Answer:
x + y = 220
25 x + 15 y = 4650
The above system can be used to find the number of adult and student tickets sold.
Step-by-step explanation:
Let us assume the number of Adult's ticket sold = x
And The number of student's tickets sold = y
Now, total number of tickets sold = 220
⇒ x + y = 220
Cost of 1 adult Ticket = $ 25
⇒ Cost of x adult tickets = x ( $25) = 25 x
Cost of 1 student Ticket = $ 15
⇒ Cost of y student tickets = y ( $15) = 15 x
Also, the total amount made from selling all tickets = $4650
⇒ 25 x + 15 y = $4650
Here, the given system of equations are:
x + y = 220
and 25 x + 15 y = 4650
The above system can be used to find the number of adult and student tickets sold.
Answer:
10^10
Step-by-step explanation:
Anything to the power of 0 is 1. 10^10 *1 is still 10^10
Divide the first one by the second and the answer will be B.
Answer:

Step-by-step explanation:
Since this is mathematics, significant figures are not necessary. Thus, we can just take 0.58 and delete the 0s. Now,
.
<span>280
I'm assuming that this question is badly formatted and that the actual number of appetizers is 7, the number of entres is 10, and that there's 4 choices of desserts. So let's take each course by itself.
You can choose 1 of 7 appetizers. So we have
n = 7
After that, you chose an entre, so the number of possible meals to this point is
n = 7 * 10 = 70
Finally, you finish off with a dessert, so the number of meals is:
n = 70 * 4 = 280
Therefore the number of possible meals you can have is 280.
Note: If the values of 77, 1010 and 44 aren't errors, but are actually correct, then the number of meals is
n = 77 * 1010 * 44 = 3421880
But I believe that it's highly unlikely that the numbers in this problem are correct. Just imagine the amount of time it would take for someone to read a menu with over a thousand entres in it. And working in that kitchen would be an absolute nightmare.</span>