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3 years ago

15

$17.00. If the CD is 20% off, and sales tax is 8%, what is the total price of the CD, including tax?

1 answer:

Over [174]3 years ago

4 0

Divide 17 by .2 and multiply by .8

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choose the equation in point slope form for the line that is perpendicular to the given line and passing through the given point

alexdok [17]

We want to determine the equation in point slope form for the line that is perpendicular to the given line and passing through the point (5.6) .

The equation and the point is;

$y=\frac{2}{3}x+9,(5,6)$

We know that for two lines to be perpendicular, the product of their slopes should be -1.

Therefore, the slope of the perpendicular should be;

$m=-\frac{1}{(\frac{2}{3})}=-\frac{3}{2}$

The second condition is that the line must pass through the point (5,6) , to do thid, we write the equation of the line in point slope form which is;

$y-y_1=m(x-x_{1_{}})$

Inserting all values, we have,

$y-6=.-\frac{3}{2}(x-5)$

That is the final answer.

6 0

1 year ago

In 14 days, the daily high temperature in Cedar Hills dropped by 16.8 degrees Fahrenheit. What is the average daily rate of chan

lesya [120]

Answer:

-16.8 / 14 = - 1.2 per day

Step-by-step explanation:

4 0

2 years ago

PLEASE HELP ME!

salantis [7]

Answer:

the last one

Step-by-step explanation:

3 0

2 years ago

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

Find f(2) given f(x) = -3x^3 + x^2 – 3

aleksley [76]

Answer:

D

Step-by-step explanation:

f(x) = -3x^3 + x^2 – 3 f(2) means that wherever you see a x, put in a 2.

f(2)= -3(2)^3 + (2)^2 - 3

f(2) = -3*8 + 4 - 3

f(2) = - 24 + 1

f(2) = - 23

6 0

2 years ago

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