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3 years ago

6

In 1990 sausage cost an average of $2.42 per pound. In 1994 it cost 51.35 per pound. What was the percent of depreciation (perce

nt of decrease)?
*(show your work)*

1 answer:

Katena32 [7]3 years ago

5 0

Answer:

Cumulative price change 105.96%

Average inflation rate 2.36%

Converted amount ($100 base) $205.96

Price difference ($100 base) $105.96

CPI in 1990 130.700

Step-by-step explanation:

You might be interested in

- 1 2 (x + 5) = -10 I need helpppppp

Anna007 [38]

Answer:

See below

Step-by-step explanation:

Do you mean -12(x+5)=-10?

Divide both sides by -12 -> x+5=10/12

Subtract 5 on both sides -> x=10/12-5 -> x=-4 1/6

So x=-4 1/6

Let me know if this wasn't the right equation

7 0

3 years ago

Read 2 more answers

HELP PLEASE WITH THIS ASAP ASAP

Georgia [21]

Answer:

52

Step-by-step explanation:

62 - 42 = 20

20 / 2 = 10

42 + 10 = 52

52... i think :\

5 0

1 year ago

On a number line, a number, b, is located the same distance from 0 as another number, a, but in the opposite direction. The numb

sdas [7]

Answer:

b=-a

Step-by-step explanation:

8 0

3 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

2. Sarah's craft project uses pieces of yarn that are

Lilit [14]

Answer:

She will be able to cut approximately 2 yards off and still have 1 yard.

Step-by-step explanation:

Because 3-2=1 so there would be 1 yard left.

4 0

3 years ago