((r²+7r+10)/3) × (3r-30)/(r²-5r=50)
= Factoring the expression (r²+7r+10) we get (r+2)(r+5)
Then multiply it with the other numerator, we get
= (r+2)(r+5)(3(r-10)
Then we factorise
= (r²-5r-50) = (r+5)(r-10)
Multiplying it with the other denominator, we have;
= 3(r+5)(r-10)
Therefore dividing
(r+2)(r+5)(3(r-10)/(3(r+5)(r-10) we get;
= r+2
Therefore, the correct answer is r+2
29.8 would be the number, but rounded would be 30
Answer:
(3a-1)(3a+1)
Step-by-step explanation:
We can quickly see with this problem that it is the difference of two squares as 9a^2 is (3a)^2 and 1 is 1^2 and therefore can factorise quickly using this rule.
x^2-y^2 = (x-y)(x+y) where x = 3a and y = 1
The x-intercept of f(x) = (x + 6)(x - 3)
f(x) = 0 → (x + 6)(x - 3) = 0 → x + 6 =0 or x - 3 =0
x = -6 or x = 3
Therefore the x-intercepts are: (-6; 0) and (3; 0)
Your answer is (-6; 0)
No because when one number is odd and the other one is even it can't be simplified (unless they are multiples/factors of each other) ex: 7/14
