Stack 1 : 2^0 = 1 quarter.
Stack 2 : 2^1 = 2 quarters.
Stack 3: 2^2 = 4 quarters.
Stack 4: 2^3 = 8 quarters.
Stack 5 : 2^4 = 16 quarters.
Stack 6: 2^5 = 32 quarters.
1 + 2 + 4 + 8 + 16 + 32 = 63
Answer: She used 63 quarters.
We plug the value of x into the equation:
2x+1
2(5)+1
10+1
11
c = number of children tickets sold
a = number of adult tickets sold.
we know that 178 total tickets were sold, thus whatever "c" and "a" are, c + a = 178.
the price of a child ticket is $5.4, so the price for all "c" tickets must be 5.4c.
the price of a adult ticket is $9, so the price for all "a" tickets must be ac.
we also know the total sales for all tickets combined is $1310.4, so then 5.4c + 9a = 1310.4.
![\bf \begin{cases} c+a=178\implies \boxed{c}=178-c\\ 5.4c+9a=1310.4\\[-0.5em] \hrulefill\\ 5.4c+9\left( \boxed{178-c} \right)=1310.4 \end{cases} \\\\\\ 5.4c+1602-9c=1310.4\implies -3.6c+1602=1310.4\implies -3.6c=-291.6 \\\\\\ c=\cfrac{-291.6}{3.6}\implies c=81](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%20c%2Ba%3D178%5Cimplies%20%5Cboxed%7Bc%7D%3D178-c%5C%5C%205.4c%2B9a%3D1310.4%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%205.4c%2B9%5Cleft%28%20%5Cboxed%7B178-c%7D%20%5Cright%29%3D1310.4%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5C%5C%205.4c%2B1602-9c%3D1310.4%5Cimplies%20-3.6c%2B1602%3D1310.4%5Cimplies%20-3.6c%3D-291.6%20%5C%5C%5C%5C%5C%5C%20c%3D%5Ccfrac%7B-291.6%7D%7B3.6%7D%5Cimplies%20c%3D81)
Take the common denominator and divide by 3 then youll get 58
Arithmetic, nth = a + (n - 1)d, common difference, d = 5
a₁₂ = a + (12 -1)d = 63
<span>a + 11d = 63
</span>
a + 11*5 = 63
a = 63 - 11*5
a = 63 - 55
a = 8
So the nth term = a + (n - 1)d = 8<span> + (n - 1)5
8 + 5*(n - 1)
8 + 5n - 5
8 - 5 + 5n
3 + 5n
<span>So nth term = 3 + 5n</span></span>