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11Alexandr11 [23.1K]
3 years ago
8

Explain why you would make one of the addends a tens number when solving an addition problem

Mathematics
1 answer:
leonid [27]3 years ago
4 0
<span> You refer to a multiple of ten. As a multiple of ten you can just add it to another number easier than a "ones number".</span>
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(6x10^2)/(3x10-5) help what is this in standard form?
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<u>Answer: </u>

The standard form of \frac{6 \times 10^{2}}{3 \times 10^{-5}} is 20,00,0000

<u>Solution: </u>

Given that \frac{6 \times 10^{2}}{3 \times 10^{-5}} ---- eqn 1

To write\frac{6 \times 10^{2}}{3 \times 10^{-5}} in standard form,

We know that \bold{\frac{1}{a^{-m}} = a^{m}} .So \frac{1}{10^{-5}}  becomes 10^{5}.

Now eqn 1 becomes,

= \frac{6 \times 10^{2}}{3} \times 10^{5} ----- eqn 2

We know that \bold{a^{m} \times a^{n}=a^{m+n}}, so 10^{2} \times 10^{5} = 10^{7}

Now eqn 2 becomes,

= \frac{6}{3} \times 10^{7}

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Here 10 is the base term and 7 is the exponent value. So base term 10 is multiplied by itself 7 times.

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Now eqn 3 becomes,

= 2 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10 \times 10

= 20,00,0000  

Hence the standard form of \frac{6 \times 10^{2}}{3 \times 10^{-5}} is 20,00,0000

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