2 years ago

37,916,037,349 + 6,966,839

Mathematics

1 answer:

lubasha [3.4K]2 years ago

3 0

Answer:

37923004188

Step-by-step explanation:

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Step-by-step explanation:

Raise 17 to the power of 2

Is this what you are asking for

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kondor19780726 [428]

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Answer:

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Step-by-step explanation:

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So AB = CD and AD = BC

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and

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Answer

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How many zeros should have 5 to have 72 dividers?

xenn [34]

Hello,

nice as problem.

$| If\ a\ number\ x=a_1^{p_1}*a_2^{p_2}*a_3^{p_3}*....*a_n^{p_n}\\ the\ number\ of\ his\ dividers is:\\{p_1+1}*{p_2+1}*{p_3+1}*....*{p_n+1}$
Here x has a power of 5 as divider and a power of 2 as divider (in order to make a power of 10.

$5=5^1*2^0$

$50=5^1*(5^1*2^1)=5^2*2^1$

$500=5^1*(5^2*2^2)=5^3*2^2$

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