hybrik
  • hybrik
Find x+y+z if x(square root of y)=108 y(square root of z)=80(square root of 3) z(square root of x)=225(square root of 3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
hybrik
  • hybrik
in other terms: \[x \sqrt{y}=108, y \sqrt{z}=80\sqrt{3}, and, z \sqrt{x}=225\sqrt{3}\]
hybrik
  • hybrik
@ganeshie8
hybrik
  • hybrik
@Hero

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hybrik
  • hybrik
@welshfella
hybrik
  • hybrik
any ideas?
hybrik
  • hybrik
@undeadknight26
undeadknight26
  • undeadknight26
This is mind boggling o.o
hybrik
  • hybrik
I got 6 minutes then i have lunch.
undeadknight26
  • undeadknight26
Getting confused from just looking at it.
hybrik
  • hybrik
its a algebra 2 question i believe
undeadknight26
  • undeadknight26
You pretty much have to do them all at once...And the x(sqrt y) thing confuses me.
hybrik
  • hybrik
i got lunch, i was thinking to square the whole equations
hybrik
  • hybrik
any ideas?
freckles
  • freckles
\[x \sqrt{y}=108 \\ y \sqrt{z}=80 \sqrt{3} \\ z \sqrt{x}=225 \sqrt{3}\] this is right?
hybrik
  • hybrik
ye
freckles
  • freckles
and is x and y and z integers are real?
hybrik
  • hybrik
I don't know :|
freckles
  • freckles
and is x and y and z integers or real? *
hybrik
  • hybrik
I just want to learn how to do these questions, because I like to learn.
freckles
  • freckles
I understand that but I was asking if you knew anything about the variables
freckles
  • freckles
I guess we will assume they are real
hybrik
  • hybrik
ok
freckles
  • freckles
\[x^2 y=108^2 \\ y^2 z=80^2(3) \\ z^2 x=225^2 (3) \\ \text{ here I just squared both sides of equation equation } \] \[y=\frac{108^2}{x^2} \text{ solved first equation for } y \\ \text{ replacing } y \text{ \in second equation with this } \\ (\frac{108^2}{x^2})^2z=80^2(3) \\ \text{ by law of exponents you have } \frac{108^4}{x^4} z=80^2 (3) \\ \text{ multiplying both sides by } \frac{x^4}{108^4} \\ z=\frac{80^2(3)x^4}{108^4} \\ \text{ so replacing the } z \text{ \in the last equation with this } \\ (\frac{80^2(3)x^4}{108^4})^2x=225^2(3) \text{ you can solve this one for } x \text{ now }\] this seems like a pretty long way there might be a shorter way but you can solve the last equation for x then go back and find z and then go back and find y then add them up to find x+y+z
hybrik
  • hybrik
This is taking a bit of time, need 2 minutes.
freckles
  • freckles
@ganeshie8 do you see any kind of short cut here
hybrik
  • hybrik
x=27
hybrik
  • hybrik
y=16
freckles
  • freckles
cool I got x=27 too
hybrik
  • hybrik
Can't find Z though..
hybrik
  • hybrik
5 radical 3
freckles
  • freckles
\[z=\frac{80^2 (3)x^4}{108^4}\]
hybrik
  • hybrik
x+y+z = 42+\[5\sqrt{3}\]
freckles
  • freckles
\[y=\frac{108^2}{x^2} \\ y=\frac{108^2}{(27)^2}=16 \\ z=\frac{80^2(3)x^4}{108^4} \\ z=\frac{80^2(3)(27)^4}{108^4}\]
hybrik
  • hybrik
o
hybrik
  • hybrik
75
freckles
  • freckles
cool stuff
freckles
  • freckles
x+y+z=27+16+75
hybrik
  • hybrik
42+75=127
freckles
  • freckles
so i guess they were integers after all
hybrik
  • hybrik
What if they weren't what would you do?
