Answer:
blank 1 : 2
blank 2: 4
blank 3: 6
blank 4: 8
blank 5: 3
blank 6: 6
blank 7: 9
blank8: 12
Step-by-step explanation:
domain = first number
range = second number
Answer:
to solve use the distributive property and multiply
Step-by-step explanation:
I'm sorry I do not have the product
2y^3 – 2y – 10y + 10 + y^2 – 1 < 0 [the terms are simply reorganized again]
factor 2y from the first two terms, -10 from the second two terms
2y (y^2 -1) – 10 (y-1) + y^2 – 1 < 0
2y (y+1)(y–1) – 10 (y-1) + (y+1)(y–1) < 0 [ because y^2 – 1 = (y+1)(y–1) ]
factor out (y-1) from all the terms
(y-1) [2y(y+1)-10+ y+1] < 0
(y-1) [(y+1) (2y+1) - 10] < 0
Let us simplify (y+1) (2y+1) - 10 < 0 now
(y-1) (2y^2+y+2y+1-10) < 0
(y-1) (2y^2 +3y -9 < 0
(y-1) (2y^2 +6y -3y - 9) < 0 [ because 3y = 6y -3y] j
