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3 years ago

The coordinates of a square ar 1,8 7,8 7,2 ?,?

1 answer:

6 0

8- | a b
7- |
6- |
5- |
4- |
3- |
2- | x c
1- |______________________________
0 1 2 3 4 5 6 7 8 9

x,x=1,2

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FinnZ [79.3K]

Answer:

divide d by 2 r should be 5

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3 years ago

ololo11 [35]

It's 2/11 because all the others round to 1/2.

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3 years ago

mel-nik [20]

Answer:

Last choice is correct.

Step-by-step explanation:

$\left(\sqrt{10x^4}-x\sqrt{5x^2}\right)\left(2\sqrt{15x^4}+\sqrt{3x^3}\right)$

$\left(x^2\sqrt{10}-x\cdot x\sqrt{5}\right)\left(2\cdot x^2\sqrt{15}+x\sqrt{3x}\right)$

$\left(x^2\sqrt{10}-x^2\sqrt{5}\right)\left(2x^2\sqrt{15}+x\sqrt{3x}\right)$

$x^2\sqrt{10}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)-x^2\sqrt{5}\left(2x^2\sqrt{15}+x\sqrt{3x}\right)$

$2x^4\sqrt{150}+x^3\sqrt{30x}-2\sqrt{75}x^4-x^3\sqrt{15x}$

$2x^4\cdot5\sqrt{6}+x^3\sqrt{30x}-2\cdot5\sqrt{3}x^4-x^3\sqrt{15x}$

$10x^4\sqrt{6}+x^3\sqrt{30x}-10\sqrt{3}x^4-x^3\sqrt{15x}$

$10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}$

Hence final answer is $10x^4\sqrt{6}+x^3\sqrt{30x}-10x^4\sqrt{3}-x^3\sqrt{15x}$

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3 years ago

Likurg_2 [28]

Answer:

Anything to the 0th power is 1.

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3 years ago

BaLLatris [955]

Answer:

Where is the graph to use

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3 years ago