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Reika [66]
3 years ago
15

If (m+n)^2=m^2+n^2, what is (3^m)^n?

Mathematics
1 answer:
svp [43]3 years ago
4 0

Answer:

(3^m)^n=1

Step-by-step explanation:

(m+n)^2=m^2+n^2\\m^2+2mn+n^2=m^2+n^2\\(m^2+n^2)-(m^2+n^2)+2mn=0\\2mn=0\\This\ equation\ has\ two\ solutions\ : m=0,n=0\\\\But\ either\ way\ will\ result\ in\ the\  same\ answer.\\If\ m=0\ then,\\(3^0)^n\\=1^n\\=1\\\\If\ n=0 \ then,\\(3^m)^0\\=1

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Answer:

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Explanation:

Let Arlo be "x"

Then,

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Donnie = \frac{x+20}{2} (Because Donnie is half of Bert's weight)

Average = 57

Step by step Calculations:

57 = \frac{(x) + (x+20) + (x+20-5) + (\frac{x+20}{2} )}{4}  (The whole divided by 4 because the total people is 4)

57*4   =  x + x + 20 + x + 20 - 5 + \frac{x+20}{2}

228 = 3x + 35 + \frac{x+20}{2}

228 - 3x - 35 = \frac{x+20}{2}

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386 - 20 = x + 6x

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