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What is the x and y equal if x - 3y equals 3?
There are an infinite number of them. Here are a few: x= -2, y= -5/3 x= -1, y= -4/3 x= 0, y= -1 x= 1, y= -2/3 x= 2, y= -1/3 x= 3, y= 0 x= 12, y= 3
If x exceeds y by 1 and y exceeds z by 3 how are x and z related?
x + 1 = y y + 3 = z z = y + 3 = (x + 1) + 3 = x + 4 Or: x = y - 1 = (z - 3) - 1 = z - 4 Which results in the same: x exceeds z by 4.
What is x when log base 3 x - 2log base x 3 equals 1?
log3x - 2logx3 = 1 so 1/logx3 - 2logx3 = 1 Let y = logx3 then 1/y - 2y = 1 multiply through by y: 1 - 2y2 = y that is, 2y2 + y - 1 = 0 which factorises to (y + 1)(2y -1) = 0 so that y = -1 or y = 1/2 y = -1 implies logx3 = -1 so that x-1 = 3 ie 1/x = 3 or x= 1/3 y = 1/2 implies logx3 = 1/2 so that x1/2 = 3 or x = 32 = 9
What are y equals x plus 2 y equals -x plus 4?
y = x + 2 y = -x + 4 x + 2 = -x + 4 2x + 2 = 4 2x = 2 x = 1 y = x + 2 y = 1 + 2 y = 3 (1, 3)
Use the graph to solve the system of linear equations x - y=4 4x + y=1?
x=4+y 4x+y=1 = 4(4+y)+y=1 16+4y+y=1 16+5y=1 5y=-15 y=-3 x=4+y = x=4+(-3) x=1
How do you solve y -x 1 if x -3?
4
Who do solve FLT some lines?
I like back into multidimensional space where were born to be sleeping forever thousands autumn. Pierre De Fermat's last theorem. The conditions. x,y,z,n are the integers >0 and n>2. z^n=/x^n+y^n. Assumptions z^3=x^3+y^3. Therefore z=(x^3+y^3)^1/3. I define . F(x,y)=(x^3+y^3}^1/3 - [ (x-x-1)^3+(y-x-1)^3]^1/3. Therefore [z-F(x,y)]^3={ (x^3+y^3)^1/3 - (x^3+y^3}^1/3 + [ (x-x-1)^3+(y-x-1)^3] ^1/3 }^3={ [(x-x-1)^3+(y-x-1)^3]^1/3 }^3 =(x-x-1)^3+(y-x-1)^3= (y-x-1)^3-1. Because [z-F(x,y)]^3=(y-x-1)^3-1 . Attention [(y-x-1)^3-1 ] is an integer , [(y-x-1)^3-1 ]^1/3 is an irrational number therefore [(y-x-1)^3-1 ]^2/3 is an irrational number too. Example (2^3-1) is an integer , (2^3-1)^1/3 is an irrational number and (2^3-1)^2/3 is an irrational number too. Because z-F(x,y)=[y-x-1)^3-1]^1/3. Therefore z=F(x,y)+[(y-x-1)^3-1] ^1/3. Therefore z^3=[F(x,y)]^3.+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+3F(x,y)*[(y-x-1)-1]^2/3+[(y-x-1)^3-1]. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3+[F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3 =0 Because. z-F(x,y)=[ (y-x-1)^3-1]^1/3. Therefore F(x,y)=z - [(y-x-1)^3-1]^1/3. Therefore 3F(x,y)*[(y-x-1)^3-1]^2/3 =3z*[(y-x-1)^3-1]^2/3 -3[(y-x-1)^3-1]. Therefore. 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+3[F(x,y)]^2*[y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3=0.. Named [F(x,y)]^3+3[F(x,y)]^2*[(y-x-1)^3-1]^1/3+[(y-x-1)^3-1] - z^3= W. We have 3z*[(y-x-1)^3-1]^2/3 is an irrational number because z is an integer and had proved [(y-x-1)^3-1]^2/3 is an irrational number. And 3[(y-x-1)^3-1] is an integer because x,y are the integers. And 3z*[(y-x-1)^3-1]^2/3 - 3[(y-x-1)^3-1]+[F(x,y)]^3+W=0. Therefore an irrational number - an integer+W=0. Therefore W is an complex irrational number. Named 3z*[(y-x-1)^3-1]^2/3 is 3z*B And Named 3[(y-x-1)^3-1] is C . Therefore 3z*B - C +W=0. Therefore 3z*B=C-W. Because z is an integer. B is an irrational number=[(y-x-1)^3-1]^2/3 Attention (an integer)^2/3 and (an integer)^2/3 is an irrational number. W is an complex irrational number. C is an integer. Therefore. an integer*an irrational number=an complex irrational number + an integer. Unreasonable. Therefore. z^3=/x^3+y^3 Similar z^n=/x^n+y^n. ISHTAR.
