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hybrik
 one year ago
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)
hybrik
 one year ago
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)

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hybrik
 one year ago
Best ResponseYou've already chosen the best response.1in other terms: \[x \sqrt{y}=108, y \sqrt{z}=80\sqrt{3}, and, z \sqrt{x}=225\sqrt{3}\]

undeadknight26
 one year ago
Best ResponseYou've already chosen the best response.0This is mind boggling o.o

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1I got 6 minutes then i have lunch.

undeadknight26
 one year ago
Best ResponseYou've already chosen the best response.0Getting confused from just looking at it.

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1its a algebra 2 question i believe

undeadknight26
 one year ago
Best ResponseYou've already chosen the best response.0You pretty much have to do them all at once...And the x(sqrt y) thing confuses me.

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1i got lunch, i was thinking to square the whole equations

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[x \sqrt{y}=108 \\ y \sqrt{z}=80 \sqrt{3} \\ z \sqrt{x}=225 \sqrt{3}\] this is right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4and is x and y and z integers are real?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4and is x and y and z integers or real? *

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1I just want to learn how to do these questions, because I like to learn.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I understand that but I was asking if you knew anything about the variables

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I guess we will assume they are real

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[x^2 y=108^2 \\ y^2 z=80^2(3) \\ z^2 x=225^2 (3) \\ \text{ here I just squared both sides of equation equation } \] \[y=\frac{108^2}{x^2} \text{ solved first equation for } y \\ \text{ replacing } y \text{ \in second equation with this } \\ (\frac{108^2}{x^2})^2z=80^2(3) \\ \text{ by law of exponents you have } \frac{108^4}{x^4} z=80^2 (3) \\ \text{ multiplying both sides by } \frac{x^4}{108^4} \\ z=\frac{80^2(3)x^4}{108^4} \\ \text{ so replacing the } z \text{ \in the last equation with this } \\ (\frac{80^2(3)x^4}{108^4})^2x=225^2(3) \text{ you can solve this one for } x \text{ now }\] 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
 one year ago
Best ResponseYou've already chosen the best response.1This is taking a bit of time, need 2 minutes.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4@ganeshie8 do you see any kind of short cut here

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[z=\frac{80^2 (3)x^4}{108^4}\]

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1x+y+z = 42+\[5\sqrt{3}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[y=\frac{108^2}{x^2} \\ y=\frac{108^2}{(27)^2}=16 \\ z=\frac{80^2(3)x^4}{108^4} \\ z=\frac{80^2(3)(27)^4}{108^4}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so i guess they were integers after all

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1What if they weren't what would you do?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4isn't 27+16 equal to 43?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4well we just assumed they were real numbers in the beginning

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1So thats why I got a 145/150 on this Mu Alpha contest

freckles
 one year ago
Best ResponseYou've already chosen the best response.4that includes the integers and nonintegers

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I was thinking that there should be a short cut for some reason but I do not see one

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Two awards, [Fan medal], https://www.youtube.com/watch?v=jHmiNEnIgE

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1One more question maybe?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4ok I will try to answer

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Given 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
 one year ago
Best ResponseYou've already chosen the best response.4\[a^2_1+a^2_2+a^2_3=7 \\ \text{ find max of } 3a_1+2a_2+a_3\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I'm thinking that we can try to use that one thing it starts with an L lebniz?

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Leibniz, I am in 6th grade, I dont know this stuff well, explian more of it.

freckles
 one year ago
Best ResponseYou've already chosen the best response.4oh I was thinking of a calculus thing

freckles
 one year ago
Best ResponseYou've already chosen the best response.4i'm so dumb it was another l word

freckles
 one year ago
Best ResponseYou've already chosen the best response.4http://tutorial.math.lamar.edu/Classes/CalcIII/LagrangeMultipliers.aspx this is what it is

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[g(z,y,z)=x^2+y^2+z^2=7 \\ f(x,y,z)=3x+2y+z \\ f_x=3 \\ f_y=2 \\ f_z=1 \\ g_x=2x \\ g_y=2y \\ g_z=2z \\ \text{ so we have the equations } \\ 3= \lambda 2x \\ 2= \lambda 2y \\ 1 =\lambda 2z\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4we need to solve that bottom system for (x,y,z)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4and we also have that first equation

