Question

B) - 2x^2+32=0 C) 36y^2=1

Answer 1

Ответ:

B)x=8

C)y=1/72(дробь)

Объяснение:

B)-2x×2+32=0

-4x+32=0

-4x=-32

x=0

C)36y×2=1

72y=1

y=1/73(дробь)

Answer 2

It seems like you've provided two mathematical equations. I can help you with them.

B) - 2x^2 + 32 = 0:

To solve this equation for x, you can follow these steps:

1. Start by moving the constant term to the other side of the equation: -2x^2 = -32

2. Divide both sides by -2 to isolate x^2: x^2 = 16

3. Take the square root of both sides. Remember to consider both the positive and negative square roots: x = ±4

So, the solutions to this equation are x = 4 and x = -4.

C) 36y^2 = 1:

To solve this equation for y, follow these steps:

1. Divide both sides by 36: y^2 = 1/36

2. Take the square root of both sides. Remember to consider both the positive and negative square roots: y = ±(1/6)

So, the solutions to this equation are y = 1/6 and y = -1/6.