sure what is it about maybe I can help you if it's possible
In this case the temperature depends on the hour. So temperature is your X and hour is your Y variable. When h=0, x=0. So the 21*C is you x=0 or y-int point.
If we are trying to match the equation for a line of y=mx+b, 21*C=b.
Our rate of change is -4*C/h, which is our m variable.
So y=(-4)x+(21)
OR in this case, t=-4h+21.
Answer D.
Answer:
(-18,9)
Step-by-step explanation (work shown in picture attached):
1) The first step in solving by substitution is to isolate the y-term in one of the equations if it is not already isolated. Isolate y in the first equation by dividing both sides by -2. This leaves us with the equation 
2) Now, substitute 
for the y in the second equation. Simplify and isolate x. (You can multiply everything by -2 to get rid of the fraction.) This leaves us with x = -18.
3) We've found the x-value of the solution, -18. Now, substitute -18 for x back into one of the original equations (I chose the first one because it was simple) and isolate y. This gives us y = 9. Thus, in point form, the answer is (-18,9).
This expression is called the Discriminant, also shown as Δ.
It is equal to b² - 4ac. This is a very important part of the quadratic formula as it determines whether x will have two values, one repeated value or no real values. Here are a few examples.
a) x² - 2x - 1. a is equal to 1 since 1x² = x². b = -2, c = -1
The discriminant will be (-2)² - 4×1×-1 = 4 + 4 = 8.
Since Δ > 0, there are two x values. Graphed, the parabola sinks below the x axis.
b) x². a = 1, b = 0 (0x = 0), c = 0
The discriminant will be 0² - 4×1×0 = 0 - 0 = 0.
Since Δ = 0, there is only one x value. Graphed, the parabola touches the x axis at only one point.
c) x² + 1. a = 1, c = 1.
The discriminant will be 0² - 4×1×1 = 0 - 4 = -4
Since Δ < 0, there are no real x values. Graphed, the parabola floats above the x axis.
Hope this helps!
