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garri49 [273]
3 years ago
9

36d^3+24 solve please.

Mathematics
2 answers:
Oduvanchick [21]3 years ago
5 0
The answer is:

12 (3d^3 + 2)

pls mark as brainlist!!
Sergeeva-Olga [200]3 years ago
3 0

Solution, 36d^3+24:\quad 12\left(\sqrt[3]{3}d+\sqrt[3]{2}\right)\left(3^{\frac{2}{3}}d^2-\sqrt[3]{6}d+2^{\frac{2}{3}}\right)

Steps:

36d^3+24

\mathrm{Factor\:out\:common\:term\:}12,\\12\left(3d^3+2\right)

\mathrm{Factor}\:3d^3+2,\\12\left(\sqrt[3]{3}d+\sqrt[3]{2}\right)\left(\left(\sqrt[3]{3}\right)^2d^2-\sqrt[3]{2}\sqrt[3]{3}d+\left(\sqrt[3]{2}\right)^2\right)

\mathrm{Refine},\\12\left(\sqrt[3]{3}d+\sqrt[3]{2}\right)\left(3^{\frac{2}{3}}d^2-\sqrt[3]{6}d+2^{\frac{2}{3}}\right)

Hope\:This\:Helps!!!

-Austint1414

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<h3>Answer:   10000 in base 5</h3>

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

Explanation:

4+1 = 5 in base 10

But in base 5, the digit "5" does not exist.

The only digits in base five are: 0, 1, 2, 3, 4

This is similar to how in base ten, the digits span from 0 to 9 with the digit "10" not being a thing (rather it's the combination of the digits "1" and "0" put together).

----------------

Anyways let's go back to base 5.

Instead of writing 4+1 = 5, we'd write 4+1 = 10 in base 5. The first digit rolls back to a 0 and we involve a second digit of 1.

Think how 9+1 = 10 in base 10.

Similarly,

44+1 = 100 in base 5

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and so on.

----------------

Here are the first few numbers in base 5, when counting up by 1 each time.

0, 1, 2, 3, 4,

10, 11, 12, 13, 14,

20, 21, 22, 23, 24,

30, 31, 32, 33, 34,

40, 41, 42, 43, 44,

100, 101, 102, 103, ...

Notice each new row is when the pattern changes from what someone would expect in base 10. This is solely because the digit "5" isn't available in base 5.

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3

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Graph B

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Step-by-step explanation:

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