Using the quadratic formula to solve 5x = 6x2 – 3, what are the values of x? StartFraction 5 plus-or-minus 3 StartRoot 11 EndRoo
t Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction StartFraction 5 plus-or-minus StartRoot 47 EndRoot Over 12 EndFraction StartFraction negative 5 plus-or-minus StartRoot 97 EndRoot Over 12 EndFraction
2 answers:
Answer:
Using the quadratic formula to solve 5x = 6x^2 – 3, what are the values of x?
5x = 6x^2 – 3
Subtract 5x from both sides:
0 = 6x^2 – 5x – 3
a = 6, b = -5, c = -3
x = (-b ± √(b^2 - 4ac))/(2a)
x = (-(-5) ± √((-5)^2 - 4(6)(-3)))/(2(6))
x = (5 ± √(25 + 72))/12
x = (5 ± √(97))/12
Step-by-step explanation:
Answer:
your answer is C
Step-by-step explanation:
thank me later;)
You might be interested in
Twelve point nine plus eight equals twenty point nine (12.9 + 8 = 20.9)
Hope this helped :)
Answer: A not a triangle
Step-by-step explanation:
Answer:
a. 62
Step-by-step explanation:
use the Bodmas method
=3×9-5+20×2
=<u>62</u><u> </u>
Answer:
1728
Step-by-step explanation:
2
r h
2
(11) × 14
968
B=
B=
11^2 =121
S=2
rh + 2
S=968 +2 x 121
S=968 +242
= 1728