Assuming that the pool was drained at a constant rate, the speed at which it was drained can be expressed as a function of time. In this case, the pool level will be expressed in feet per hour.
The time changed by 4 hours (6-2), and the level of the pool changed by -8 feet (2-10). Diving the feet by the hours to get the rate of decreasing depth, we find that the rate equals -2 feet/hour.
Answer:
The answer to the problem is J
1st:
f(x) = x^2.
f(x +1): Anywhere we see x we replace it with (x+1)
f(x+1) = (x+1)^2 , Expand
(x+1)^2 = (x+1)(x+1) = x^2 + 2x + 1= C.
2nd:
Noting general equation, y = mx + c
m = slope, c = vertical intercept.
From the graph, slope = (2,0) and (0, -4)
= (-4 -0) / (0 - 2) = -4/-2 = 2.
Vertical intercept, c = -4.
y = mx + c , y = 2x + (-4), y = 2x - 4 = D.
3rd:
f(x) = x^2 - x, f(-4). we would substitute x = -4.
f(-4) = (-4)^2 - (-4) = 16 + 4 = 20 = D.
Cheers.
Answer:
a=110 b=35 c= 10 d=55 e=14 f=10
Step-by-step explanation:
a = 11c , 180-70=110
b=2*35 =70
c= 11*10=110
2d=110 , d=55
e=5*14=70
7f=70 , f=10
