anonymous
  • anonymous
Find the value of each variable. If your answer is not an integar, express it in simplest radical form.
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
|dw:1362695099329:dw|
anonymous
  • anonymous
I need to know step by step on how to solve this.
anonymous
  • anonymous
@jim_thompson5910 Please help. :(

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jim_thompson5910
  • jim_thompson5910
is this a right triangle?
anonymous
  • anonymous
|dw:1362695278150:dw|
jim_thompson5910
  • jim_thompson5910
ok thank you
jim_thompson5910
  • jim_thompson5910
|dw:1362695556789:dw|
jim_thompson5910
  • jim_thompson5910
this is because all angles must add to 180 notice how 45 + 45 + 90 = 180
jim_thompson5910
  • jim_thompson5910
so we have a 45-45-90 triangle which means that the two legs are congruent
jim_thompson5910
  • jim_thompson5910
|dw:1362695629297:dw|
jim_thompson5910
  • jim_thompson5910
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
  • jim_thompson5910
|dw:1362695698593:dw|
anonymous
  • anonymous
if you already figured out the hypotenuse's are the same, why'd you do all that work up there?
jim_thompson5910
  • jim_thompson5910
the legs are the same, not the hypotenuses
anonymous
  • anonymous
oh. Should I put down the equations you did to get a decent grade?
jim_thompson5910
  • jim_thompson5910
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 |dw:1362695974272:dw|
jim_thompson5910
  • jim_thompson5910
also, if you know the legs, you can find the hypotenuse using the pythagorean theorem, which is a^2 + b^2 = c^2
anonymous
  • anonymous
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
  • jim_thompson5910
alright, did you see how I figured out how x = sqrt(2) ?
jim_thompson5910
  • jim_thompson5910
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
  • jim_thompson5910
with me so far?
anonymous
  • anonymous
can you write it down, like.."step 1, step 2" to make it more simple?
jim_thompson5910
  • jim_thompson5910
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
  • jim_thompson5910
|dw:1362696451678:dw|
anonymous
  • anonymous
would I write that down on a paper?
jim_thompson5910
  • jim_thompson5910
yeah if you wanted to, you could also use the template above as a shortcut
jim_thompson5910
  • jim_thompson5910
this template |dw:1362696580301:dw|
anonymous
  • anonymous
Ugh..I'll show you what I have on my paper and you say if it's correct or not..
jim_thompson5910
  • jim_thompson5910
ok
anonymous
  • anonymous
|dw:1362696649159:dw|
jim_thompson5910
  • jim_thompson5910
ok good so far, you just have shown that the missing angle is 45 degrees
jim_thompson5910
  • jim_thompson5910
oh and you threw in x = sqrt(2)
jim_thompson5910
  • jim_thompson5910
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)
anonymous
  • anonymous
okay
jim_thompson5910
  • jim_thompson5910
the next thing you need to show is how to get y
jim_thompson5910
  • jim_thompson5910
there are 2 ways to do it: use the template above or the pythagorean theorem
jim_thompson5910
  • jim_thompson5910
both ways are shown in this post
anonymous
  • anonymous
can you show them again. you can copy them
jim_thompson5910
  • jim_thompson5910
here is the pythagorean theorem method
jim_thompson5910
  • jim_thompson5910
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
  • jim_thompson5910
sound good?
anonymous
  • anonymous
Yeah. Sorry, I was off the computer. Taking a break. :P
anonymous
  • anonymous
Wait..Y=2? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
yes, y = 2
jim_thompson5910
  • jim_thompson5910
|dw:1362701799670:dw|
anonymous
  • anonymous
could I make it look like this?|dw:1362701962485:dw|
anonymous
  • anonymous
from the one thing you showed me? instead of putting down "sqrt"?
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
yes you could do it that way (it's a bit cleaner looking)
anonymous
  • anonymous
just a question.. How does it equal y?
jim_thompson5910
  • jim_thompson5910
what do you mean?
anonymous
  • anonymous
nevermind I think I understand.
jim_thompson5910
  • jim_thompson5910
are you sure?
jim_thompson5910
  • jim_thompson5910
alright, I'm glad it's (sorta) making sense
anonymous
  • anonymous
so, x= sqrt2? It can't be simplified?
jim_thompson5910
  • jim_thompson5910
no it cannot
jim_thompson5910
  • jim_thompson5910
that's as simple as it gets for x
anonymous
  • anonymous
okay, can you help me with 2 more? Or no?
jim_thompson5910
  • jim_thompson5910
sure I can help
jim_thompson5910
  • jim_thompson5910
what's your question
anonymous
  • anonymous
the second one explains the same thing, but the problem is different.
anonymous
  • anonymous
I'll draw it
jim_thompson5910
  • jim_thompson5910
alright
anonymous
  • anonymous
|dw:1362702888077:dw|
anonymous
  • anonymous
:/
jim_thompson5910
  • jim_thompson5910
the legs are equal (we have another 45-45-90 triangle)
jim_thompson5910
  • jim_thompson5910
so y = x
anonymous
  • anonymous
would I write down that y=x?
