2x 10^1 cm this will be the answer
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
Puting it into the quadrait equation ((-b+-[b^2-4ac]^(1/2)/2a), we get that x= 31^(1/2)-5 or 31^(1/2)+5
#4 is 47, 48, 49
47+48+49=144
<span>Find an equation for the line parallel to 2y+4x = 8 and goes through the point (-3,3). Write your answer in the form y=mx+b.</span>