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3 years ago

What is the ones digit of a number if its cube ends in 7

Mathematics

1 answer:

Ostrovityanka [42]3 years ago

3 0

Well, I have never seen a question posed this way, but let's check it out by trial and error.

1^3 = 3

2^3 = 8

3^3 = 27 Hey! There's one. And the ones digit ends in 3.

Let's try another number that ends in 3 and see if it works as well.

13^3 = 2197 Wow. It works again. I never noticed this before, so you taught me something new.

I will test one more.

33^3 = 35937 Bingo. I think we have a winner.

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3 years ago

PLEASE HELP

max2010maxim [7]

Hello!

So, this is quite the complex question, and here are the following steps:

What is the quotient of $\frac{6-3\sqrt[3]{6}}{\sqrt[3]{9}}$ ?

$\frac{(6 - 3\sqrt[3]{6})\sqrt[3]{9^{2}}}{9}$ (rationalize the denominator)

$\frac{3(2 -\sqrt[3]{6})\sqrt[3]{9^{2}}}{9}$ (factor 3 from the expression)

$\frac{(2-\sqrt[3]{6})\sqrt[3]{9^{2}}}{3}$ (reduce the fraction with 3)

$\frac{2\sqrt[3]{9^{2}}-\sqrt[3]{9^{2}}}{3}$ (distributive property)

$\frac{2\sqrt[3]{81}-\sqrt[3]{486}}{3}$ (simplify 3 · 9²)

$\frac{6\sqrt[3]{3}-3\sqrt[3]{18}}{3}$ (simplify the radical)

$\frac{3(2\sqrt[3]{3}-3\sqrt[3]{18})}{3}$ (factor 3 from the expression)

$2\sqrt[3]{3}-\sqrt[3]{18}$ (reduce the fraction)

The answer, is simply, choice A, $2\sqrt[3]{3} -\sqrt[3]{18}$ ≈ 0.263758.

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3 years ago