What are the possible values of x in 6x^2 + 432 = 0

x = ±6i sqrt(2)

Step-by-step explanation:

6x^2 + 432 = 0

Subtract 432 from each side

6x^2 =- 432

Divide by 6

6x^2 /6 =- 432/6

x^2 =-72

Take the square root of each side

sqrt(x^2) = ±sqrt( -72)

x = ±sqrt(-1) sqrt(2*36)

We know the square root of -1 = i

x = ±i sqrt(2*36)

x = ±i sqrt(36) sqrt(2)

x = ±6i sqrt(2)

hello from mrbilldoesmath!

answer:

25x^2 + 90x + 84

discussion:

mn(x) = m(n(x)) = m( 5x + 9) =

(5x + 9)^2 + 3 =>

25x^2 + 90x + 9^2 + 3 =

25x^2 + 90x + 84                 (as 9^2 = 9*9 = 81)

which is the second choice

regards,  

mrb

p.s.   i'll be on vacation from friday, dec 22 to jan 2, 2019. have a great new year!

answer: 270

step-by-step explanation:

you have to times 200 by 1.35 to get your answer