1. 343x³ - 8 = 0
343x³ + 98x² - 98x² + 28x - 28x - 8 = 0
343x³ + 98x² + 28x - 98x² - 28x - 8 = 0
7x(49x²) + 7x(14x) + 7x(4) - 2(49x²) - 2(14x) - 2(4) = 0
7x(49x² + 14x + 4) - 2(49x² + 14x + 4) = 0
(7x - 2)(49x² + 14x + 4) = 0
7x - 2 = 0 or 49x² + 14x + 4 = 0
+ 2 + 2 x = -(14) ± √((14)² - 4(49)(4))
7x = 2 2(49)
7 7 x = -14 ± √(196 - 784)
x = ²/₇ 98
x = -14 ± √(-588)
-98
x = -14 ± 14i√(3)
-98
x = -14 + 14i√(3) or x = -14 - 14i√(3)
-98 -98
x = ¹/₇ - ¹/₇i√(3) or x = ¹/₇ + ¹/₇√(3)
x = ¹/₇ ± ¹/₇i√(3)
Solution Set: {¹/₇ ± ¹/₇i√(3), ²/₇}
2. 64x³ = 0
64 64
x³ = 0
∛(x³) = ∛(0)
x = 0
Solution Set: {0}
Answer:
he spent $34.67 that is the answer
Marty went to the store he bought M&M's and there were that were in rows of 12 and 4 in each row. he sees a smaller pack with four in one.How many M&M's does Marty have.Explain your answer
Answer:
16/81
Hope this Helped!!
Answer: yes
Step-by-step explanation:
Lets look at side AC and BC.
BC = 10
AC = 15
10/15 = 2/3
Now side CE and CD.
CD = 8
CE = 12
8/12 = 2/3
SO YES!
