Yes, as x-y2=0
No. A simple example of this is y = x2; the inverse is x = y2, which is not a function.
Yes
x2 - y2 = (x + y)(x - y) is an identity, not a function.
No, because there is more than one solution: y2 = x2 y = ±(x2)1/2 y = ±x Because there are multiple solutions for a single value of x, this does not qualify as a function.
Yes, as x-y2=0
No. A simple example of this is y = x2; the inverse is x = y2, which is not a function.
Need two points. m = slope. (X1, Y1) and (X2, Y2) m = Y2 - Y1/X2 - X1 ==============Or, if function is in this form...... Y =mX + b ======== Read off of function, or get function is this form.
Y2 = Xtake square root each sideY = (+/-) sqrt(X)=============now it is a function as both these equation pass the vertical line test
Yes
x2 - y2 = (x + y)(x - y) is an identity, not a function.
yes no maybe so yes no maybe so
No, because there is more than one solution: y2 = x2 y = ±(x2)1/2 y = ±x Because there are multiple solutions for a single value of x, this does not qualify as a function.
The only function that can do this is f(x)=0, or y=0 What about y2-x2=0
y2 + y2 = 2y2
The formula: distance=sqrt(((x1-x2)*(x1-x2))+((y1-y2)*(y1-y2))+((z1-z2)*(z1-z2))) In DarkBASIC it's: function distance3D(x1,y1,z1,x2,y2,z2) x=x1-x2 y=y1-y2 z=z1-z2 result=sqrt((x*x)+(y*y)+(z*z)) endfunction result In classic BASIC I think it's: FUNCTION distance3D(x1,y1,z1,x2,y2,z2) x=x1-x2 y=y1-y2 z=z1-z2 result=SQRT((x*x)+(y*y)+(z*z)) RETURN result END FUNCTION
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