Emily778
Find the value of each variable. If your answer is not an integar, express it in simplest radical form.
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Emily778
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Emily778
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I need to know step by step on how to solve this.
Emily778
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@jim_thompson5910 Please help. :(
jim_thompson5910
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is this a right triangle?
Emily778
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|dw:1362695278150:dw|
jim_thompson5910
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ok thank you
jim_thompson5910
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|dw:1362695556789:dw|
jim_thompson5910
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this is because all angles must add to 180
notice how 45 + 45 + 90 = 180
jim_thompson5910
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so we have a 45-45-90 triangle which means that the two legs are congruent
jim_thompson5910
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jim_thompson5910
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if the leg is x, then the hypotenuse is x*sqrt(2)
So
hypotenuse = x*sqrt(2)
hypotenuse = sqrt(2)*sqrt(2)
hypotenuse = (sqrt(2))^2
hypotenuse = 2
jim_thompson5910
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|dw:1362695698593:dw|
Emily778
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if you already figured out the hypotenuse's are the same, why'd you do all that work up there?
jim_thompson5910
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the legs are the same, not the hypotenuses
Emily778
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oh. Should I put down the equations you did to get a decent grade?
jim_thompson5910
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and I did all that work based on the fact that if you have a 45-45-90 triangle, then you'll have this basic template
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jim_thompson5910
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also, if you know the legs, you can find the hypotenuse using the pythagorean theorem, which is
a^2 + b^2 = c^2
Emily778
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you're kinda confusing me. Like, I'm understanding it more, but the way you're explaining it. You're just kind of throwing it out randomly. I need a step by step procedure.
jim_thompson5910
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alright, did you see how I figured out how x = sqrt(2) ?
jim_thompson5910
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the legs of a 45-45-90 triangle are congruent
x is a leg and so is the sqrt(2) length, so that means x = sqrt(2)
jim_thompson5910
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with me so far?
Emily778
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can you write it down, like.."step 1, step 2" to make it more simple?
jim_thompson5910
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the legs of a 45-45-90 triangle are congruent
so x must be sqrt(2) units long since the other leg is sqrt(2) units long
this means x = sqrt(2)
---------------------------------------------
we know both legs are sqrt(2) units, so let's use them to find the hypotenuse y
a^2 + b^2 = c^2
( sqrt(2) )^2 + ( sqrt(2) )^2 = y^2
2 + 2 = y^2
4 = y^2
y^2 = 4
y = sqrt( 4 )
y = 2
So the hypotenuse is 2 units
jim_thompson5910
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|dw:1362696451678:dw|
Emily778
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would I write that down on a paper?
jim_thompson5910
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yeah if you wanted to, you could also use the template above as a shortcut
jim_thompson5910
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this template
|dw:1362696580301:dw|
Emily778
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Ugh..I'll show you what I have on my paper and you say if it's correct or not..
Emily778
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|dw:1362696649159:dw|
jim_thompson5910
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ok good so far, you just have shown that the missing angle is 45 degrees
jim_thompson5910
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oh and you threw in x = sqrt(2)
jim_thompson5910
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I would say why x is sqrt(2) and you can say that the legs are congruent (because it's a 45-45-90 triangle)
Emily778
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okay
jim_thompson5910
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the next thing you need to show is how to get y
jim_thompson5910
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there are 2 ways to do it: use the template above or the pythagorean theorem
jim_thompson5910
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both ways are shown in this post
Emily778
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can you show them again. you can copy them
jim_thompson5910
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here is the pythagorean theorem method
jim_thompson5910
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the legs of a 45-45-90 triangle are congruent
so x must be sqrt(2) units long since the other leg is sqrt(2) units long
this means x = sqrt(2)
---------------------------------------------
we know both legs are sqrt(2) units, so let's use them to find the hypotenuse y
a^2 + b^2 = c^2
( sqrt(2) )^2 + ( sqrt(2) )^2 = y^2
2 + 2 = y^2
4 = y^2
y^2 = 4
y = sqrt( 4 )
y = 2
So the hypotenuse is 2 units
jim_thompson5910
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sound good?
Emily778
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Yeah. Sorry, I was off the computer. Taking a break. :P
Emily778
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Wait..Y=2? @jim_thompson5910
jim_thompson5910
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yes, y = 2
jim_thompson5910
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|dw:1362701799670:dw|
Emily778
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could I make it look like this?|dw:1362701962485:dw|
Emily778
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from the one thing you showed me? instead of putting down "sqrt"?
