Answer:
a
b
![x(t) = x_o e^{\frac{-\alpha y_o }{\beta }[e^{-\beta t} - 1] }](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%20x_o%20e%5E%7B%5Cfrac%7B-%5Calpha%20y_o%20%7D%7B%5Cbeta%20%7D%5Be%5E%7B-%5Cbeta%20t%7D%20-%201%5D%20%7D)
c
Step-by-step explanation:
From the question we are told that
Now integrating both sides
Now taking the exponent of both sides
=> 
Let 
So
Now from the question we are told that
Hence
=> 
So
From the question we are told that
substituting for y
=> 
Now integrating both sides
Now taking the exponent of both sides
=> 
Let 
=> 
Now from the question we are told that
So
=> 
divide both side by 
=> 
So
=> 
=> ![x(t) = x_o e^{\frac{\alpha y_o }{\beta }[e^{-\beta t} - 1] }](https://tex.z-dn.net/?f=x%28t%29%20%3D%20%20x_o%20e%5E%7B%5Cfrac%7B%5Calpha%20y_o%20%7D%7B%5Cbeta%20%7D%5Be%5E%7B-%5Cbeta%20t%7D%20-%201%5D%20%7D)
Generally as t tends to infinity , 
tends to zero
so
Since the graph is a straight line, we know that this equation follows some sort of y = mx + b format.
Let's take the points that are on the axes (-4,0) and (0,-3)
use them to find the slope = (0 - (-3))/(-4 - 0) = 3/(-4) = -3/4
this is our m in the general equations above
we now have y = (-3/4)x + b
b is the y-intercept which is where x = 0 we already have that point in (0,-3)
plug the -3 in for b to give as a final answer:
y = (-3/4)x - 3
Just write down the negative values of each withdrawal and add up those values for the first two.
1. -25 + -45 + -75 = -145
2.-35 + -55 + -65 = -155
As for the third question, the brother took out $10 more than Julie, so he withdrew $165. A possible equation with possible values could be:
3. -15 + -65 + -85 = -165
Keep in mind that there are multiple ways to answer the third question and that's only one of them. Hope this helped!
Answer:
-4 and -3
A negative times a negative results to a positive number, but when two negative numbers are added to each other, it still results to a negative number.
Hope I helped!
12xy i think lol :) hope that helps a bit
