<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:
D
Step-by-step explanation:
The equations are
● 4x + 2y = 10 (1)
● 4x - 2y = -10 (2)
● 4x + 2y = 10
Add - 4x to both sides
● 4x + 2y -4x = 10 -4x
● 2y = 10 -4x
Divide both sides by 2
● 2y/2 = (10 - 4x)/2
● y = 5 - 2x
● y = -2x + 5 (1)
● 4x - 2y = -10
Add -4x to both sides
● 4x -2y -4x = -10 - 4x
● -2y = -10 - 4x
Divide both sides by -2
● -2y/-2 = (-10 -4x)/-2
● y = 10 + 2x
● y = 2x + 5 (2)
So the equation are
● y = 2x + 5
● y = -2x + 5
Graph them
The lines intersect at (0,5) but aren't perpendicular
So the answer is d
[tex]Domain:x\geq0\\\\\sqrt{x}\leq5\ \ \ \ |square\ both\ sides\\\\x\leq25[tex]
Answer: B. 0 ≤ x ≤ 25
Answer:
First choice is the correct one
Explanation:
The given is:
[(x+5) / (x+2)] - [(x+1) / x(x+2)]
First, we will need to have a common denominator and then we will solve the subtraction normally. To get a common denominator, we will have to multiply both numerator and denominator of first term by x.
Therefore:
[(x+5) / (x+2)] - [(x+1) / x(x+2)] = [x(x+5) / x(x+2)] - [(x+1) / x(x+2)]
= [x(x+5)-(x+1)] / [x(x+2)]
= (x^2 + 5x - x - 1) / [x(x+2)]
= [(x^2 + 4x - 1)] / [x(x+2)]
Hope this helps :)
