The cost of children’s ticket is $ 5
<h3><u>Solution:</u></h3>
Let "c" be the cost of one children ticket
Let "a" be the cost of one adult ticket
Given that adult ticket to a museum costs 3$ more than a children’s ticket
<em>Cost of one adult ticket = 3 + cost of one children ticket</em>
a = 3 + c ------ eqn 1
<em><u>Given that 200 adult tickets and 100 children tickets are sold, the total revenue is $2100</u></em>
200 adult tickets x cost of one adult ticket + 100 children tickets x cost of one children ticket = 2100

200a + 100c = 2100 ------ eqn 2
<em><u>Let us solve eqn 1 and eqn 2 to find values of "a" and "c"</u></em>
Substitute eqn 1 in eqn 2
200(3 + c) + 100c = 2100
600 + 200c + 100c = 2100
600 + 300c = 2100
300c = 1500
<h3>c = 5</h3>
Thus the cost of children’s ticket is $ 5
Answer:
Thus # 2 is correct (5 (y + 4))/2
Step-by-step explanation:
Simplify the following:
(5 y (y^2 - 16))/(2 y (y - 4))
(5 y (y^2 - 16))/(2 y (y - 4)) = y/y×(5 (y^2 - 16))/(2 (y - 4)) = (5 (y^2 - 16))/(2 (y - 4)):
(5 (y^2 - 16))/(2 (y - 4))
y^2 - 16 = y^2 - 4^2:
(5 (y^2 - 4^2))/(2 (y - 4))
Factor the difference of two squares. y^2 - 4^2 = (y - 4) (y + 4):
(5 (y - 4) (y + 4))/(2 (y - 4))
Cancel terms. (5 (y - 4) (y + 4))/(2 (y - 4)) = (5 (y + 4))/2:
Answer: (5 (y + 4))/2
Answer:
d = (sn - 3)/9
Step-by-step explanation:
Answer:11x+4
Step-by-step explanation:
11x+4