Answer: The required solution is 
Step-by-step explanation:
We are given to solve the following differential equation :

where k is a constant and the equation satisfies the conditions y(0) = 50, y(5) = 100.
From equation (i), we have

Integrating both sides, we get
![\int\dfrac{dy}{y}=\int kdt\\\\\Rightarrow \log y=kt+c~~~~~~[\textup{c is a constant of integration}]\\\\\Rightarrow y=e^{kt+c}\\\\\Rightarrow y=ae^{kt}~~~~[\textup{where }a=e^c\textup{ is another constant}]](https://tex.z-dn.net/?f=%5Cint%5Cdfrac%7Bdy%7D%7By%7D%3D%5Cint%20kdt%5C%5C%5C%5C%5CRightarrow%20%5Clog%20y%3Dkt%2Bc~~~~~~%5B%5Ctextup%7Bc%20is%20a%20constant%20of%20integration%7D%5D%5C%5C%5C%5C%5CRightarrow%20y%3De%5E%7Bkt%2Bc%7D%5C%5C%5C%5C%5CRightarrow%20y%3Dae%5E%7Bkt%7D~~~~%5B%5Ctextup%7Bwhere%20%7Da%3De%5Ec%5Ctextup%7B%20is%20another%20constant%7D%5D)
Also, the conditions are

and

Thus, the required solution is 
Answer:
the answer is 7.33
Step-by-step explanation:
I used a calculator
Answer:
No, this is not a linear function
Step-by-step explanation:
If you graph each point, you will notice that none of the three points line up together in one line with a consitstant slope.
Divide the pie chart into 10 pieces, then estimate to the best of your ability what percentage out of 100 that part is. Once completed multiply all of these percentages by 10 and this will be how many students in algebra, calculus, and geometry.
Answer:
170
Step-by-step explanation:
The given relations can be used to write and solve an equation for the number of stickers Peter has.
<h3>Setup</h3>
Let p represent the number of stickers Peter has. That is twice as many as Joe, so Joe has (p/2) stickers. Joe has 40 more stickers than Emily, so the number of stickers Emily has is (p/2 -40).
The total number of stickers is 300:
p +p/2 +(p/2 -40) = 300
<h3>Solution</h3>
2p = 340 . . . . . . . . . . . . . . add 40, collect terms
p = 170 . . . . . . . . . . . divide by 2
Peter has 170 stickers.
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<em>Additional comment</em>
Joe has 170/2 = 85 stickers. Emily has 85-40 = 45 stickers.
We could write three equations in three unknowns. Solving those using substitution would result in substantially the same equation that we have above. Or, such a system of equations could be solved using a calculator's matrix functions, as in the attachment.
p +j +e = 300
p -2j +0e = 0
0p +j -e = 40