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[3=\lambda 2x \\ 2=\lambda 2y \\ 1=\lambda 2z \\ x^2+y^2+z^2=7\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4well 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
 one year ago
Best ResponseYou've already chosen the best response.1I honestly don't know what that is :(, I read it but it doesnt make my brain into order

freckles
 one year ago
Best ResponseYou've already chosen the best response.4so maybe we need another way to solve besides lagrange multiplers then

freckles
 one year ago
Best ResponseYou've already chosen the best response.4that is the only one I have unfortunately

freckles
 one year ago
Best ResponseYou've already chosen the best response.4that 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
 one year ago
Best ResponseYou've already chosen the best response.1It is hard to learn pre cal in 6th grade :(

freckles
 one year ago
Best ResponseYou've already chosen the best response.4yeah I think lagrange multipler is a cal 3 method

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Let x and y be real numbers satisfying 2/x = y/3 = x/y. Determine x^3

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Im assuming 0, but that just seems wrong

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\] maybe we can play with these equations

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0\[x^3+y^3+z^3=(x+y+z)^23(x^2y+y^2x+y^2z+z^2y+z^2x+x^2z)9xyz\]

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1@imqwerty the question i just asked is that the simplified part of it or the one before ">_<"

imqwerty
 one year ago
Best ResponseYou've already chosen the best response.0is there any way to find\[x^3+y^3+z^3\]?

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1@freckles Where did you get the values to equal from (6)

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[6=xy \\ 2y=x^2 \ \text{ note multiply both sides by } x \text{ and then use first equation }\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\frac{2}{x}=\frac{y}{3} \\ \text{ multiply both sides by } 3 \\ \text{ multiply both sides by } x \\ \text{ this is how I got } 6=xy \] Is this the equation with 6 you are talking about @hybrik

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Or did you cross multiply 2 * 3 and x*y

freckles
 one year ago
Best ResponseYou've already chosen the best response.4that is what some people like to call it

freckles
 one year ago
Best ResponseYou've already chosen the best response.4that is what I did for each pair of fractions that were equal

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I only needed the first two equations though

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\ \] 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
 one year ago
Best ResponseYou've already chosen the best response.1Im just going with my method, proceed your explanation

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you just replace xy with 6

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1Huh? I dont remember 2xy=x^3?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4did you read anything I said?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4that isn't meant as a rude question by the way

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[\frac{2}{x}=\frac{y}{3} \implies 6=xy \\ \frac{2}{x}=\frac{x}{y} \implies 2y=x^2 \\ \\ \frac{y}{3}=\frac{x}{y} \implies y^2=3x\] So you see the second equation? 2y=x^2 ?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4I multiplied both sides by x 2y(x)=x^2(x) 2xy=x^3

freckles
 one year ago
Best ResponseYou've already chosen the best response.4the first equation says xy=6

freckles
 one year ago
Best ResponseYou've already chosen the best response.4replace xy with 6 so you have 2(6)=x^3

hybrik
 one year ago
Best ResponseYou've already chosen the best response.112 = x^3, x = 2 radical 3

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you don't need to find x

freckles
 one year ago
Best ResponseYou've already chosen the best response.4\[x^3=12 \text{ also this doesn't give a solution of } x=2 \sqrt{3}\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.4you do have to show work on these right?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4lol what does that mean?

freckles
 one year ago
Best ResponseYou've already chosen the best response.4since 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
 one year ago
Best ResponseYou've already chosen the best response.4oops 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

freckles
 one year ago
Best ResponseYou've already chosen the best response.4what kind of face is that? an agreeing face? or a " I don't know" face?

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1its a 3 and a type of colon 3;

freckles
 one year ago
Best ResponseYou've already chosen the best response.4well I might go to sleep now then goodnight

hybrik
 one year ago
Best ResponseYou've already chosen the best response.1me too 1:24 am is hurting me head

phi
 one year ago
Best ResponseYou've already chosen the best response.1Find 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
 one year ago
Best ResponseYou've already chosen the best response.1I can see what you did, very easy. @phi
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