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creativ13 [48]
3 years ago
6

Evaluate 9p^2 for p=4 . When p=4,9p^2=?

Mathematics
1 answer:
Licemer1 [7]3 years ago
6 0

Answer:

  144

Step-by-step explanation:

Put the number where the variable is and do the arithmetic.

  9p^2 = 9·4^2 = 9·16

  9p^2 = 144 . . . . when p=4

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Plz write this on paper help me and send it❤️
vovangra [49]

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}

6.

(-8c^{9} d^{-3} )^{\frac{1}{3} } *(6c^{-1}d^{4})^{2} =\sqrt[3]{-8} c^{3} d^{-1} 36c^{-2} d^{8} \\\\-2c^{3} d^{-1} 36c^{-2} d^{8}=-72cd^{7}

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