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3 years ago

What's a single digit odd number you'll get from dividing a two digit number with 2?

Mathematics

2 answers:

STatiana [176]3 years ago

7 0

One of the answers would probably be 5

Andreas93 [3]3 years ago

6 0

Answer:

A single-digit odd number you'll get from dividing a two-digit number with 2 is 7.

Step-by-step explanation:

If you were to divide 14 by 2, you would get 7, which is a single-digit odd number.

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Write in simplest form:<br> <img src="https://tex.z-dn.net/?f=%5Csqrt%5B3%5D%7B24a%5E%7B10%7Db%5E%7B6%7D%7D" id="TexFormula1" ti

Charra [1.4K]

Answer:

$2a^3b^2\sqrt[3]{3a}$

Step-by-step explanation:

Use the following rules for exponents:

$a^m*a^n=a^{m+n}\\\\\sqrt[3]{x^3}=x$

Simplify 24. Find two factors of 24, one of which should be a perfect cube:

$8*3=24\\\\2^3=8$

Insert:

$\sqrt[3]{2^3*3a^{10}b^6}$

Now split the exponents. Split 10 into as many 3's as possible:

$10=3+3+3+1$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^6}$

Split 6 into as many 3's as possible:

$6=3+3$

Insert as exponents:

$\sqrt[3]{2^3*3*a^3*a^3*a^3*a^1*b^3*b^3}$

Now simplify. Any terms with an exponent of 3 will be moved out of the radical (rule #2):

$2\sqrt[3]{3*a^3*a^3*a^3*a^1*b^3*b^3}\\\\\\2*a*a*a\sqrt[3]{3*a^1*b^3*b^3}\\\\\\2*a*a*a*b*b\sqrt[3]{3*a^1}$

Simplify:

$2a^3b^2\sqrt[3]{3a}$

:Done

6 0

3 years ago