How do you completely expand an expression?
140923
The other property of logarithms is that Log_x A - Log_x B = Log_x (A/B)\n\n\nso you can combine this rule with the other one i told you about in your other question...\n\nthe only thing is that u are working in reverse order\n\n\nLn 2y/x= Ln 2+ Ln y - Ln x\n\n\nThese are properties, they should be tought to you as principals, without them you cant solve logs:D hope this helps ;) if you think it does pls gimme 10 pts:P
600427
It is already factored, you can write it more concisely as (x-y)^2.\n\nThe EXPANDED form is x^2-2xy+y^2.
771896
expand x^2-6x+9 = 5\nor x^2-6x+4 =0
813834
It didn't disappear.  Expand (x + b/2a)^2 and you will find it.
195396
What's with the ... ? Is that a space for a number, are you supposed to guess whether to use the + or the x ? I don't get it, I think I'm gonna cry :~(
376630
You could use either "bc" or "expr", but there's\nnot much point in using both.\n\nHere's a Bourne shell example:\n\n#!/bin/sh\n\nif [ $# -ne 2 ] ; then\n    echo "You need to enter two numeric arguments"\n    exit 1\nfi\n\nsum=`/bin/expr $1 + $2`\necho $sum\n\n\n\nAnd then to test it:\n\n$ add 1 2 3\nYou need to enter two numeric arguments\n\n\n$ add 123 456\n579
786690
u can do it by\n(n/2)(a+l)\nwhere \nn is no. of terms\na is first term\nl is last term
590233
Hint:\nDo a simple substitution u = x-5\nWhich means x = u + 5\n\nYou should get\n\n(2/5)(x-5)^(5/2) + 2(x-5)^(3/2) + C
906761
Try this:\n\nvoid addIt(int* begin, int* end, int x)\n{\nint* p;\nfor (p=begin; p!=end; ++p) {\n*p = *p + x;\n}
1091569
could you expand the question please?
589015
on here? try  & nbsp; (without the space between the ampresand and n, and paste like 5 or 6)\nelsewhere? hit tab\n\n\n       .
853794
Evaluate the expression after the ∑ for each of k = 1, 2, ..., up to 6, and add them:\n\n[ 1 + (-1)^1 ] 2^-1 +\n[ 1 + (-1)^2 ] 2^-2 +\n[ 1 + (-1)^3 ] 2^-3 +\n[ 1 + (-1)^4 ] 2^-4 +\n[ 1 + (-1)^5 ] 2^-5 +\n[ 1 + (-1)^6 ] 2^-6\n\nNow use (-1)^1 = -1, (-1)^2 = (-1) × (-1) = 1, ... to get\n\n[ 1 + (-1) ] 2^-1 +\n[ 1 + (1) ] 2^-2 +\n[ 1 + (-1) ] 2^-3 +\n[ 1 + (1) ] 2^-4 +\n[ 1 + (-1) ] 2^-5 +\n[ 1 + (1) ] 2^-6\n\n=\n\n[ 0 ] 2^-1 +\n[ 2 ] 2^-2 +\n[ 0 ] 2^-3 +\n[ 2 ] 2^-4 +\n[ 0 ] 2^-5 +\n[ 2 ] 2^-6\n\n= 2 × 2^-2 + 2 × 2^-4 + 2 × 2^-6\n\nUsing a^m × a^n = a^(m + n), simplify the exponents, to get\n\n2^-1 + 2^-3 + 2^-5 \n= 1 / 2 + 1 / 8 + 1 / 32\n= 16 / 32 + 4 / 32 + 1 / 32 \n= 21 / 32
531494
Ask your teacher.\n\nBut usually, you are supposed to simplify.
768707
The first equation is incorrect. It should read:\n12(1+x) = 12 + 12x (the expanded form); if x=1, the 12(1+1)=12 +12 or 12(2)=24 \n\n3(x+1)= 3x + 3 (the expanded form) ; if x=1, then 3(1+1)= 3 +3 or 3(2) = 6
379357
It is called "extention".\n\nSo, you need an extension list.\nOne of them (very wide) is attached. You can obtain there an alphabetically list of extensions as well as associated extentions with some program.
1031190
&+hearts+;\n\nPut them together without the plus signs  ♥ Or just copy and paste them.
833635
2)\nx and y can increase infinitely as long as the value of x is 1 more than the value of y