What is the x and y equal if x - 3y equals 3?
There are an infinite number of them. Here are a few: x= -2, y= -5/3 x= -1, y= -4/3 x= 0, y= -1 x= 1, y= -2/3 x= 2, y= -1/3 x= 3, y= 0 x= 12, y= 3
Differentiate 3 Sq root x?
y = 3√x y = 3x^(1/2) y' = 3(1/2)x^(1/2 -1) y'= (3/2)x^(-1/2) y' = 3/[2x^(1/2)] y' = 3/(2√x)
What is the second derivative of xsqrtx-1?
let y= xsqrt(x) -1 y= x^(3/2) -1 ---- since xsqrt(x) is the same as x^(3/2) y' = (3/2) x^(3/2-1) y' = (3/2) x^(1/2) y'' = (3/2) (1/2) x^(1/2-1) y'' =(3/2)(1/2) x^(-1/2) y'' = 3/4x^(1/2) y'' = 3 / 4sqrt(x)
Which line running through (3,4) is perpendicular to y=-3x+4?
The slope of y = − 3 x 4 y=−3x+4 is -3. The negative reciprocal of -3 is 1 / 3 1/3. Therefore, a line perpendicular to y = − 3 x 4 y=−3x+4 and passing through (3, 4) would have a slope of 1 / 3 1/3 and can be expressed in the form y = m x b y=mx+b where m = 1 / 3 m=1/3 and ( 3 , 4 ) (3,4) satisfies the equation. To find the equation, substitute m = 1 / 3 m=1/3 and the point ( 3 , 4 ) (3,4) into the point-slope form: y − y 1 = m ( x − x 1 ) y−y 1 =m(x−x 1 ), y − 4 = ( 1 / 3 ) ( x − 3 ) y−4=(1/3)(x−3), y − 4 = ( 1 / 3 ) x − 1 y−4=(1/3)x−1, y = ( 1 / 3 ) x 3 y=(1/3)x+3. Therefore, the line perpendicular to y = − 3 x 4 y=−3x+4 and passing through (3, 4) is y = ( 1 / 3 ) x 3 y=(1/3)x+3
What is the answer for y equals x plus 3?
there are 4 possible answers. X= 1 , Y=2 Y=1 , X=2 X=0 , Y-3 Y=0 , X=3
If x exceeds y by 1 and y exceeds z by 3 how are x and z related?
x + 1 = y y + 3 = z z = y + 3 = (x + 1) + 3 = x + 4 Or: x = y - 1 = (z - 3) - 1 = z - 4 Which results in the same: x exceeds z by 4.
What is inverse function for fx equals x divided by x minus 3?
y = x/(x - 3) so y(x - 3) = x xy - 3y = x xy - x = 3y x(y - 1) = 3y x = 3y/(y - 1) for y ≠1 So, the inverse function is f(y) = 3y/(y - 1) where y ≠1
What two numbers multiply to be 3 and add to -4?
-1
What is the derivative of 3x?
Let y = 3x y' = 3(x)' y' = 3(x1)' y' = 3[1x(1-1)] y' = 3(1x0) y' = 3(1 x 1) y' = 3 In general: y = xn y' nx(n-1)
What is the answer to 3x y equals 3 for y?
y=1 and x=1 so 3(1)(1)=3 also you could say that xy=1 and that x=y
