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1700 has 16 Factors. The number 1700 has 16 factors not counting 1 or itself makeing it composite. Prime Factors for the number 1700 = (22 × 52 × 17) or(2 × 2 × 5 × 5 × 17). The math used by the server to come up with that prime factorization can be found in the right column next to the true factors on the left.

found it off of google

GuestMay 24, 2017

#1

+130555

+1

I don't have a lot of experience at this, but I'll give it a go

Arrange the data in order from lowest to highest

0, 1, 1, 2, 2, 2, 2, 2, 3, 3, 4, 4, 5, 5, 6

find the median = 2

find the median of the data set below the median = 2 = first quartile value

find the median of the data set above the median = 4 = third quartile value

Find the inter-quartile range (IQR) = third quartile value - first quartile value = 4 - 2 = 2

To find any possible lower outliers we find all the values less than

first quartile value - 1.5 (IQR) = 2 - (1.5)(2) = 2 - 3 = -1

There are no values less than -1

To find any possible highest outliers, we find all values greater than

third quartile value + 1.5 (IQR) = 4 + (1.5)(2) = 4 + 3 = 7

Tthere are no values greater than 7

So.....no outliers exist

CPhillMay 24, 2017

#1

+80

0

What is i?

ummm.... what?

mathguardian23May 24, 2017

#4

+80

0

I thought that this was a MATH website, not a talk all you want about non-related topics site. That's what Facebook, Snapchat, Instagram, and other sites are for.

mathguardian23May 24, 2017

#4

0

"Mathematics is a game played according to certain simple rules with meaningless marks on paper"

David hilbert, one of the most important mathematicians of the 20's century.

"Philosophy is a game with objectives and no rules. Mathematics is a game with rules and no objectives."

unknown.

Math has no point. Some people find math beautiful, others find it annoying and hard. Math can achieve many things, but the truh is, math at its purest form- has no "point".

GuestMay 24, 2017

#1

+80

0

Pi is a nimber that never ends and starts with the numbers 3.1415926535897932.

mathguardian23May 24, 2017

#6

+9491

+1

\(\lim_{n\rightarrow \infty}(\frac{\sqrt{2n^3}-1}{n^{\frac32}}) \\~\\ = \lim_{n\rightarrow \infty}(\frac{ \sqrt{2n^3}}{n^{\frac32}}-\frac{1}{n^{\frac32}}) \\~\\ = \lim_{n\rightarrow \infty}( \frac{ 2^{\frac12}n^{\frac32} }{ n^{\frac32}}-\frac{1}{n^{\frac32}}) \\~\\ = \lim_{n\rightarrow \infty}( 2^{\frac12}-\frac{1}{n^{\frac32}} ) \\~\\ =2^{\frac12}-0 \\~\\ =\sqrt2\)

.

hectictarMay 24, 2017

#1

+130555

+2

This forms a trapezoid.....the bases are 5 and 8 units and the height is 6

So.....the area = (1/2)height [ sum of the bases ] = (1/2) (6) [5 + 8 ] =

3 [ 13 ] = 39 square units