This is a linear relationship. The slope-intercept form of the linear equation is:
Where,
- m is the rate of change (positive if increasing rate and negative if decreasing rate)
- c is the y intercept, or the initial value
<u><em>Rate of change is 2.5 quarts per minute (it is decreasing so m is -2.5)</em></u>
<u><em>Initial value is 50 quarts, so c is 50.</em></u>
Plugging in the values we get 
.
Changing variables to w instead of y and t instead of x, gives us,
. Where w is warts of water left in the tub and t is the time in minutes.
There is no viable solution when t=30 because at t=20, w=0 (
). It means after 20 minutes, there is no water left. So t=30 minutes doesn't make sense.
ANSWER:
Modelling equation:
When 
, there is no VIABLE solution
I hope this helps you
1=3
2=4
1 and 2
3 and 4 adjacent
First one:
cos(A)=AC/AB=3/4.24
cos(B)=BC/AB=3/4.24
Cos(A)/cos(B)=AC/AB / (BC/AB) = AC/AB * AB/BC = AC/BC=3/3=1
Second one:
To solve this problem, we have to ASSUME AFE is a straight line, i.e. angle EFB is 90 degrees. (this is not explicitly given).
If that's the case, AE is a transversal of parallel lines AB and DE.
And Angle A is congruent to angle E (alternate interior angles).
Therefore sin(A)=sin(E)=0.5
1/3, 1/8, 1/24
I'm pretty sure
