d=square root of (square root 2 -5 sr 2)^2 +(- sr 3 - seven sr 3)^2

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- anonymous

d=square root of (square root 2 -5 sr 2)^2 +(- sr 3 - seven sr 3)^2

- jamiebookeater

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- anonymous

all right. the equations is.
\[d = \sqrt{2 - 5 \sqrt{2}} + (-\sqrt{3} - 7 * \sqrt{3})^2\]

- anonymous

is that right?

- anonymous

well, the 1st sr is over the whole problem....

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## More answers

- anonymous

nm.....duh

- anonymous

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.

- anonymous

sorry can you try rewriting the problem

- anonymous

there are a lot of sqrt's and I can't tell which numbers of grouped under square roots and not, o_0...

- anonymous

\[ d=\sqrt (\sqrt 2-5\sqrt2)^2+(-\sqrt 3 -7\sqrt3)^2\]

- anonymous

It sqrt's the whole equation.... make sence? I hope

- anonymous

ok

- anonymous

so, let's start with each paranthetical group, focusing on the left one with root(2)

- anonymous

can you evaluate \[\sqrt{2} - 5\sqrt{2}\]

- anonymous

got lost...yep

- anonymous

-4\[-4\sqrt 2\]

- anonymous

now square it

- anonymous

\[-8\sqrt3\]

- anonymous

thats where I'm lost....how do you ^2 a sqrt?

- anonymous

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.

- anonymous

\[do you 16*2 then ^2 or do you ^ the 4 and the 2 separate?\]

- anonymous

square root of 2 is actually a value less than two. if you have multiply \[\sqrt{2} * \sqrt{2}\] you'll get 2

- anonymous

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 = \[\sqrt{2}\]

- anonymous

so you get 16 * 2 then square?

- anonymous

16 *2 is the result of \[-4\sqrt{?}\] sqaured

- anonymous

sorry, that ? is a 2

- anonymous

i'm confused as to why you square it again?

- anonymous

you've solved the left paranthetical group, so solve the right

- anonymous

\[d=\sqrt (x2 -x1)^2 +(y2-y1)^2\] right?

- anonymous

this is the distance equation, yes

- anonymous

did you solve the problem?

- anonymous

1024+36864 right?

- anonymous

sqrt? right?

- anonymous

so you have 16*2 on the left and what on the right? i get 64 * 3 on the right

- anonymous

so that is 16*2 + 64*3

- anonymous

squareroot( 16 *2 + 64*3 ), unless there are other operations missing in the original equation...

- anonymous

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?

- anonymous

i think the numbers you have are way too big, how did you get 1024?

- anonymous

I squared the 16*2 and 64*3....?

- anonymous

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?

- anonymous

why did you square them?

- anonymous

you already did, that's how you have 16*2 and not -4*sqrt(2)

- anonymous

look at the prob again.....in th () you have to get the sqrt down to be squared...

- anonymous

If not then how do you ^ 2 a sqrt?

- anonymous

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

- anonymous

\[(-4*\sqrt{2})^2 = 16 * 2\]

- anonymous

right there you squared a square root

- anonymous

oh, ok....

- anonymous

rt(2) squared is 2

- anonymous

so how do I do the -8 sqrt 3 ^2?

- anonymous

the same way as you did with the left, ;)
so what is -8 squared?

- anonymous

64

- anonymous

what is \[\sqrt{3}\] squared?

- anonymous

3?

- anonymous

right

- anonymous

so you have 16*2 + 64*3

- anonymous

so i sqrt 160?

- anonymous

you have 32 + 192

- anonymous

that is the value you square root

- anonymous

the 32 is negative?

- anonymous

no.....

- anonymous

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

- anonymous

So d=14.96?

- anonymous

right, \[\sqrt{204}\]

- anonymous

So you are in Georgia IT?

- anonymous

sorry, that wwas poor adding

- anonymous

\[\sqrt{224}\]

- anonymous

yes, i am

- anonymous

was this helpful?

- anonymous

where are you from?

- anonymous

Great Help!! Thanx more than you'll ever know!! Good ol' Tennessee

- anonymous

right on, no problem, :)

- anonymous

Maybe you can help me again soon!! God Bless!

- anonymous

of course, would be glad too

- anonymous

thanks

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