(z+2)= (z-3)(5/-3)
z+2 = 5z/-3 +5
-3= 2z/-3
9= 2z
9/2 = z
Consider the charge for parking one car for t hours.
If t is more than 1, then the function is y=3+2(t-1), because 3 $ are payed for the first hour, then for t-1 of the left hours, we pay 2 $.
If t is one, then the rule y=3+2(t-1) still calculates the charge of 3 $, because substituting t with one in the formula yields 3.
75% is 75/100 or 0.75.
For whatever number of hours t, the charge for the first car is 3+2(t-1) $, and whatever that expression is, the price for the second car and third car will be
0.75 times 3+2(t-1). Thus, the charge for the 3 cars is given by:
3+2(t-1)+0.75[3+2(t-1)]+0.75[3+2(t-1)]=3+2(t-1)+<span>0.75 × 2[3 + 2(t − 1)].
Thus, the function which total parking charge of parking 3 cars for t hours is:
</span><span>f(t) = (3 + 2(t − 1)) + 0.75 × 2(3 + 2(t − 1))
Answer: C</span>
3 x^{2} - 15x - 42
(x-7)(x+2)
x-7=0 x+2=0
x=7,-2
So it can be known anywhere. So we use all of the same measurements
