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3 years ago

15

I need help on my math

1 answer:

Kruka [31]3 years ago

4 0

Answer:

B

Step-by-step explanation:

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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.

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

7 0

3 years ago

100 POINTS solve the system using linear combination. show all work.{ 5x + 3y = 41 {3x - 6y = 9

PIT_PIT [208]

Answer:

y = 2

x = 7

Explanation:

5x + 3y = 41

3x - 6y = 9

<u>Make the coefficient of "y" in equation 1 same</u>

{ 5x + 3y = 41 } * 2 → 10x + 6y = 82 ........equation 1

→ 3x - 6y = 9 ........equation 2

<u>solving them using linear combination</u>

10x + 6y = 82

3x - 6y = 9

============

13x = 91

x = 7

Find y:

5x + 3y = 41

5(7) + 3y = 41

3y = 41 - 35

3y = 6

y = 2

4 0

2 years ago

Read 2 more answers

I really just need help with part C, Also this is 50 points!

ziro4ka [17]

2.418 x 10^8 is the answer.

6 0

3 years ago

Read 2 more answers

How do you find the 2nd graph point?

krok68 [10]

Can you make the picture a little clearer

7 0

3 years ago

Read 2 more answers

✓<br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B19%20%2B%20%20%5Csqrt%7B30%20%2B%20%20%5Csqrt%7B32%20%2B%20x%20%20%7D%20%7D%

Setler79 [48]

We'll have to repeatedly square both sides of the equation, in order to get rid of the square roots. Squaring a first time yields

$19+\sqrt{30+\sqrt{32+x}}=25$

Move the 19 to the right hand side:

$\sqrt{30+\sqrt{32+x}}=6$

And square again:

$30+\sqrt{32+x}=36 \iff \sqrt{32+x}=6$

Square one last time:

$32+x=36 \iff x=36-32=4$

Let's check the solutions: all these squaring might have created external solutions:

$\sqrt{19+\sqrt{30+\sqrt{32+4}}}=\sqrt{19+\sqrt{30+6}}=\sqrt{19+6}=\sqrt{25}=5$

So, $x=4$ is a feasible solution.

8 0

3 years ago