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Illusion [34]
2 years ago
12

The function C gives the cost, in dollars, to shred w pounds of confidential documents of a company.

Mathematics
1 answer:
SSSSS [86.1K]2 years ago
8 0
C’(500)=80 means at 500 pounds of documents, the cost is increasing at a rate of $80 dollars/pound
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Step-by-step explanation:

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From the homogeneous part of the ODE, we can get two fundamental solutions. The characteristic equation is

3r^2-6r+6=0\iff r^2-2r+2=0

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y_1=e^x\cos x
y_2=e^x\sin x

The particular solution is easiest to obtain via variation of parameters. We're looking for a solution of the form

y_p=u_1y_1+u_2y_2

where

u_1=-\displaystyle\frac13\int\frac{y_2e^x\sec x}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\frac13\int\frac{y_1e^x\sec x}{W(y_1,y_2)}\,\mathrm dx

and W(y_1,y_2) is the Wronskian of the fundamental solutions. We have

W(e^x\cos x,e^x\sin x)=\begin{vmatrix}e^x\cos x&e^x\sin x\\e^x(\cos x-\sin x)&e^x(\cos x+\sin x)\end{vmatrix}=e^{2x}

and so

u_1=-\displaystyle\frac13\int\frac{e^{2x}\sin x\sec x}{e^{2x}}\,\mathrm dx=-\int\tan x\,\mathrm dx
u_1=\dfrac13\ln|\cos x|

u_2=\displaystyle\frac13\int\frac{e^{2x}\cos x\sec x}{e^{2x}}\,\mathrm dx=\int\mathrm dx
u_2=\dfrac13x

Therefore the particular solution is

y_p=\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x

so that the general solution to the ODE is

y=C_1e^x\cos x+C_2e^x\sin x+\dfrac13e^x\cos x\ln|\cos x|+\dfrac13xe^x\sin x
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3 years ago
Shellie spent $10.56 on oranges that cost $0.88 per orange. How many oranges did she purchase?
vlada-n [284]
You have to do 10.56 plus 88 and you will have your answer
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3 years ago
On hot, sunny, summer days, Jane rents inner tubes by the river that runs through her town. Based on her past experience, she ha
Sveta_85 [38]

Answer:

a. P(X=50)= 0.36

b. P(X≤75) =  0.9

c. P(X>50)=  0.48

d. P(X<100) = 0.9

Step-by-step explanation:

The given data is

x                      25        50       75        100          Total

P(x)                0.16       0.36    0.38    0.10          1.00

Where X is the variable and P(X) = probabililty of that variable.

From the above

a. P(X=50)= 0.36

We add the probabilities of the variable below and equal to 75

b. P(X≤75) = 0.16+ 0.36+ 0.38= 0.9

We find the probability of the variable greater than 50 and add it.

c. P(X>50)= 0.38+0.10=  0.48

It can be calculated in two ways. One is to subtract the probability of 100 from total probability of 1. And the other is to add the probabilities of all the variables less than 100 . Both would give the same answer.

d. P(X<100)= 1- P(X=100)= 1-0.1= 0.9

8 0
3 years ago
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