM, 5PM, 7PM, 9PM

2nd bell: very 3 hours:

9AM, 12AM, 3PM, 6PM, 9PM

3rd bell: every 4 hours:

9AM, 1PM, 5PM, 9PM

So All 3 bells will ring at 9PM again

7 0

3 years ago

Read 2 more answers