answer: it would be one solution, becasue the variables are different and the constants dont matter, since the variables are different
Group and factor
undistribute then undistribute again
remember
ab+ac=a(b+c)
this is important
6d^4+4d^3-6d^2-4d
undistribute 2d
2d(3d^3+2d^2-3d-2)
group insides
2d[(3d^3+2d^2)+(-3d-2)]
undistribute
2d[(d^2)(3d+2)+(-1)(3d+2)]
undistribute the (3d+2) part
(2d)(d^2-1)(3d+2)
factor that difference of 2 perfect squares
(2d)(d-1)(d+1)(3d+2)
77.
group
(45z^3+20z^2)+(9z+4)
factor
(5z^2)(9z+4)+(1)(9z+4)
undistribuet (9z+4)
(5z^2+1)(9z+4)
Answer:
16/3
Step-by-step explanation:
12 1/4 - 3 3/5 =
49/4 - 18/5 =
245/20 - 72/20 =
173/20 =
8 13/20
If 3, 6, 9, 12, 15, and 18 are multiples of 3 between 1 and 20, and 5, 10, 15 and 20 are multiples of 5 between 1 and 20, then you add the amount of multiples together to get 10.
10 over 20 as a fraction can be simplified to 1 over 2 which is 50%.
This means that there is a 50% chance of drawing out a multiple of 3 or 5.
I hope this helps
