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Find the value of each variable. If your answer is not an integar, express it in simplest radical form.

Mathematics
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|dw:1362695099329:dw|
I need to know step by step on how to solve this.
@jim_thompson5910 Please help. :(

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Other answers:

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

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