1002, 1005, 1008, 1011, 1014, 8151, 8154, 8157, 8160, 8163 don’t really know how to explain that
First, let's make these two into equations.
The first plan has an initial fee of $40 and costs an additional $0.16 per mile driven.
Our equation would then be
C = 40 + 0.16m
where C is the total cost, and m is the number of miles driven.
The second plan has an initial fee of $51 and costs an additional $0.11 per mile driven.
So, the equation is
C = 51 + 0.11m
where C is the total cost, and m is the number of miles driven.
Now, your question seems to be asking for one mileage for both, equalling one cost. I would go through all the steps I've taken to try and find this for you, but it would probably take hours to type out and read. In short, I'm not entirely sure that an answer like that is possible in this situation, simply because of the large difference in the initial fee of the two plans, along with the sparse common multiples between the two mileage costs.
The answer is D because...
(4)^-2 is...
1/16
(1/16)×2 is...
2/16 or 1/8
Hope this helps!
Answer:
The answer is: The adult ticket costs $15 and the child ticket costs $10.
Step-by-step explanation:
Let a = the dollar amount of adult tickets and c = the dollar amount for child tickets. The child ticket cost $5 less than the adult ticket. Then:
c = a - 5
The number of adult tickets times the adult cost plus the number of child tickets times the cost for child tickets is equal to the total dollar amount. Set up the equation:
8a + 2c = $140
Substitute:
8a + 2(a-5) = 140
8a + 2a - 10 = 140
10a = 150
a = 150 / 10 = 15, so the adult ticket is $15
Solve for c:
c = a - 5
c = 15 - 5 = 10, so the child ticket is $10
Proof:
8a + 2c = 140
8(15) + 2(10) = 140
120 + 20 = 140
140 =140
Answer:
36
Step-by-step explanation:
Rewrite this as
(2)(3)(√3)(√12), or
6 * √36, or
6 * 6 = 36