dpflan
d=square root of (square root 2 -5 sr 2)^2 +(- sr 3 - seven sr 3)^2
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dpflan
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all right. the equations is.
\[d = \sqrt{2 - 5 \sqrt{2}} + (-\sqrt{3} - 7 * \sqrt{3})^2\]
dpflan
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is that right?
mspicer
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well, the 1st sr is over the whole problem....
mspicer
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nm.....duh
mspicer
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I've been doing this hw since yesterday, I'm in a dark zone....I am just feed up with gettin to a prob. then stuck, which thats been all 87 prob.
dpflan
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sorry can you try rewriting the problem
dpflan
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there are a lot of sqrt's and I can't tell which numbers of grouped under square roots and not, o_0...
mspicer
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\[ d=\sqrt (\sqrt 2-5\sqrt2)^2+(-\sqrt 3 -7\sqrt3)^2\]
mspicer
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It sqrt's the whole equation.... make sence? I hope
dpflan
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ok
dpflan
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so, let's start with each paranthetical group, focusing on the left one with root(2)
dpflan
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can you evaluate \[\sqrt{2} - 5\sqrt{2}\]
mspicer
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got lost...yep
mspicer
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-4\[-4\sqrt 2\]
dpflan
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now square it
mspicer
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\[-8\sqrt3\]
mspicer
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thats where I'm lost....how do you ^2 a sqrt?
dpflan
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ok, so a square root is just this: you have a number, x, that you multiply by itself, x * x, to yield another value, x*x = y, so the square root of y is x.
mspicer
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\[do you 16*2 then ^2 or do you ^ the 4 and the 2 separate?\]
dpflan
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square root of 2 is actually a value less than two. if you have multiply \[\sqrt{2} * \sqrt{2}\] you'll get 2
dpflan
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so in that case you had, y = x*x, so the square root of y is x, here you could say that y =2, therefore x = \[\sqrt{2}\]
mspicer
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so you get 16 * 2 then square?
dpflan
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16 *2 is the result of \[-4\sqrt{?}\] sqaured
dpflan
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sorry, that ? is a 2
dpflan
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i'm confused as to why you square it again?
dpflan
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you've solved the left paranthetical group, so solve the right
mspicer
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\[d=\sqrt (x2 -x1)^2 +(y2-y1)^2\] right?
dpflan
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this is the distance equation, yes
dpflan
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did you solve the problem?
mspicer
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1024+36864 right?
mspicer
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sqrt? right?
dpflan
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so you have 16*2 on the left and what on the right? i get 64 * 3 on the right
dpflan
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so that is 16*2 + 64*3
dpflan
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squareroot( 16 *2 + 64*3 ), unless there are other operations missing in the original equation...
mspicer
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Sorry! I keep gettin stuck with trin to post my comments to you!! I got 1024 + 36864 then sqrt 37288? am I right so far?
dpflan
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i think the numbers you have are way too big, how did you get 1024?
mspicer
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I squared the 16*2 and 64*3....?
mspicer
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Are you still with me? If not then thanx for all your help...if you are then am I right or wher am I going wrong?
dpflan
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why did you square them?
dpflan
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you already did, that's how you have 16*2 and not -4*sqrt(2)
mspicer
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look at the prob again.....in th () you have to get the sqrt down to be squared...
mspicer
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If not then how do you ^ 2 a sqrt?
dpflan
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a sqrt is just a number that if you square that number, you get another number...so if you have \[\sqrt{2}\], if you square that you get 2. if you "^2 a sqrt" those two operations are cancelling eachother out...so you did this when you found that
dpflan
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\[(-4*\sqrt{2})^2 = 16 * 2\]
dpflan
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right there you squared a square root
mspicer
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oh, ok....
dpflan
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rt(2) squared is 2
mspicer
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so how do I do the -8 sqrt 3 ^2?
dpflan
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the same way as you did with the left, ;)
so what is -8 squared?
mspicer
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64
dpflan
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what is \[\sqrt{3}\] squared?
mspicer
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3?
dpflan
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right
dpflan
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so you have 16*2 + 64*3
mspicer
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so i sqrt 160?
dpflan
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you have 32 + 192
dpflan
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that is the value you square root
mspicer
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the 32 is negative?
mspicer
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no.....
dpflan
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both value are positive, they were calculated from negative values that you then squared, and when you multiply two negative numbers you get a positive
mspicer
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So d=14.96?
dpflan
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right, \[\sqrt{204}\]
mspicer
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So you are in Georgia IT?
dpflan
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sorry, that wwas poor adding
dpflan
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\[\sqrt{224}\]
dpflan
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yes, i am
dpflan
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was this helpful?
dpflan
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where are you from?
mspicer
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Great Help!! Thanx more than you'll ever know!! Good ol' Tennessee
dpflan
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right on, no problem, :)
mspicer
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Maybe you can help me again soon!! God Bless!
dpflan
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of course, would be glad too
dpflan
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thanks