3 years ago

Solve (X-3)^2/3=9 (the ^ is an exponent)

Mathematics

1 answer:

Olenka [21]3 years ago

4 0

Step-by-step explanation:

To answer the question you have t the square of the differences.

So this means that (x-3)^2= x^2-9x+9

Next youmultiply 3 on both sides which says…

x^2-9x+9=27

x^2-9x=18

So, x equals

9/2+3/2 square root 17

9/2-3/2 square root 17

You might be interested in

Rewrite the rational exponent as a radical by extending the properties of integer exponents.

Aloiza [94]

$\bf ~~~~~~~~~~~~\textit{negative exponents} \\\\ a^{-n} \implies \cfrac{1}{a^n} \qquad \qquad \cfrac{1}{a^n}\implies a^{-n} \qquad \qquad a^n\implies \cfrac{1}{a^{-n}} \\\\\\ \textit{also recall that }a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} \qquad \qquad \sqrt[ m]{a^ n}\implies a^{\frac{ n}{ m}} \\\\ -------------------------------$

$\bf \cfrac{11^{\frac{2}{5}}}{11^{\frac{1}{4}}}\implies 11^{\frac{2}{5}}\cdot 11^{-\frac{1}{4}}\implies 11^{\frac{2}{5}-\frac{1}{4}}\implies 11^{\frac{8-5}{20}}\implies 11^{\frac{3}{20}} \\\\\\ \sqrt[20]{11^3}\implies \sqrt[20]{1331}$