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4 years ago

Use the rules for long division to divide 262 by 9.

Mathematics

2 answers:

ExtremeBDS [4]4 years ago

7 0

Answer:

A. 29r1 is the answer you looking for

Step-by-step explanation:

TEA [102]4 years ago

3 0

______2___9_______
9| 2 6<span> </span>2
-1 8
---------
8 2
-1 8
----------
1

So, 29 with a remainder of 1, meaning the answer is A.

You might be interested in

Find the length of side x in simplest radical form with rational denominator.

DaniilM [7]

Answer:

sqrt(20)

Step-by-step explanation:

The two perpendicular side lengths are equal, so using Pythagoras theorem:

x = sqrt( sqrt(10)^2 + sqrt(10)^2 )

x = sqrt(10 + 10)

x = sqrt(20)

8 0

3 years ago

The drama club is selling tickets to their play to raise money for the show's expenses. Each student ticket sells for $4 and eac

ki77a [65]

Answer:

79 people can only be in this so we have 53 kids so 53 * 4 = 212

Step-by-step explanation:

However, there is 53 kids to find how many adults you will have to do 79 - 53 which that = 26 so 26 adults were there. so 26 * 8.50 = 221.

To fine how much money they earn you will just have to add them up so

$212 + $ 221 = $433 so they didn't really earn the amount of money they really want which is $ 460.

4 0

3 years ago

In 2000, Florida's total population was 15,982,378 people and 183 out of every 1000 residents were 65 years and older. Determine

dmitriy555 [2]

Hello! There were 15,982,378 people living in Florida in 2000 and 183 out of 1,000 were ages 65 and over. That's 18.3% of all Florida residents, because 183/1,000 is 0.183 and 0.183 * 1,000 is 18.3. What you could do is multiply the amount of people living in Florida by the percentage, in this case, 18.3% (0.183). When you do that, you get 2,924,775.174 or 2,924,775 when rounded to the nearest whole number. Either way, you can't have part of a person. In the year 2000, there were 2,924,775 Florida residents 65 and older.

5 0

4 years ago

Find a particular solution to <img src="https://tex.z-dn.net/?f=%20x%5E%7B2%7D%20%20%5Cfrac%7B%20d%5E%7B2%7Dy%20%7D%7Bd%20x%5E%7

Digiron [165]

$y=x^r$
$\implies r(r-1)x^r+6rx^r+4x^r=0$
$\implies r^2+5r+4=(r+1)(r+4)=0$
$\implies r=-1,r=-4$

so the characteristic solution is

$y_c=\dfrac{C_1}x+\dfrac{C_2}{x^4}$

As a guess for the particular solution, let's back up a bit. The reason the choice of $y=x^r$ works for the characteristic solution is that, in the background, we're employing the substitution $t=\ln x$ , so that $y(x)$ is getting replaced with a new function $z(t)$ . Differentiating yields

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1x\dfrac{\mathrm dz}{\mathrm dt}$
$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\dfrac1{x^2}\left(\dfrac{\mathrm d^2z}{\mathrm dt^2}-\dfrac{\mathrm dz}{\mathrm dt}\right)$

Now the ODE in terms of $t$ is linear with constant coefficients, since the coefficients $x^2$ and $x$ will cancel, resulting in the ODE

$\dfrac{\mathrm d^2z}{\mathrm dt^2}+5\dfrac{\mathrm dz}{\mathrm dt}+4z=e^{2t}\sin e^t$

Of coursesin, the characteristic equation will be $r^2+6r+4=0$ , which leads to solutions $C_1e^{-t}+C_2e^{-4t}=C_1x^{-1}+C_2x^{-4}$ , as before.

Now that we have two linearly independent solutions, we can easily find more via variation of parameters. If $z_1,z_2$ are the solutions to the characteristic equation of the ODE in terms of $z$ , then we can find another of the form $z_p=u_1z_1+u_2z_2$ where

$u_1=-\displaystyle\int\frac{z_2e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$
$u_2=\displaystyle\int\frac{z_1e^{2t}\sin e^t}{W(z_1,z_2)}\,\mathrm dt$

where $W(z_1,z_2)$ is the Wronskian of the two characteristic solutions. We have

$u_1=-\displaystyle\int\frac{e^{-2t}\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_1=\dfrac23(1-2e^{2t})\cos e^t+\dfrac23e^t\sin e^t$

$u_2=\displaystyle\int\frac{e^t\sin e^t}{-3e^{-5t}}\,\mathrm dt$
$u_2=\dfrac13(120-20e^{2t}+e^{4t})e^t\cos e^t-\dfrac13(120-60e^{2t}+5e^{4t})\sin e^t$

$\implies z_p=u_1z_1+u_2z_2$
$\implies z_p=(40e^{-4t}-6)e^{-t}\cos e^t-(1-20e^{-2t}+40e^{-4t})\sin e^t$

and recalling that $t=\ln x\iff e^t=x$ , we have

$\implies y_p=\left(\dfrac{40}{x^3}-\dfrac6x\right)\cos x-\left(1-\dfrac{20}{x^2}+\dfrac{40}{x^4}\right)\sin x$