8.3333333333333333333333333 ( recurring/ infinite 3 )
Greatest common factor (or denominator) is 6,
least common multiple is 180
hmm
(note:
I spent like 30 mins trying to use a math only of finding the values
but it didn't work so I did a force brute and elimination method
explained below)
so
a and b must be multiples of 6
so list all the multiples of 6
wait
180=6*30
and 30's factors are 1,2,3,5,6,10,15,30 so only list the numbers that
are the result of multiplying 6 and any of those numbers in that list
(so we can have the lcm of 180)
so
6*1=6
6*2=12
6*3=18
6*5=30
6*6=36
6*10=60
6*15=90
6*30=180
these are our possible candidates for the 2 numbers
now we must find the pair that has a GCD of only 6
doing the math is long and tedious so do it yourself (trial and error)
we see that our choices that fulfill both requirements (GCD of 6 and LCM of 180) are
90&12
60&18
30&36
sum them to find the least one
90+12=102
60+18=78
30+36=66
the least possible sum is 66
Answer:
(-3/2,2)
Step-by-step explanation:
Answer:
p² -16pq + 36q²
Step-by-step explanation:
Given
(- 
p + 6q)²
= (- p + 6q)(- p + 6q)
Each term in the second factor is multiplied by each term in the first factor, that is
- p(- p + 6q) + 6q(- p + 6q)
= p² - 8pq - 8pq + 36q² ← collect like terms
= p² - 16pq + 36q²
