Complete the square to form a true equation; x^2-3/4x+ = (x-)^2

Mathematics

2 answers:

timofeeve [1]3 years ago

6 0

Step-by-step explanation:

(x^2-3/4x+y^2)=(x-y)2

(x-3/4*1/2)2

(x-3/8)(x-3/8)

(x^2-3x/8-3x/8+9/64)

(x^2-6x/8+9/64)

(x-3/8)2

Free_Kalibri [48]3 years ago

4 0

172273-3-$=$==8399993939384888884-%"%/=%=%8%

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I don't know how to do the problem 6^4 x 10^2 over 6^2 x 2^4. Do you think you could help?

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The answer to your question is: Yes, I think I could help".
Thank you for the points.

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

(6^4 x 10^2) / (6^2 x 2^4)

Divide the numerator and denominator by 6^2 ,

(6^2 x 10^2) / (2^4)

6^2 = 36
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Plug these equivalents into the fraction:

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3 years ago

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