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3 years ago

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Amy has set aside $150 a month for her cell phone bill. The cost for a monthly cell phone plan is a $65 access fee plus $20 per

gigabyte of data used. Which expression represents the amount of money Amy has left from her monthly budget after she pays her cell phone bill?

1 answer:

Lapatulllka [165]3 years ago

3 0

65+20=85 150-85=65

She will have 65$ left over

The equation can be
65+20=85
Or
150-85
Or
65+20=85 150-85=65
It depends on the option. I hope this helped

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The volume of a cylinder can is 1.54 litre and area of base is 77cm^2 Find its height

goblinko [34]

Answer:

The height is 20 cm.

Step-by-step explanation:

First, we have to know that the volume formula is V = πr²h and the base area of cylinder is a circle. So we can let πr² be 77 cm² . Then we have to substitute the following values into the formula :

$v = 1.54 \: l$

$v = 1540 \: {cm}^{3}$

$v = \pi \times {r}^{2} \times h$

Let πr² be 77,

Let v be 1540,

$1540 = 77 \times h$

$77h = 1540$

$h = 1540 \div 77$

$h = 20 \: cm$

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Don't Answer If You Don't Know Yo! Lenny has a check for $450.00. He wants to deposit half of the money into his checking accoun

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Answer:

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Step-by-step explanation:

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2 years ago

A stock market fell 75 points over a period of 5 days. What was the average change in the stock market each day?

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3 years ago

Read 2 more answers

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

3 years ago