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dpflanBest ResponseYou've already chosen the best response.0
all right. the equations is. \[d = \sqrt{2  5 \sqrt{2}} + (\sqrt{3}  7 * \sqrt{3})^2\]
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
well, the 1st sr is over the whole problem....
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
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.
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
sorry can you try rewriting the problem
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
there are a lot of sqrt's and I can't tell which numbers of grouped under square roots and not, o_0...
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
\[ d=\sqrt (\sqrt 25\sqrt2)^2+(\sqrt 3 7\sqrt3)^2\]
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
It sqrt's the whole equation.... make sence? I hope
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
so, let's start with each paranthetical group, focusing on the left one with root(2)
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
can you evaluate \[\sqrt{2}  5\sqrt{2}\]
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
thats where I'm lost....how do you ^2 a sqrt?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
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.
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
\[do you 16*2 then ^2 or do you ^ the 4 and the 2 separate?\]
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
square root of 2 is actually a value less than two. if you have multiply \[\sqrt{2} * \sqrt{2}\] you'll get 2
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
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 = \[\sqrt{2}\]
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
so you get 16 * 2 then square?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
16 *2 is the result of \[4\sqrt{?}\] sqaured
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
i'm confused as to why you square it again?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
you've solved the left paranthetical group, so solve the right
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
\[d=\sqrt (x2 x1)^2 +(y2y1)^2\] right?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
this is the distance equation, yes
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
did you solve the problem?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
so you have 16*2 on the left and what on the right? i get 64 * 3 on the right
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
squareroot( 16 *2 + 64*3 ), unless there are other operations missing in the original equation...
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
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?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
i think the numbers you have are way too big, how did you get 1024?
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
I squared the 16*2 and 64*3....?
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
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?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
why did you square them?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
you already did, that's how you have 16*2 and not 4*sqrt(2)
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
look at the prob again.....in th () you have to get the sqrt down to be squared...
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
If not then how do you ^ 2 a sqrt?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
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
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
\[(4*\sqrt{2})^2 = 16 * 2\]
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
right there you squared a square root
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
so how do I do the 8 sqrt 3 ^2?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
the same way as you did with the left, ;) so what is 8 squared?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
what is \[\sqrt{3}\] squared?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
that is the value you square root
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
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
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
So you are in Georgia IT?
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
sorry, that wwas poor adding
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
Great Help!! Thanx more than you'll ever know!! Good ol' Tennessee
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
right on, no problem, :)
 3 years ago

mspicerBest ResponseYou've already chosen the best response.0
Maybe you can help me again soon!! God Bless!
 3 years ago

dpflanBest ResponseYou've already chosen the best response.0
of course, would be glad too
 3 years ago
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