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Mathematics 7 Online
OpenStudy (anonymous):

The number 2012 x 2013 x 2014 + 2013 is the cubed of: A) 2012 B) 2013 C) 2014 D) 2112 E) 2113

OpenStudy (lgbasallote):

actually it's (2013 x 2012 x 2014) + 1

OpenStudy (lgbasallote):

solve that...then get the cube root to get your answer

OpenStudy (lgbasallote):

\[\huge \sqrt[3]{2012 \times 2013 \times 2014 + 1} = \text{answer}\] any questions?

OpenStudy (anonymous):

Is the answer 2013? @lgbasallote

OpenStudy (lgbasallote):

wait...let me check in wolfram

OpenStudy (anonymous):

And yeah I'll put up more questions that I don't understand (:

OpenStudy (lgbasallote):

i meant follow up qustion lol...anyway wolfram gave me 2012.99

OpenStudy (lgbasallote):

so yeah i guess it can be rounded off to 2013

OpenStudy (anonymous):

limited precision arithmetic

OpenStudy (anonymous):

Oh my bad and okayy thanks again

OpenStudy (lgbasallote):

welcome welcome

OpenStudy (anonymous):

2013

OpenStudy (anonymous):

(2012 2013 2014+2013)^(1/3)

OpenStudy (anonymous):

But is there a faster way to do it without using a calculator

OpenStudy (lgbasallote):

....manual? o.O

OpenStudy (anonymous):

Yeah

OpenStudy (lgbasallote):

...i do not think so....

OpenStudy (anonymous):

this is the fastermethod http://www.wolframalpha.com/input/?i=cube+root+of+%282012+x+2013+x+2014+%2B+2013%29

OpenStudy (anonymous):

Because I spent a long time multiplying the numbers and curbing some of the answers

OpenStudy (rsadhvika):

2012 x 2013 x 2014 + 2013 (2013-1) x 2013 x (2013+1) + 2013 2013 [ 2013^2 - 1^2 + 1] 2013^3

OpenStudy (lgbasallote):

but i can approximate it to be 2013 because it's 2012*2013*2014 2013 is like the middle...and it's multiplication....so it's like 2013*2013*2013 does that make sense? lol

OpenStudy (anonymous):

|dw:1343909084110:dw|

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