Someone help if you can

Four more than three times a number is greater than twenty-two

Verizon [17]

If u'r looking for the equation
3x+5 > 22

5 0

3 years ago

How do I solve this problem?

Sedaia [141]

Answer:the value of x will be -4 and the value of y will be 40

Step-by-step explanation:-7x+12=-3x+8

3 0

3 years ago

Read 2 more answers

You're leading s fundraiser for a local after school program. You have 236 cookies donated by parents, and your goal is to raise

Art [367]

Answer: .64 cents, 1 dollar

.64: would sell all 236 cookies

1 dollar: sell 150 cookies

Step-by-step explanation:

150/236 = .64

1 * 150 = 150

5 0

3 years ago

A company prints 4580 copies of the same book. The number of pages in the book, including the cover pages, is 79. What is the to

tigry1 [53]

Answer: ‭361,820‬ pages

Step-by-step explanation:

The company prints 4,580 copies of the same book and it has 79 pages including cover pages.

The total number of pages the company prints therefore is:

= 4,580 * 79

= ‭361,820‬ pages

3 0

3 years ago

Solve the initial value problem 2ty" + 10ty' + 8y = 0, for t > 0, y(1) = 1, y'(1) = 0.

Eva8 [605]

I think you meant to write

$2t^2y''+10ty'+8y=0$

which is an ODE of Cauchy-Euler type. Let $y=t^m$ . Then

$y'=mt^{m-1}$

$y''=m(m-1)t^{m-2}$

Substituting $y$ and its derivatives into the ODE gives

$2m(m-1)t^m+10mt^m+8t^m=0$

Divide through by $t^m$ , which we can do because $t\neq0$ :

$2m(m-1)+10m+8=2m^2+8m+8=2(m+2)^2=0\implies m=-2$

Since this root has multiplicity 2, we get the characteristic solution

$y_c=C_1t^{-2}+C_2t^{-2}\ln t$

If you're not sure where the logarithm comes from, scroll to the bottom for a bit more in-depth explanation.

With the given initial values, we find

$y(1)=1\implies1=C_1$

$y'(1)=0\implies0=-2C_1+C_2\implies C_2=2$

so that the particular solution is

$\boxed{y(t)=t^{-2}+2t^{-2}\ln t}$

# # #

Under the hood, we're actually substituting $t=e^u$ , so that $u=\ln t$ . When we do this, we need to account for the derivative of $y$ wrt the new variable $u$ . By the chain rule,

$\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{\mathrm dy}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm dy}{\mathrm du}$

Since $\frac{\mathrm dy}{\mathrm dt}$ is a function of $t$ , we can treat $\frac{\mathrm dy}{\mathrm du}$ in the same way, so denote this by $f(t)$ . By the quotient rule,

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\mathrm d}{\mathrm dt}\left[\dfrac ft\right]=\dfrac{t\frac{\mathrm df}{\mathrm dt}-f}{t^2}$

and by the chain rule,

$\dfrac{\mathrm df}{\mathrm dt}=\dfrac{\mathrm df}{\mathrm du}\dfrac{\mathrm du}{\mathrm dt}=\dfrac1t\dfrac{\mathrm df}{\mathrm du}$

where

$\dfrac{\mathrm df}{\mathrm du}=\dfrac{\mathrm d}{\mathrm du}\left[\dfrac{\mathrm dy}{\mathrm du}\right]=\dfrac{\mathrm d^2y}{\mathrm du^2}$

so that

$\dfrac{\mathrm d^2y}{\mathrm dt^2}=\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm dy}{\mathrm du}}{t^2}=\dfrac1{t^2}\left(\dfrac{\mathrm d^2y}{\mathrm du^2}-\dfrac{\mathrm dy}{\mathrm du}\right)$

Plug all this into the original ODE to get a new one that is linear in $u$ with constant coefficients:

$2t^2\left(\dfrac{\frac{\mathrm d^2y}{\mathrm du^2}-\frac{\mathrm d y}{\mathrm du}}{t^2}\right)+10t\left(\dfrac{\frac{\mathrm dy}{\mathrm du}}t\right)+8y=0$

$2y''+8y'+8y=0$

which has characteristic equation

$2r^2+8r+8=2(r+2)^2=0$

and admits the characteristic solution

$y_c(u)=C_1e^{-2u}+C_2ue^{-2u}$

Finally replace $u=\ln t$ to get the solution we found earlier,

$y_c(t)=C_1t^{-2}+C_2t^{-2}\ln t$

4 0

4 years ago