Use ac method
for ax²+bx+c=0 when a≠1
multiply a and c, let's call the product r
what 2 numbesr multiply to get r and add to get b
2y²+9y+9=0
2*9=18
wat 2 numbers multiply to get 18 and add to get 9
3 and 6
split it up to that
2y²+3y+6y+9=0
group
(2y²+3y)+(6y+9)=0
factor
y(2y+3)+3(2y+3)=0
undistribute
(y+3)(2y+3)=0
set to zero
y+3=0
y=-3
2y+3=0
2y=-3
y=-3/2
y=-3 and -3/2
it's factored form is (x+3)(2x+3)
Answer:
cosA = 0.6
Step-by-step explanation:
Using the Pythagorean identity
sin²A + cos²A = 1 ( subtract sin²A from both sides )
cos²A = 1 - sin²A ( take the square root of both sides )
cosA = ± 
Since only the positive value is required , then
cosA = 
= 
= 
= 0.6
Answer:
(3(6)-1)/(2(6)+1)
Step-by-step explanation:
See attached image.
Hey there,
The first step in solving this problem would have to be to add the two binomials together:
Knowing this now we find the GCF which would have to be 2x which goes into both terms.
The factored form would have to be 2x(5x - 2).
The GCF is 2x and the factored form is 2x(5x - 2).
Hope I helped,
Amna