Answer:
r≈6ft
Step-by-step explanation:
Using the formula
C=2πr
Solving forr
r=C
2π=37.68
2·π≈5.99696ft
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<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:
E) 1 and 2
Step-by-step explanation:
We are given that there are two integers (s and t) and they are factors of another integer (n). For example if s = 3 and t = 2, we can have n = 6.
Thus:
n^(st) = 6^(2*3) = 6^6 = (2^6)(3^6)
For the first condition: s^t = 3^2 is a factor of (2^6)(3^6)
For the second condition: (st)^2 = (3*2)^2 = 6^2 is a factor of 6^6
For the third condition: s+t = 3+2 = 5 is not a factor of 6^6 or (2^6)(3^6)
Therefore, only 1 and 2 are factors of n^(st)