Emily778
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@jim_thompson5910
jim_thompson5910
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yes you could do it that way (it's a bit cleaner looking)
Emily778
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just a question.. How does it equal y?
jim_thompson5910
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what do you mean?
Emily778
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nevermind I think I understand.
jim_thompson5910
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are you sure?
jim_thompson5910
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alright, I'm glad it's (sorta) making sense
Emily778
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so, x= sqrt2? It can't be simplified?
jim_thompson5910
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no it cannot
jim_thompson5910
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that's as simple as it gets for x
Emily778
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okay, can you help me with 2 more? Or no?
jim_thompson5910
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sure I can help
jim_thompson5910
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what's your question
Emily778
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the second one explains the same thing, but the problem is different.
Emily778
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I'll draw it
jim_thompson5910
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alright
Emily778
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|dw:1362702888077:dw|
Emily778
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:/
jim_thompson5910
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the legs are equal (we have another 45-45-90 triangle)
jim_thompson5910
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so y = x
Emily778
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would I write down that y=x?
jim_thompson5910
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use the pythagorean theorem to get
|dw:1362703031520:dw|
jim_thompson5910
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then you plug in y = x
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jim_thompson5910
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and you solve for x
Emily778
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is that " X^2 + X^2"?
Emily778
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where did the y go?
jim_thompson5910
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y = x
so y and x are the same
Emily778
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ohh. I get ya. How do I solve for X for this?
jim_thompson5910
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so you can replace y with x
Emily778
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How do i solve for x?
jim_thompson5910
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well the two x^2 terms add up to 2x^2
jim_thompson5910
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so you will have
2x^2 = 450
then what?
Emily778
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well, I'd think you'd divide the 450 by 2 but it's squared..so, I don't know..
jim_thompson5910
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yes,
2x^2 = 450
x^2 = 450/2
x^2 = 225
x = sqrt(225)
x = 15
jim_thompson5910
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so both x and y are 15 (since y = x)
Emily778
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Oh, I get it. Thankyou. :)
Emily778
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here's the 3rd one..
Emily778
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|dw:1362704401057:dw|
jim_thompson5910
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this is another 45-45-90 triangle (the legs are congruent)
Emily778
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Okay
jim_thompson5910
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so what does that mean for the value of y?
Emily778
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it's bigger
jim_thompson5910
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yes and what else
jim_thompson5910
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how can you find the value of y?
Emily778
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isn't it a^2 +b^2=c^2?
jim_thompson5910
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very good
jim_thompson5910
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replace a with sqrt(5)
replace b with sqrt(5)
and you want to solve for c
Emily778
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I know. :)
jim_thompson5910
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ok great
Emily778
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Do I actually solve sqrt5^2 + sqrt5^2 in the calculator?
jim_thompson5910
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well the square of the square roots cancel each other out
jim_thompson5910
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so (sqrt(5))^2 = 5
jim_thompson5910
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so you are left with 5 + 5 which turns into 10
jim_thompson5910
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so c^2 = 10
Emily778
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I'm confused. How did you get 5+5?
jim_thompson5910
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|dw:1362706218096:dw|
jim_thompson5910
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so that's how I got 2 five terms being added
Emily778
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wouldn't I replace C with Y?
jim_thompson5910
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you would
jim_thompson5910
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so y^2 = 10
jim_thompson5910
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making
|dw:1362706377944:dw|
Emily778
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is that it for that problem?
Emily778
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@jim_thompson5910
jim_thompson5910
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yes, you found y, so you're done
Emily778
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Oh, I'm so sorry. but I don't get this other one. Can you help me with it? c:
Emily778
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|dw:1362707591687:dw|
Emily778
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@jim_thompson5910
Emily778
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hullo?
jim_thompson5910
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sry was afk for a sec
jim_thompson5910
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were you able to get anywhere with this at all?
Emily778
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No. :/
Emily778
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I think Y = 2 sqrt 3?