jim_thompson5910
  • jim_thompson5910
use the pythagorean theorem to get |dw:1362703031520:dw|
jim_thompson5910
  • jim_thompson5910
then you plug in y = x |dw:1362703073963:dw|
jim_thompson5910
  • jim_thompson5910
and you solve for x
anonymous
  • anonymous
is that " X^2 + X^2"?
anonymous
  • anonymous
where did the y go?
jim_thompson5910
  • jim_thompson5910
y = x so y and x are the same
anonymous
  • anonymous
ohh. I get ya. How do I solve for X for this?
jim_thompson5910
  • jim_thompson5910
so you can replace y with x
anonymous
  • anonymous
How do i solve for x?
jim_thompson5910
  • jim_thompson5910
well the two x^2 terms add up to 2x^2
jim_thompson5910
  • jim_thompson5910
so you will have 2x^2 = 450 then what?
anonymous
  • anonymous
well, I'd think you'd divide the 450 by 2 but it's squared..so, I don't know..
jim_thompson5910
  • jim_thompson5910
yes, 2x^2 = 450 x^2 = 450/2 x^2 = 225 x = sqrt(225) x = 15
jim_thompson5910
  • jim_thompson5910
so both x and y are 15 (since y = x)
anonymous
  • anonymous
Oh, I get it. Thankyou. :)
anonymous
  • anonymous
here's the 3rd one..
jim_thompson5910
  • jim_thompson5910
yw
anonymous
  • anonymous
|dw:1362704401057:dw|
jim_thompson5910
  • jim_thompson5910
this is another 45-45-90 triangle (the legs are congruent)
anonymous
  • anonymous
Okay
jim_thompson5910
  • jim_thompson5910
so what does that mean for the value of y?
anonymous
  • anonymous
it's bigger
jim_thompson5910
  • jim_thompson5910
yes and what else
jim_thompson5910
  • jim_thompson5910
how can you find the value of y?
anonymous
  • anonymous
isn't it a^2 +b^2=c^2?
jim_thompson5910
  • jim_thompson5910
very good
jim_thompson5910
  • jim_thompson5910
replace a with sqrt(5) replace b with sqrt(5) and you want to solve for c
anonymous
  • anonymous
I know. :)
jim_thompson5910
  • jim_thompson5910
ok great
anonymous
  • anonymous
Do I actually solve sqrt5^2 + sqrt5^2 in the calculator?
jim_thompson5910
  • jim_thompson5910
well the square of the square roots cancel each other out
jim_thompson5910
  • jim_thompson5910
so (sqrt(5))^2 = 5
jim_thompson5910
  • jim_thompson5910
so you are left with 5 + 5 which turns into 10
jim_thompson5910
  • jim_thompson5910
so c^2 = 10
anonymous
  • anonymous
I'm confused. How did you get 5+5?
jim_thompson5910
  • jim_thompson5910
|dw:1362706218096:dw|
jim_thompson5910
  • jim_thompson5910
so that's how I got 2 five terms being added
anonymous
  • anonymous
wouldn't I replace C with Y?
jim_thompson5910
  • jim_thompson5910
you would
jim_thompson5910
  • jim_thompson5910
so y^2 = 10
jim_thompson5910
  • jim_thompson5910
making |dw:1362706377944:dw|
anonymous
  • anonymous
is that it for that problem?
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
yes, you found y, so you're done
anonymous
  • anonymous
Oh, I'm so sorry. but I don't get this other one. Can you help me with it? c:
jim_thompson5910
  • jim_thompson5910
sure
anonymous
  • anonymous
|dw:1362707591687:dw|
anonymous
  • anonymous
anonymous
  • anonymous
hullo?
jim_thompson5910
  • jim_thompson5910
sry was afk for a sec
jim_thompson5910
  • jim_thompson5910
were you able to get anywhere with this at all?
anonymous
  • anonymous
No. :/
anonymous
  • anonymous
I think Y = 2 sqrt 3?
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
sry i had to go afk, but I'm back
jim_thompson5910
  • jim_thompson5910
y is actually 2 the legs are not congruent
jim_thompson5910
  • jim_thompson5910
x = 2*y x = 2*2 x =4
jim_thompson5910
  • jim_thompson5910
the reason how I know y is 2 is from this template |dw:1362709753724:dw|
anonymous
  • anonymous
um, 60 degrees is down below and down below, its 2 sqrt3
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
well i flipped the triangle around, but if it were oriented the way the problem has it, then you would have |dw:1362710229246:dw|
anonymous
  • anonymous
|dw:1362710294844:dw|
anonymous
  • anonymous
you missed the 3?
anonymous
  • anonymous
I mean 2
jim_thompson5910
  • jim_thompson5910
the 2*sqrt(3) is above though isnt it?