freckles
  • freckles
isn't 27+16 equal to 43?
freckles
  • freckles
and isn't 43+75=118
hybrik
  • hybrik
bad math confirmed.
freckles
  • freckles
well we just assumed they were real numbers in the beginning
hybrik
  • hybrik
So thats why I got a 145/150 on this Mu Alpha contest
freckles
  • freckles
that includes the integers and non-integers
freckles
  • freckles
I was thinking that there should be a short cut for some reason but I do not see one
hybrik
  • hybrik
Two awards, [Fan medal], https://www.youtube.com/watch?v=jH-miNEnIgE
hybrik
  • hybrik
One more question maybe?
freckles
  • freckles
ok I will try to answer
hybrik
  • hybrik
Given that a^2 sub 1 + a^2 sub 2 + a^2 sub 3= 7, find the maximum possible value of 3a sub 1 + 2a sub 2 + a sub 3
freckles
  • freckles
\[a^2_1+a^2_2+a^2_3=7 \\ \text{ find max of } 3a_1+2a_2+a_3\]
hybrik
  • hybrik
yeah
freckles
  • freckles
I'm thinking that we can try to use that one thing it starts with an L lebniz?
hybrik
  • hybrik
Leibniz, I am in 6th grade, I dont know this stuff well, explian more of it.
freckles
  • freckles
oh I was thinking of a calculus thing
hybrik
  • hybrik
I do pre calculus
freckles
  • freckles
i'm so dumb it was another l word
freckles
  • freckles
lagrange multiplers
hybrik
  • hybrik
What is that exactly?
freckles
  • freckles
http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx this is what it is
hybrik
  • hybrik
K then?
freckles
  • freckles
\[g(z,y,z)=x^2+y^2+z^2=7 \\ f(x,y,z)=3x+2y+z \\ f_x=3 \\ f_y=2 \\ f_z=1 \\ g_x=2x \\ g_y=2y \\ g_z=2z \\ \text{ so we have the equations } \\ 3= \lambda 2x \\ 2= \lambda 2y \\ 1 =\lambda 2z\]
freckles
  • freckles
we need to solve that bottom system for (x,y,z)
freckles
  • freckles
and we also have that first equation
freckles
  • freckles
\[3=\lambda 2x \\ 2=\lambda 2y \\ 1=\lambda 2z \\ x^2+y^2+z^2=7\]
hybrik
  • hybrik
3x+2y+z = what number?
freckles
  • freckles
well if you can solve that system of equations above for (x,y,z) you can find out you might need to find lambda first though
hybrik
  • hybrik
I honestly don't know what that is :(, I read it but it doesnt make my brain into order
freckles
  • freckles
so maybe we need another way to solve besides lagrange multiplers then
freckles
  • freckles
any ideas out there
freckles
  • freckles
that is the only one I have unfortunately
freckles
  • freckles
that lagrange multipler way is pretty cute I found lambda and it wasn't too bad it was just a quadratic equation and then as result I was able to find x,y, and z :(
hybrik
  • hybrik
It is hard to learn pre cal in 6th grade :(
freckles
  • freckles
yeah I think lagrange multipler is a cal 3 method
freckles
  • freckles
or cal 2
hybrik
  • hybrik
:_:
hybrik
  • hybrik
I have, one more >_<
hybrik
  • hybrik
Let x and y be real numbers satisfying 2/x = y/3 = x/y. Determine x^3
hybrik
  • hybrik
Im assuming 0, but that just seems wrong
freckles
  • freckles
\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\] maybe we can play with these equations
imqwerty
  • imqwerty
\[x^3+y^3+z^3=(x+y+z)^2-3(x^2y+y^2x+y^2z+z^2y+z^2x+x^2z)-9xyz\]
hybrik
  • hybrik
@imqwerty the question i just asked is that the simplified part of it or the one before ">_<"
imqwerty
  • imqwerty
is there any way to find\[x^3+y^3+z^3\]?