Emily778
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@jim_thompson5910
jim_thompson5910
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sry i had to go afk, but I'm back
jim_thompson5910
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y is actually 2
the legs are not congruent
jim_thompson5910
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x = 2*y
x = 2*2
x =4
jim_thompson5910
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the reason how I know y is 2 is from this template
|dw:1362709753724:dw|
Emily778
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um, 60 degrees is down below and down below, its 2 sqrt3
Emily778
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@jim_thompson5910
jim_thompson5910
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well i flipped the triangle around, but if it were oriented the way the problem has it, then you would have
|dw:1362710229246:dw|
Emily778
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|dw:1362710294844:dw|
Emily778
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you missed the 3?
Emily778
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I mean 2
jim_thompson5910
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the 2*sqrt(3) is above though isnt it?
Emily778
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I just showed you the diagram. :P
jim_thompson5910
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but in this one, the 2*sqrt(3) is above
|dw:1362710579138:dw|
Emily778
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Are you supposed to change what I showed you?
jim_thompson5910
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no i just copied what you drew
Emily778
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you're mixing up everything I just drew and you know that. ._.
Emily778
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look up at the picture again..
jim_thompson5910
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yeah i see the new one, but you drew a totally different one
jim_thompson5910
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or is it a case the second one is the correct one and the first one was a typo?
Emily778
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OH I'm sorry!
jim_thompson5910
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yeah i was like...what in the world lol
Emily778
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that was my fault.
jim_thompson5910
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its fine, so which one is it?
Emily778
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the second one is correct.
jim_thompson5910
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ok one sec
Emily778
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k
Emily778
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are you gonna help? I don't have all day.
jim_thompson5910
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|dw:1362712584316:dw|
jim_thompson5910
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the value of x is exactly half that of the hypotenuse
Emily778
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That's correct. xP Lol.
jim_thompson5910
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yeah just wanted to make sure and put it up again
jim_thompson5910
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so x = 2*sqrt(3)/2 = sqrt(3)
jim_thompson5910
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which means
|dw:1362712746325:dw|
Emily778
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wait stop
jim_thompson5910
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ok whats up
Emily778
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is it 2 times sqrt 3/2?
jim_thompson5910
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the 2 terms cancel
Emily778
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so I write this down? x = 2*sqrt(3)/2 = sqrt(3)
jim_thompson5910
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yes and that will give you the value of x
Emily778
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Just to make sure we're on the same topic..|dw:1362713238955:dw|
Emily778
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is that how it should be shown?
jim_thompson5910
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yes you have it correct, so now that you have x, you can find y
jim_thompson5910
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y = x*sqrt(3)
y = sqrt(3)*sqrt(3)
y = 3
Emily778
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so, Y= 3 and X= Sqrt 3
Emily778
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?
jim_thompson5910
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yes good
Emily778
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wanna help me with one more please? :D
jim_thompson5910
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ok one more
Emily778
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:) Thanks!
jim_thompson5910
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sure thing
Emily778
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|dw:1362714107413:dw|
jim_thompson5910
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|dw:1362714258167:dw|
jim_thompson5910
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x = 2*12
x = 24
since the hypotenuse is double the shorter leg in a 30-60-90 triangle
jim_thompson5910
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y = 12*sqrt(3) because of the template given above
Emily778
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where'd you get the 2 in the first one?
jim_thompson5910
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in general, with any 30-60-90 triangle
the hypotenuse is twice as long as the shorter leg
Emily778
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oh, so that's why it's 2 x 24..Okay,
jim_thompson5910
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2 * 12 = 24, but yeah you have it
Emily778
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what's Y?
jim_thompson5910
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y = 12*sqrt(3)
jim_thompson5910
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sry mixed up the 30 and the 60
jim_thompson5910
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|dw:1362714824453:dw|
Emily778
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What is the answer for Y? I got 20..:/
jim_thompson5910
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20 is incorect
jim_thompson5910
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incorrect*
jim_thompson5910
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|dw:1362714907609:dw|
Emily778
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you can't simplify it?
jim_thompson5910
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that's as simplified as it gets
Emily778
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how did you find y?
jim_thompson5910
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in the template above, do you see how the longer leg is x*sqrt(3) ?
Emily778
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ohhh.. now I get it.
Emily778
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Thankyou ! :)
jim_thompson5910
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ok I would write that template down, it's very handy
jim_thompson5910
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also the 45-45-90 template as well
Emily778
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thankyou for helping me
jim_thompson5910
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you're welcome, sry for being afk a lot lol