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:
6 < x < 8
Step-by-step explanation:
The compound inequality in this instance represent the "and" condition, not the "or" condition.* We might solve it like this.
-47 > 1 -8x > -63 . . . . given
Multiply by -1:
47 < -1 +8x < 63
Add 1:
48 < 8x < 64
Divide by 8:
6 < x < 8
______
* You can tell this is the case by looking at the ends of the given statement:
-47 > -63 . . . . a true statement, so the solution set will be an intersection, not a union.
Answer:
Una relación lineal es de la forma:
y = a*x + b.
donde a es la pendiente y b es la ordenada al origen.
en este caso, y es el precio de la camioneta, x es el numero de años que pasaron, a es la razon de depreciación de la camioneta y b es el precio inicial de la camioneta, b = $42,000.
Sabemos que después de 5 años, el precio de la camioneta es 21,000, entonces podemos resolver:
$21,000 = a*5 + $42,000
a*5 = $21,000 - $42,000 = -$21,000
a = -$21,000/5 = -$4,200
Esto significa que el precio decae $4,200 por año
Answer:
D. y²/5² - x²/8² = 1
Step-by-step explanation:
A and B are both incorrectly oriented, and D is the only hyperbola that contains the points (0,5) and (0,-5).
Verification (0,5) and (0,-5) are in the hyperbola:
First replace x and y with corresponding x and y values (We will start with x=0 and y=5)

Then simplify.



If the result is an equation (where both sides are equal to each other) then the original x and y values inputted are valid. The same is true with x and y inputs x=0 and y=-5, or any other point along the hyperbola.