If you know the formular a^3+b^3=(a+b)(a^2-ab+b^2), you can solve this problem.
8 is 2 cubed, so x^3+2^3=(x+2)(x^2-2x+4)
so the other quadratic factor is x^2-2x+4
Answer:
6 (y + 8) (y - 1)
Step-by-step explanation:
Factor the following:
6 y^2 + 42 y - 48
Factor 6 out of 6 y^2 + 42 y - 48:
6 (y^2 + 7 y - 8)
The factors of -8 that sum to 7 are 8 and -1. So, y^2 + 7 y - 8 = (y + 8) (y - 1):
Answer: 6 (y + 8) (y - 1)
PS: it's really helpful to pose the question correctly 6 y^2 + 42 y - 48 NOT
6y 2 - 48 + 42y
he paid = 4*10 = $40
48 eggs = $40
so, waste money would be 6/48 * 100 = 12.5%
Answer:
- Q6. x = 17
- Q8. Yes, parallel
- Q9. No, not parallel
Step-by-step explanation:
Q6. Solve for x
<u>Use ratios of similar sides</u>
- 28/(82 - 28) = (2x + 8)/(5x - 4)
- 28/54 = (2x + 8)/(5x - 4)
- 14/27 = (2x + 8)/(5x - 4)
- 14(5x - 4) = 27(2x + 8)
- 70x - 56 = 54x + 216
- 70x - 54x = 56 + 216
- 16x = 272
- x = 272/16
- x = 17
Q8
If ratios are same, the lines are parallel
- 41.8/33 = (102 - 45)/45
- 41.8/33 = 57/45
- 41.8*45 = 33*57
- 1881 = 1881
Yes, lines are parallel
Q9
Incorrect, lines are not parallel
Answer:
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