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>
P=Ae^kt
225 = 210 * e^(k*(1990-1980)
225/210=e^10k ln(225/210)=10k
k=ln(225/210)/10=0.0069
P = 210*e^(0.0069t)
for 2000 ===> t = 2000-1980=20
P = 210*e^(0.0069*20)
P=241.0749=241
Answer:
∠ B = 116°
Step-by-step explanation:
The sum of the 3 angles in a triangle = 180° , that is
∠ B + 28° + 36° = 180°
∠ B + 64° = 180° ( subtract 64° from both sides )
∠ B = 116°
This is infinity solutions -
33x-33+99=33x-33-99
99=-99
Since 99 __equal -99 the equation has many solutions
