The Answer Is 104.50 Because You Have To Multiply The Two Adults By The Admission ( 21.75×2 ) And Also Multiply The Four Kids By The Admission ( 15.25×4 ) And Add Both Of Those Answers Together ( 43.50+61.00=104.50)
<u>Question 8</u>
a^2 + 7a + 12
= (a+3)(a+4)
When factorising a quadratic, the product of the two factors should equal the constant term (12), and the sum of the two factors should equal the linear term (7). To find the two factors, list out the factors of 12 (1x12, 2x6, 3x4) and identify the pair that adds up to 7 (3+4).
An alternative method if you get stuck during your exam would be to solve it algebraically using the quadratic formula and then write it in the factorised form.
a = (-7 +or- sqrt(7^2 - 4(1)(12)) / 2(1)
= (-7 +or- sqrt(1))/2
= -3 or -4
These factors are the negative of the values that would go in the brackets when written in factorised form, as when a = -3 the factor (a+3) would equal 0. (If it were positive 3 instead, then in the factorised form it would be a-3).
<u>Question 10</u>
-3(x - y)/9 + (4x - 7y)/2 - (x + y)/18
Rewrite each fraction with a common denominator so you can combine the fractions into one.
= -6(x - y)/18 + 9(4x - 7y)/18 - (x + y)/18
= (-6(x - y) + 9(4x - 7y) - (x + y)) /18
Expand the brackets and collect like terms.
= (-6x + 6y + 36x - 63y - x - y)/18
= (29x - 58y)/18
= 29/18 x - 29/9 y
Answer:
385-13x
Step-by-step explanation:
that is my answer
13 its not hard aslo give n if not no one can figure it out
Answer:
a) The arithmetic sequence with common difference 2 that has 8 as the first term.
b) The arithmetic sequence of common difference -5 and first term 15.
Step-by-step explanation:
Let's use for example the arithmetic sequence with common difference 2 that has 8 as the first term. Then the first two terms of this sequence are:
8, and (8+2) = 10 Therefore the second term is 10.
Another arithmetic sequence of common difference -5 and first term 15. The firs two terms of this sequence are:
15, and (15 - 5) = 10. Therefore again a 10 as second term.