hybrik
  • hybrik
@freckles Where did you get the values to equal from (6)
freckles
  • freckles
\[6=xy \\ 2y=x^2 \ \text{ note multiply both sides by } x \text{ and then use first equation }\]
imqwerty
  • imqwerty
:O ok :)
freckles
  • freckles
\[\frac{2}{x}=\frac{y}{3} \\ \text{ multiply both sides by } 3 \\ \text{ multiply both sides by } x \\ \text{ this is how I got } 6=xy \] Is this the equation with 6 you are talking about @hybrik
hybrik
  • hybrik
Or did you cross multiply 2 * 3 and x*y
freckles
  • freckles
that is what some people like to call it
freckles
  • freckles
that is what I did for each pair of fractions that were equal
freckles
  • freckles
I only needed the first two equations though
freckles
  • freckles
\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\ \] Do you see multiplying the second one on both sides by x gives 2xy=x^3 and you see that xy is given as 6 in the first equation?
hybrik
  • hybrik
Im just going with my method, proceed your explanation
freckles
  • freckles
my method is done :p
freckles
  • freckles
you just replace xy with 6
freckles
  • freckles
2(6)=x^3
hybrik
  • hybrik
Huh? I dont remember 2xy=x^3?
freckles
  • freckles
did you read anything I said?
freckles
  • freckles
that isn't meant as a rude question by the way
freckles
  • freckles
\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\] So you see the second equation? 2y=x^2 ?
freckles
  • freckles
I multiplied both sides by x 2y(x)=x^2(x) 2xy=x^3
freckles
  • freckles
the first equation says xy=6
freckles
  • freckles
replace xy with 6 so you have 2(6)=x^3
hybrik
  • hybrik
12 = x^3, x = 2 radical 3
freckles
  • freckles
you don't need to find x
hybrik
  • hybrik
ok
freckles
  • freckles
\[x^3=12 \text{ also this doesn't give a solution of } x=2 \sqrt{3}\]
hybrik
  • hybrik
ik that was x^2 :/
freckles
  • freckles
you do have to show work on these right?
hybrik
  • hybrik
i guess
freckles
  • freckles
lol what does that mean?
hybrik
  • hybrik
yes 3:
freckles
  • freckles
since you have to show work, I will show you the answer for the second one under one condition you come back and show me your method that you were suppose to use. ok how about this then... I show you the lagrange multipler way on that second one and show you my result... then you come up with the right method for your level and compare our answers
freckles
  • freckles
oops I didn't realize I was repetitive in that paragraph I was doing some cutting and pasting and rearranging and forgot I said some stuff when I did :p
hybrik
  • hybrik
3;
freckles
  • freckles
what kind of face is that? an agreeing face? or a " I don't know" face?
hybrik
  • hybrik
its a 3 and a type of colon 3;
freckles
  • freckles
well I might go to sleep now then goodnight
hybrik
  • hybrik
me too 1:24 am is hurting me head
phi
  • phi
Find x+y+z if \[ x\sqrt{y}=108 \\ y\sqrt{z}=80\sqrt{3} \\ z\sqrt{x}=225\sqrt{3}\] I would solve it this way: A \(y\sqrt{z}=2^4 5^1\sqrt{3} \) z has exactly one factor of 3 B \( z\sqrt{x}=3^2 5^2 \sqrt{3}\) z has exactly one factor of 3 , so \( \sqrt{x}\) must contain \( 3\sqrt{3}\) and x has the factor 9*3=27. Also, 2 is not a factor of x C \( x\sqrt{y}=2^2 3^3\) we know from B, 2 is not a factor of x, but 27 is. That leaves \( \sqrt{y}=4 \) and y=16 D \( z\sqrt{x}=3^2 5^2 \sqrt{3} \) with x=27 , we find z= 75 x+y+z= 27+16+75= 118
hybrik
  • hybrik
I can see what you did, very easy. @phi

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