3 years ago

(2x^0y^2)^-3 multiply 2yx^3

Mathematics

1 answer:

hodyreva [135]3 years ago

3 0

Answer:

$\frac{x^{3} }{4y^{5} }$

Step-by-step explanation:

Simplifica la expresión

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Step-by-step explanation:

Given that,

A quadratic equation,

2x(x + 1.5) = -1

We need to solve the quadratic equation. Firstly we need to simplify the above equation to form it as $ax^2+bx+c=0$ .

So,

$2x^2+3x=-1\\\\2x^2+3x+1=0$

Here, a = 2, b = 3 and c = 1

The roots of the given equation can be given by :

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Putting all the values we get :

$x=\dfrac{-b\pm \sqrt{b^2+4ac} }{2a}, \dfrac{-b\pm \sqrt{b^2-4ac} }{2a}\\\\x=\dfrac{-3\pm \sqrt{3^2+4(2)(1)} }{2(2)}, \dfrac{-3\pm \sqrt{3^2-4(2)(1)} }{2(2)}\\\\x=\dfrac{-3+1}{4}, \dfrac{-3-1}{4}\\\\x=-\dfrac{1}{2}, -1$

So, the roots of the given equation is -1/2 and -1.

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