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2 years ago

9

Suppose FLX =ZAQ Which other congruency statements are correct

1 answer:

Stella [2.4K]2 years ago

5 0

LFX=AZQ
XLF=QAZ
FXL=ZQA
XFL=QZA

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Answer:

1. $27^{\frac{2}{3} } =9$

2. $\sqrt{36^{3} } =216$

3. $(-243)^{\frac{3}{5} } =-27$

4. $40^{\frac{2}{3}}=4\sqrt[3]{25} =4325$

5. Step 4: $(\frac{343}{27}) ^{-1} =\frac{27}{343}$

6. $D. -72cd^{7}$

Step-by-step explanation:

Use the following properties:

$a^{\frac{x}{y} } =\sqrt[x]{a^{y} }$

$\sqrt[n]{ab} =\sqrt[n]{a} \sqrt[n]{b}$

$a^{-n} =\frac{1}{a^{n} }$

$(xy)^{z} =x^{z} y^{z} \\\\$

$(x^{y}) ^{z} =x^{yz}$

$x^{y} x^{z} =x^{y+z}$

So:

1. $27^{\frac{2}{3} } =\sqrt[3]{27^{2}} =\sqrt[3]{729} }=9$

2. $\sqrt{36^{3} } =\sqrt{36*36*36} =\sqrt{36} \sqrt{36} \sqrt{36} =6*6*6=216$

3. $(-243)^{\frac{3}{5} } =\sqrt[5]{-243^{3} } =\sqrt[5]{-14348907} =-27$

4. $40^{\frac{2}{3}}=\sqrt[3]{40^{2} } =\sqrt[3]{2^{6} 5^{2} } =\sqrt[3]{2^{6} } \sqrt[3]{5^{2} } =2^{\frac{6}{3} } 5^{\frac{2}{3} } =4 *5^{\frac{2}{3} } =4\sqrt[3]{5^{2} } =4\sqrt[3]{25}=4325$

5. $(\frac{343}{27}) ^{-1} =\frac{1}{\frac{343}{27} } =\frac{27}{343}$

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