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2 years ago

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Caden owns a popular retail website. The initial number of transactions that took place in the first hour of day is 10 and the n

umber of transactions doubles every hour after that.
Which statement best describes the initial number of transactions in this situation?

A.
The initial number of transactions is an independent variable.
B.
The initial number of transactions is a dependent variable.
C.
The initial number of transactions is both an independent and a dependent variable.
D.
The initial number of transactions is neither an independent nor a dependent variable.

1 answer:

harkovskaia [24]2 years ago

3 0

Is that all the information privided? Im tring to figure it out but I feel like im missing something.

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Apply the method of undetermined coefficients to find a particular solution to the following system.wing system.

jarptica [38.1K]

$y''-y'+y=\sin x$

The corresponding homogeneous ODE has characteristic equation $r^2-r+1=0$ with roots at $r=\dfrac{1\pm\sqrt3}2$ , thus admitting the characteristic solution

$y_c=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x$

For the particular solution, assume one of the form

$y_p=a\sin x+b\cos x$

${y_p}'=a\cos x-b\sin x$

${y_p}''=-a\sin x-b\cos x$

Substituting into the ODE gives

$(-a\sin x-b\cos x)-(a\cos x-b\sin x)+(a\sin x+b\cos x)=\sin x$

$-b\cos x+a\sin x=\sin x$

$\implies a=1,b=0$

Then the general solution to this ODE is

$\boxed{y(x)=C_1e^x\cos\dfrac{\sqrt3}2x+C_2e^x\sin\dfrac{\sqrt3}2x+\sin x}$

$y''-3y'+2y=e^x\sin x$

$\implies r^2-3r+2=(r-1)(r-2)=0\implies r=1,r=2$

$\implies y_c=C_1e^x+C_2e^{2x}$

Assume a solution of the form

$y_p=e^x(a\sin x+b\cos x)$

${y_p}'=e^x((a+b)\cos x+(a-b)\sin x)$

${y_p}''=2e^x(a\cos x-b\sin x)$

Substituting into the ODE gives

$2e^x(a\cos x-b\sin x)-3e^x((a+b)\cos x+(a-b)\sin x)+2e^x(a\sin x+b\cos x)=e^x\sin x$

$-e^x((a+b)\cos x+(a-b)\sin x)=e^x\sin x$

$\implies\begin{cases}-a-b=0\\-a+b=1\end{cases}\implies a=-\dfrac12,b=\dfrac12$

so the solution is

$\boxed{y(x)=C_1e^x+C_2e^{2x}-\dfrac{e^x}2(\sin x-\cos x)}$

$y''+y=x\cos(2x)$

$r^2+1=0\implies r=\pm i$

$\implies y_c=C_1\cos x+C_2\sin x$

Assume a solution of the form

$y_p=(ax+b)\cos(2x)+(cx+d)\sin(2x)$

${y_p}''=-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x)$

Substituting into the ODE gives

$(-4(ax+b-c)\cos(2x)-4(cx+a+d)\sin(2x))+((ax+b)\cos(2x)+(cx+d)\sin(2x))=x\cos(2x)$

$-(3ax+3b-4c)\cos(2x)-(3cx+3d+4a)\sin(2x)=x\cos(2x)$

$\implies\begin{cases}-3a=1\\-3b+4c=0\\-3c=0\\-4a-3d=0\end{cases}\implies a=-\dfrac13,b=c=0,d=\dfrac49$

so the solution is

$\boxed{y(x)=C_1\cos x+C_2\sin x-\dfrac13x\cos(2x)+\dfrac49\sin(2x)}$

7 0

2 years ago

Solve for X Need help ASAP

dezoksy [38]

X = nine. Seventy five / 9.75!

6 0

2 years ago

Given that an intercepted arc has length 6pi inches and a central angle is pi/3, find the radius

MA_775_DIABLO [31]

$\bf \textit{arc's length}\\\\
s=r\theta \quad 
\begin{cases}
r=radius\\
\theta =angle~in\\
\qquad radians\\
------\\
s=6\pi \\
\theta =\frac{\pi }{3}
\end{cases}\implies 6\pi =r\left( \frac{\pi }{3} \right)\implies \cfrac{6\pi }{\frac{\pi }{3}}=r
\\\\\\
\cfrac{6\pi }{1}\cdot \cfrac{3}{\pi }=r\implies 18=r$

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2 years ago

frozen [14]

9
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46 ! :)

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2 years ago

How many times can 7 go into 50

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Answer:

7 without going over

Step-by-step explanation:

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3 years ago