y varies jointly with x and z if:
when x is held fixed, y varies with z and
when z is held fixed, y varies with x.
Bothe x and z may vary together.
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∙ 11y agoBlessie Ventigan
Blessie Ventigan
(x - y)2 - z2 is a difference of two squares (DOTS), those of (x-y) and z. So the factorisation is [(x - y) + z]*[(x - y) - z] = (x - y + z)*(x - y - z)
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PIERRE DE FERMAT's last Theorem. (x,y,z,n) belong ( N+ )^4.. n>2. (a) belong Z F is function of ( a.) F(a)=[a(a+1)/2]^2 F(0)=0 and F(-1)=0. Consider two equations F(z)=F(x)+F(y) F(z-1)=F(x-1)+F(y-1) We have a string inference F(z)=F(x)+F(y) equivalent F(z-1)=F(x-1)+F(y-1) F(z)=F(x)+F(y) infer F(z-1)=F(x-1)+F(y-1) F(z-x-1)=F(x-x-1)+F(y-x-1) infer F(z-x-2)=F(x-x-2)+F(y-x-2) we see F(z-x-1)=F(x-x-1)+F(y-x-1 ) F(z-x-1)=F(-1)+F(y-x-1 ) F(z-x-1)=0+F(y-x-1 ) give z=y and F(z-x-2)=F(x-x-2)+F(y-x-2) F(z-x-2)=F(-2)+F(y-x-2) F(z-x-2)=1+F(y-x-2) give z=/=y. So F(z-x-1)=F(x-x-1)+F(y-x-1) don't infer F(z-x-2)=F(x-x-2)+F(y-x-2) So F(z)=F(x)+F(y) don't infer F(z-1)=F(x-1)+F(y-1) So F(z)=F(x)+F(y) is not equivalent F(z-1)=F(x-1)+F(y-1) So have two cases. [F(x)+F(y)] = F(z) and F(x-1)+F(y-1)]=/=F(z-1) or vice versa So [F(x)+F(y)]-[F(x-1)+F(y-1)]=/=F(z)-F(z-1). Or F(x)-F(x-1)+F(y)-F(y-1)=/=F(z)-F(z-1). We have F(x)-F(x-1) =[x(x+1)/2]^2 - [(x-1)x/2]^2. =(x^4+2x^3+x^2/4) - (x^4-2x^3+x^2/4). =x^3. F(y)-F(y-1) =y^3. F(z)-F(z-1) =z^3. So x^3+y^3=/=z^3. n>2. .Similar. We have a string inference G(z)*F(z)=G(x)*F(x)+G(y)*F(y) equivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z)=G(x)*F(x)+G(y)*F(y) infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) we see G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=G(x)*F(-1)+G(y)*F(y-x-1 ) G(z)*F(z-x-1)=0+G(y)*F(y-x-1 ) give z=y. and G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)*F(-2)+G(y)*F(y-x-2) G(z)*F(z-x-2)=G(x)+G(y)*F(y-x-2) x>0 infer G(x)>0. give z=/=y. So G(z)*F(z-x-1)=G(x)*F(x-x-1)+G(y-x-1)*F(y) don't infer G(z)*F(z-x-2)=G(x)*F(x-x-2)+G(y)*F(y-x-2) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) don't infer G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So G(z)*F(z)=G(x)*F(x)+G(y)*F(y) is not equiivalent G(z)*F(z-1)=G(x)*F(x-1)+G(y)*F(y-1) So have two cases [G(x)*F(x)+G(y)*F(y)]=G(z)*F(z) and [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z-1)*F(z-1) or vice versa. So [G(x)*F(x)+G(y)*F(y)] - [ G(x)*F(x-1)+G(y)*F(y-1)]=/=G(z)*[F(z)-F(z-1)]. Or G(x)*[F(x) - F(x-1)] + G(y)*[F(y)-F(y-1)]=/=G(z)*[F(z)-F(z-1).] We have x^n=G(x)*[F(x)-F(x-1) ] y^n=G(y)*[F(y)-F(y-1) ] z^n=G(z)*[F(z)-F(z-1) ] So x^n+y^n=/=z^n Happy&Peace. Trần Tấn Cường.
5x3y2z3
Equality is a relationship that is REFLEXIVE: x = x SYMMETRIC: If x = y the y = x TRANSITIVE: If x = y and y = z then x = z.
135
If x = y and y = z then x = z
Commutative x + y = y + x x . y = y . x Associative x+(y+z) = (x+y)+z = x+y+z x.(y.z) = (x.y).z = x.y.z Distributive x.(y+z) = x.y + x.z (w+x)(y+z) = wy + xy + wz + xz x + xy = x x + x'y = x + y where, x & y & z are inputs.
Joint variation is a variation in which y varies jointly as x or powers of x () and y or powers of z ( ), if there is some nonzero constant k such that , where x ≠ 0, z ≠ 0, and n > 0. Source~http://www.icoachmath.com/SiteMap/JointVariation.html
There are 8 different subsets. The null set. {x} {y} {z} {x y} {x z} {y z} {x y z}
x=abs(y+z) x=+(y+z)=y+z x=-(y+z)=-y-z
As x ∝ yz2 then x = kyz2 where k is a constant. Substituting the given figures, 40 = k*20*22 = 80k Then k = ½ and the formula is, x = ½yz2 So, x = ½ * 30 *32 = 135
well on gamecube make a profile,exit,and on the main menu type in y,x,z,y,x,z,x,x,y,z,x,y for money or y,y,z,x,x,z,y,y,y,x,x,x for maximum reputation
this is very hard to write out without superscripts so it would be written out like this: y= k times x times cubed root of z if this was all reduced it would be: y=kx(z^1/3 ) i think this is right. >_< i hope this helped!
(x - y)2 - z2 is a difference of two squares (DOTS), those of (x-y) and z. So the factorisation is [(x - y) + z]*[(x - y) - z] = (x - y + z)*(x - y - z)
If x y and y z, which statement is true
xy + y = z xy = z - y (xy)/y = (z - y)/y x = (z - y)/y