The residual value is -1.14.
Plug 5 into x
y=-0.7(5)+2.36
=-1.14
<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:
you have to divide by 4 not subtracted.
Step-by-step explanation:
Answer:
It is a hard question i can help tho.
Step-by-step explanation:
