2,920 because if you multiply 365*8 you get 9,290
Factorise 4b² + 16b ⇒ 4b(b+4)
expand (a-1)(a-2) ⇒ a(a-2) - 1(a-2) ⇒ a² - 2a - 1a + 2 ⇒a² - 3a + 2
factorise x² + 8x + 7 ⇒(x + 1) (x + 7)
evaluate y⁶ / y² ⇒ y * y * y * y * y * y / y * y = y⁴
if x = -1 and y = 5, find z when z = x² + 2y² ;
z = -1² + 2(5²) ⇒ 1 + 2(25) ⇒1 + 50 = 51
Make x the subject: y = 4x - 3
y = 4x - 3
<u>+3 +3</u>
3 + y = 4x
<u>÷4 ÷4 </u>
(3+y)/4 = x
For the 1st member, you have 10 possible choices.
For the 2nd, you have 9 remaining choices.
For the 3rd: 8.
So all possible combinations would be:
10*9*8*7*6 = 10! - 5! = 30,240 combinations.
The advance tickets were 35 in number and the same day ticket will be 30 in number.
<h3>What will be the number of tickets?</h3>
To find out the number of the tickets sold we will give some notations so the notations are as follows:-
A= Cost of advance tickets=$30
S= cost of same-day ticket-=15
Number of the advance tickets
Number of same-day tickets.
The equations made by the given data will be:-

By solving the above equations:-

Hence the advance tickets were 35 in number and the same day ticket will be 30 in number.
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