10 students can join the class without exceeding the maximum
128, because it’s actually 256 but 256/ 2= 128
Supplementary angles add up to 180, so
6x+60+90 = 180 and solve for x
6x= 30
x=5
Hello :
if : a <span>≡ 5 ( mod 7) ....(1)
b </span><span>≡ 6 ( mod 7) ...(2)
(1)×(2) : ab </span><span>≡ 30 ( mod 7)
</span>but : 30 = 7×4 + 2 ..... so : 30 ≡ 2 ( mod 7)
then : ab ≡ 2 ( mod 7)
4x²+4x-35=0
factor: (2x+7)(2x-5)=0
2x+7=0, or 2x-5=0
2x=-7 or 2x=5
x=-3.5 or x=2.5
I don't see any rounding necessary in this case.
when you factor ax²+bx+c, you take the two factors of a and the two factors of c, one factor of a times one factor of c, the other factor of a times the other factors, the sum of the two products make b.
in this case, the factors of 4 is 2 and 2, the factors of -35 is -5 and 7. I line them up in the following way:
2 -5
2 7
then I multiple them diagonally, the top left 2 multiplying the bottom right 7=14, and the other 2 multiplying -5=-10, 14 and -10 make a sum of 4.
if you don't get the desired sum, switch the factors up and down till you have the right combination. Note: Do not switch left and right.
I hope this makes sense to you.
