Question

Pls help these are my teachers instructions

Answer 1

The mistake he did, was in calculating powers, when 12 is taken to numerator it becomes -12 and 6-12 is -6 and not 6 so power of x should be -6 i.e x^-6 and not x^6

So, the correct answer is $=\frac{27}{4x^{6}y^{8}}$

Step-by-step explanation:

We need to simplify:

$\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

Solving:

$\frac{4(3x^2y^4)^3}{(2x^3y^5)^4}$

Multiplying power with the terms inside the bracket.

$=\frac{4(3^3x^6y^{12})}{2^4x^{12}y^{20}}\\=\frac{4(27x^6y^{12})}{16x^{12}y^{20}}\\=\frac{108x^6y^{12}}{16x^{12}y^{20}}$

Now combining powers of same variable i.e if $\frac{x^a}{x^b}=x^{a-b}$

Solving:

$=\frac{108x^{6-12}y^{12-20}}{16}\\=\frac{108x^{-6}y^{-8}}{16}\\Divide\,\,by\,\,4:\\=\frac{27x^{-6}y^{-8}}{4}$

The exponent rule says: if $x^{-1} \,\,then\,\, \frac{1}{x}$

$=\frac{27}{4x^{6}y^{8}}$

The mistake he did, was in calculating powers, when 12 is taken to numerator it becomes -12 and 6-12 is -6 and not 6 so power of x should be -6 i.e x^-6 and not x^6

So, the correct answer is $=\frac{27}{4x^{6}y^{8}}$

Keywords: Exponents

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