Number of child tickets bought is 20
<h3><u>
Solution:</u></h3>
Given that It cost 5 dollars for a child ticket and 8 dollars for a adult ticket
cost of each child ticket = 5 dollars
cost of each adult ticket = 8 dollars
Let "c" be the number of child tickets bought
Let "a" be the number of adult tickets bought
Total tickets sold were 110 bringing in 820 dollars
<em>Number of child tickets bought + number of adult tickets bought = 110</em>
c + a = 110 ----- eqn 1
<em><u>Also we can frame a equation as:</u></em>
Number of child tickets bought x cost of each child ticket + number of adult tickets bought x cost of each adult ticket = 820

5c + 8a = 820 -------- eqn 2
Let us solve eqn 1 and eqn 2 to find values of "c" and "a"
From eqn 1,
a = 110 - c ------ eqn 3
Substitute eqn 3 in eqn 2
5c + 8(110 - c) = 820
5c + 880 - 8c = 820
-3c = - 60
c = 20
Therefore from eqn 3,
a = 110 - 20 = 90
a = 90
Therefore number of child tickets bought is 20
Answer:
subtract 6 from 27 the divide by 3
Step-by-step explanation:
Answer:
450
Step-by-step explanation:
414+36=450
414+36=450p
Answer:
k = 4 units
Step-by-step explanation:
Using Pythagoras' identity on the right triangle.
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
k² + 7.5² = 8.5²
k² + 56.25 = 72.25 ( subtract 56.25 from both sides )
k² = 16 ( take the square root of both sides )
k =
= 4
0 or 1 or more than one for zero functions