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

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

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

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

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

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

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

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

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

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

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

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

dpflan
 3 years ago
Best ResponseYou've already chosen the best response.016 *2 is the result of \[4\sqrt{?}\] sqaured

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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