Where are the images of points A, B, C, and D after the reflection across the x-axis?

Mathematics

1 answer:

dem82 [27]3 years ago

8 0

A is (-1,-1) B is (3,-1) C is (1,2) and D is (-3,2). hope this helps!

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Simplify the expression. Write your answers in positive exponents only.

Pani-rosa [81]

after solving $\left(\frac{\left2x^{-7}y^7\right}{5x^2y^{-6}}\right)^{-3}$ we get $\frac{125x^{27}}{8y^{39}}$

Step-by-step explanation:

We need to simplify the expression:

$\left(\frac{\left2x^{-7}y^7\right}{5x^2y^{-6}}\right)^{-3}$

Solving

$=\frac{1}{\left(\frac{\left2x^{-7}y^7\right}{5x^2y^{-6}}\right)^{3}}\\\\=\frac{1}{\left(\frac{\left2y^{7+6}\right}{5x^{2+7}}\right)^{3}}\\=\frac{1}{\left(\frac{\left2y^{13}\right}{5x^{9}}\right)^{3}}\\=\frac{1}{\left(\frac{\left2^3y^{13*3}\right}{5^3x^{9*3}}\right)}\\=\frac{1}{\left(\frac{\left8y^{39}\right}{125x^{27}}\right)}\\We\,\,know\,\,\frac{1}{\frac{b}{a}}=\frac{a}{b}\\Applying the rule:\\=\frac{125x^{27}}{8y^{39}}$

So, after solving $\left(\frac{\left2x^{-7}y^7\right}{5x^2y^{-6}}\right)^{-3}$ we get $\frac{125x^{27}}{8y^{39}}$

Keywords: Exponents in fractions

Learn more about Exponents in fractions at:

#learnwithBrainly

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4 years ago

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konstantin123 [22]

Answer:

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Step-by-step explanation:

We are given the function

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If Roger is in bad luck and nobody is willing to purchase a video, v=0

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3 years ago

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MrRa [10]

Answer:

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Step-by-step explanation:

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