Question

1) y2+6y=0 2) 5t2-t=0 3) 0,45p2 + 18p =0 4) x3 + 2x2 = 0

Answer 1

Ответ:

Объяснение:

1. y=0

2. t=0

3. p=0

4. x=0

Лол это легко!

Answer 2

To solve each of these quadratic and cubic equations for their respective variables, follow these steps:

1. y^2 + 6y = 0:

First, factor out the common term y: y(y + 6) = 0

Now, set each factor equal to zero and solve for y: y = 0 y + 6 = 0 y = -6

So, the solutions are y = 0 and y = -6.

2. 5t^2 - t = 0:

Factor out the common term t: t(5t - 1) = 0

Now, set each factor equal to zero and solve for t: t = 0 5t - 1 = 0 5t = 1 t = 1/5

So, the solutions are t = 0 and t = 1/5.

3. 0.45p^2 + 18p = 0:

First, factor out the common term 0.45p: 0.45p(p + 40) = 0

Now, set each factor equal to zero and solve for p: 0.45p = 0 p = 0

p + 40 = 0 p = -40

So, the solutions are p = 0 and p = -40.

4. x^3 + 2x^2 = 0:

Factor out the common term x^2: x^2(x + 2) = 0

Now, set each factor equal to zero and solve for x: x^2 = 0 x = 0

x + 2 = 0 x = -2

So, the solutions are x = 0 and x = -2.