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motikmotik
3 years ago
5

by 8:00 a.m., the snack bar had made $3,256 in sales. By noon, the snack bar's total sales were $8,821. On average how much mone

y did the snack bar take in each hour between 8:00 a.m. and noon?
Mathematics
1 answer:
sammy [17]3 years ago
7 0

Answer:

$1,391.25

Step-by-step explanation:

Noon would be 12:00 PM which would be 4 hours ahead of 8:00 AM

We subtract $8,821 by $3,256 to get the amount earned between 8:00 AM and 12:00 PM

$8,821 - $3,256

= $5,565

The find the average taken in within those 4 hours we will have to divide the money earned by the hours

$5,565/4

= $1,391.25

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Consider the system of differential equations dx/dt=−2y dy/dt=−2x. . Convert this system to a second order differential equation
Musya8 [376]

Answer:

Step-by-step explanation:

we have the following differential equations

\frac{dx}{dt}=-2y\\\frac{dy}{dt}=-2x\\

by differentiating the second equation we have

\frac{d}{dt}(\frac{dy}{dt})=-2\frac{dx}{dt}\\\frac{d^{2}y}{dt^{2}}=-2\frac{dx}{dt}\\\frac{dx}{dt}=\frac{-1}{2}\frac{d^{2}y}{dt^{2}}

and we replace dx/dt in the first equation

\frac{-1}{2}\frac{d^{2}y}{dt^{2}}=-2y\\\frac{d^{2}y}{dt^{2}}-4y=0

and by using the characteristic polynomial

m^{2}+4=0\\m=\±2i

the solution is

y(t)=Acos(2t)+Bsin(2t)

and to compute x(t) we have

\frac{dx}{dt}=-2Acos(2t)-2Bsin(2t)\\\\\int dx = \int[-2Acos(2t)-2Bsin(2t)]dt\\\\x(t)=-Asin(2t)+Bcos(2t)

and if we use x(0)=4 and y(0)=3, we can calculate the constants A and B

x(0)=B=4\\y(0)=A=3

I hope this is useful for you

regards

4 0
3 years ago
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4 years ago
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Y=(11/8)x+6
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Find the approximate side length of a square game board with an area of 128 in.²
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Answer:

2·x^5·y

Step-by-step explanation:

2x^5y

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3 years ago
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