3 years ago

Which graph best represents a system of equations that has no solution?

Mathematics

1 answer:

marishachu [46]3 years ago

4 0

Answer:

D, because parallel lines never have a solution.

Step-by-step explanation:

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Simplifying $\left(x^2+8xy+16y^2\right)^{\frac{1}{3}}.\left(x+4y\right)^{\frac{1}{3}}$ we get $\left(x+4y\right)}$

Step-by-step explanation:

We need to simplify:

$\left(x^2+8xy+16y^2\right)^{\frac{1}{3}}.\left(x+4y\right)^{\frac{1}{3}}$

For solving, we will first simplify the terms inside the exponent and then solve exponent

First Solving: $\left(x^2+8xy+16y^2\right)}.\left(x+4y\right)}$

Using the formula: $(a^2+2ab+b^2)=(a+b)^2$

$\left(x^2+8xy+16y^2\right)}.\left(x+4y\right)}\\\left((x)^2+2(x)(4y)+(4y)^2\right)}.\left(x+4y\right)}\\\left(x+4y\right)^2}.\left(x+4y\right)}$

Using exponent rule:

$a^m.a^n=a^{m+n}$

$\left(x+4y\right)^{2+1}}\\\left(x+4y\right)^3}$

Now,

$(\left(x+4y\right)^3})^{\frac{1}{3}}\\Simplifiying:\\\left(x+4y\right)}^{\frac{3}{3}}\\\left(x+4y\right)}$

So, simplifying $\left(x^2+8xy+16y^2\right)^{\frac{1}{3}}.\left(x+4y\right)^{\frac{1}{3}}$ we get $\left(x+4y\right)}$

Keywords: Solving Rational exponents:

Learn more about Solving Rational exponents at:

#learnwithBrainly

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