Solve each of the following equations. Be make sure to check your solutions. Solve for part b only.

Take a look at the following set of data: 5, 12, 24, 19, 1, 32, 194. What word describes the data point 133?

FrozenT [24]

Answer:

Outlier!! I know it sounds like a cat poster but it's true

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

Shona is buying a rug for her room. Store A has the rug for $45 with a 10% discount. Store B has the same rug for $46 and is off

Rudik [331]

Store A :
0.90(45) = 40.50......40.50 + .06(40.50) = 40.50 + 2.43 = 42.93

store B :
46 - 10 = 36......36 * .06(36) = 36 + 2.16 = 38.16

she would have to purchase the rug from store B...because she doesn't have enough money to purchase it from store A. She would have (40 - 38.16) = $ 1.84 in change.

I find this question to be a little misleading....because on a rebate, u dont get the discount right away....u first pay full price, and then later u get the discount mailed back to you. So actually, she wouldn't have enough money at either store.

8 0

3 years ago

How many solutions does this system of equations have? Explain how you know.

Nostrana [21]

Answer:

No solution

Step-by-step explanation:

The given equations are :

9x-3y=-6

3x-y=2.....(1)

5y=15x+10

5y-15x=10

or

y-3x=2 .....(2)

Equations (1) and (2) shows that the lines are parallel. We know that for parallel lines, there is no solution.

3 0

3 years ago

PLEASE HELP WILL GIVE TWENTY POINTS Which is the best approximation for the measure of angle ABC?

Shkiper50 [21]

Cos(x) = BC/AB = 10.5/20 = 0.525 ⇒ x = 58.3°

3 0

3 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

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

4 0

3 years ago