Nov 17, 2014 · $5+55+555+5555+55555+\cdots n$ terms, Sum of finite seqence of this pattern can be obtained by re-arranging this into geometric sequence. $5$ should be taken common first, we get $$5× (1+11+111+1111+11111+\cdots n)$$ $1, 11,111,\ldots$ are formed as quotients of chains of 9 and 9 like below

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Mar 5, 2013 · WE do this because when you round, generally it is with decimals that are rational or long. When there is 5.000000000000000001, it is better to round up. So, lets say you have the square root of 26 (5.09901951359) It's close to 6 than to 4. When rounding, you can't have 5.090919534243123 whatever to round down because decimals cant go down.

Apr 13, 2017 · $\begingroup$ Using modular arithmetic, you can certainly end up with basically the so-called "divisibility criteria" by $9$, $11$ and $5$ (respectively: sum of the digits divisible by $9$; sum of digits with alternate signs divisible by $11$, which will turn out to be in rather simple form; and last digit being a multiple of $5$). $\endgroup$

Sep 28, 2017 · Case in point, $3^2=(-3)^2$ and, as isyoung points out, $0.5^2=(-0.5)^2$. So, this fake-proof is designed to have one side equal to $0.5^2$ but the other side equal to $(-0.5)^2$. Specifically, the fake-proof says that $(4-4.5)^2=(5-4.5)^2$ since $4-4.5=-0.5$ and $5-4.5=0.5$. This trick lets the fake-proof have you thinking that $0.5=-0.5$.

Oct 27, 2017 · By halving 5 (the number you are rounding to) = 2.5 Then to find the upper bound you add it to the number you are rounding so 135 + 2.5 = 137.5 ( this is a multiple of 5). To get the lower bound you do the opposite so 135 - 2.5 = 132.5 …

The formula below uses the floor function. $\lfloor x\rfloor$ denotes the floor function of x. For example, $\lfloor2\rfloor=2$ while $\lfloor2.5\rfloor=2$. For more on the floor and ceiling functions, go checkout Wikipedia's page on them. $\frac{2i}{2}=i$, so $\lfloor\frac{2i}{2}\rfloor=i$.

May 21, 2019 · Then, $2^{5-y}=y \implies y \cdot 2^y=32$ Taking the log of both sides yields $$\log_2 y + y = 5$$ And that's where I'm stuck. Skip to main content Stack Exchange Network

Or one might repeatedly apply the difference operator. For example, the first few values are 0, 1, 5, 14, 30, 55... Then the first set of differences is 1, 4, 9, 16, 25... The second set of differences is 3, 5, 7, 9... The third set of differences is 2, 2, 2, 2, 2...

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