Miznatnasah Miznatnasah

02-01-2020
Mathematics

contestada

Solve for y in terms of x:
[tex]4 {x}^{2} + 8 {x}^{2}y = 8 {x}^{3} [/tex]
A.
[tex] \frac{2}{2x - 1} [/tex]
B.
[tex] \frac{2}{2x + 1} [/tex]
C.
[tex] \frac{2x - 1}{2} [/tex]
D.
[tex] \frac{2}{2x - 1} [/tex]

$Solve for y in terms of x tex4 x2 8 x2y 8 x3 texA tex frac22x 1 texB tex frac22x 1 texC tex frac2x 12 texD tex frac22x 1 tex class=$

Respuesta :

meredith48034 meredith48034

03-01-2020

Answer:

8x^2y= 8x^3 - 4x^2

y = x - 1/2

y = (2x - 1)/2

the answer is c

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