anonymous
  • anonymous
I just showed you the diagram. :P
jim_thompson5910
  • jim_thompson5910
but in this one, the 2*sqrt(3) is above |dw:1362710579138:dw|
anonymous
  • anonymous
Are you supposed to change what I showed you?
jim_thompson5910
  • jim_thompson5910
no i just copied what you drew
anonymous
  • anonymous
you're mixing up everything I just drew and you know that. ._.
anonymous
  • anonymous
look up at the picture again..
jim_thompson5910
  • jim_thompson5910
yeah i see the new one, but you drew a totally different one
jim_thompson5910
  • jim_thompson5910
or is it a case the second one is the correct one and the first one was a typo?
anonymous
  • anonymous
OH I'm sorry!
jim_thompson5910
  • jim_thompson5910
yeah i was like...what in the world lol
anonymous
  • anonymous
that was my fault.
jim_thompson5910
  • jim_thompson5910
its fine, so which one is it?
anonymous
  • anonymous
the second one is correct.
jim_thompson5910
  • jim_thompson5910
ok one sec
anonymous
  • anonymous
k
anonymous
  • anonymous
are you gonna help? I don't have all day.
jim_thompson5910
  • jim_thompson5910
|dw:1362712584316:dw|
jim_thompson5910
  • jim_thompson5910
the value of x is exactly half that of the hypotenuse
anonymous
  • anonymous
That's correct. xP Lol.
jim_thompson5910
  • jim_thompson5910
yeah just wanted to make sure and put it up again
jim_thompson5910
  • jim_thompson5910
so x = 2*sqrt(3)/2 = sqrt(3)
jim_thompson5910
  • jim_thompson5910
which means |dw:1362712746325:dw|
anonymous
  • anonymous
wait stop
jim_thompson5910
  • jim_thompson5910
ok whats up
anonymous
  • anonymous
is it 2 times sqrt 3/2?
jim_thompson5910
  • jim_thompson5910
yes
jim_thompson5910
  • jim_thompson5910
the 2 terms cancel
anonymous
  • anonymous
so I write this down? x = 2*sqrt(3)/2 = sqrt(3)
jim_thompson5910
  • jim_thompson5910
yes and that will give you the value of x
anonymous
  • anonymous
Just to make sure we're on the same topic..|dw:1362713238955:dw|
jim_thompson5910
  • jim_thompson5910
yep
anonymous
  • anonymous
is that how it should be shown?
jim_thompson5910
  • jim_thompson5910
yes you have it correct, so now that you have x, you can find y
jim_thompson5910
  • jim_thompson5910
y = x*sqrt(3) y = sqrt(3)*sqrt(3) y = 3
anonymous
  • anonymous
so, Y= 3 and X= Sqrt 3
anonymous
  • anonymous
?
jim_thompson5910
  • jim_thompson5910
yes good
anonymous
  • anonymous
wanna help me with one more please? :D
jim_thompson5910
  • jim_thompson5910
ok one more
anonymous
  • anonymous
:) Thanks!
jim_thompson5910
  • jim_thompson5910
sure thing
anonymous
  • anonymous
|dw:1362714107413:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1362714258167:dw|
jim_thompson5910
  • jim_thompson5910
x = 2*12 x = 24 since the hypotenuse is double the shorter leg in a 30-60-90 triangle
jim_thompson5910
  • jim_thompson5910
y = 12*sqrt(3) because of the template given above
anonymous
  • anonymous
where'd you get the 2 in the first one?
jim_thompson5910
  • jim_thompson5910
in general, with any 30-60-90 triangle the hypotenuse is twice as long as the shorter leg
anonymous
  • anonymous
oh, so that's why it's 2 x 24..Okay,
jim_thompson5910
  • jim_thompson5910
2 * 12 = 24, but yeah you have it
anonymous
  • anonymous
what's Y?
jim_thompson5910
  • jim_thompson5910
y = 12*sqrt(3)
jim_thompson5910
  • jim_thompson5910
sry mixed up the 30 and the 60
jim_thompson5910
  • jim_thompson5910
|dw:1362714824453:dw|
anonymous
  • anonymous
What is the answer for Y? I got 20..:/
jim_thompson5910
  • jim_thompson5910
20 is incorect
jim_thompson5910
  • jim_thompson5910
incorrect*
jim_thompson5910
  • jim_thompson5910
|dw:1362714907609:dw|
anonymous
  • anonymous
you can't simplify it?
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
that's as simplified as it gets
anonymous
  • anonymous
how did you find y?
jim_thompson5910
  • jim_thompson5910
in the template above, do you see how the longer leg is x*sqrt(3) ?
anonymous
  • anonymous
ohhh.. now I get it.
anonymous
  • anonymous
Thankyou ! :)
jim_thompson5910
  • jim_thompson5910
ok I would write that template down, it's very handy
jim_thompson5910
  • jim_thompson5910
also the 45-45-90 template as well
anonymous
  • anonymous
thankyou for helping me
jim_thompson5910
  • jim_thompson5910
you're welcome, sry for being afk a lot lol

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