<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
7x+6=3x-6
+6 +6
7x+12=3x
-3x -3x
4x+12=0
-4x -4x
divide both sides by -4 but you have to flip the sign.
-3>x
or
{x | x < -3}
Answer:
13, 14
Step-by-step explanation:
The parameters of the numbers are;
A whole number value = 2 × Another number + 6
The sum of the two numbers is less than 50
Given that the first number is equal to more than twice the second number, we have that the first number is the larger number, while the second number is the smaller number
Where 'x' represents the second number, we get;
x + 2·x + 6 < 50
Simplifying gives;
3·x + 6 < 50
x < (50 - 6)/3 = 14.
x < 14.
Therefore, the numbers for which the inequality holds true are numbers less than 14.. From the given option, the numbers are 13, and 14.
Answer:
y=-6/5x+3.
Step-by-step explanation:
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The tallest building in my area is the reunion tower which is 562 ft and the difference is 1106
