Answer:
16 rides
Step-by-step explanation:
Option 1 . Admission fee = $10
Each ride = $0.50
Option 2 . Admission fee = $6
Each ride = $0.75
Let no. of rides be x
So, cost of ride according to option 1 = 0.50x
So, total cost after having x rides according to option 1 :
= 10+0.50x ---1
Cost of ride according to option 2 = 0.75x
So, total cost after having x rides according to option 2 :
= 6+0.75x --2
Now to find the beak even point i.e. having the same cost
Equate 1 and 2
Thus for 16 rides , the two options have the same cost .
Hence the break even point is 16 rides
Answer:
36 cm^2
Step-by-step explanation:
S.A. = 2 (lw + wh + lh)
S.A. = 2 ((4)(2) + (2)(2) + (4)(2))
S.A. = 2 (8 + 4 + 6)
S.A. = 2 (18)
S.A. = 36
The astronomer has to choose 3 star images out of 8.This is a problem of combinations. This can be expressed as 8C3 i.e. combination of 8 objects taken 3 at a time.
This means, if 8 images are nearby the astronomer will have 56 choices <span>
of the three stars.</span>
Oldest = 2 times Youngest -> O = 2*Y
Middle = Youngest + 5 -> M = Y+5
All of them together is 57 -> O + M + Y = 57
So you have these three equations:
(1) O = 2*Y
(2) M = Y+5
(3) O + M + Y = 57
Now you want to reduce the number of variables. You can change the second equation to be Y = M-5 and then plug in "M-5" wherever there is currently a Y:
(4) O = 2*(M-5) = 2*M - 10
(5) O + M + (M-5) = 57
which becomes O + 2M = 62
Then you plug in the "O" equation (4) into (5) which gives you
(2M-10) + 2M = 62 which reduces to 4M = 72.
So now I know M is 18.
I can now plug that into my other equations:
(4) O = 2*18 - 10 which means O = 26.
Now I plug that into (1) from the top:
26 = 2*Y which becomes 13 = Y
So now I have O, Y, and M
Oldest is 26
Middle is 18
Youngest is 13
Reading the sentence again, you can see that this makes sense.
