Standard form would be x and y on one side of the equal sign with no coefficients. x-y+6=0 in standard form would be: x-y=-6.
Not entirely sure what standard form is, but... probably y = x + 5
x + y = 3
CANONICAL FORM: In canonical form each term (minterm or maxterm) must and should contains all literals corresponding to each variable of the function. It is obtained from truth table.Ex: F(x,y,z) = xy|z + xyz + x|y|z|Ex: F(x,y) = (x+y)(x|+y)STANDARD FORM: In standard form each term doesn't contains literals corresponding to each variable of the function. that is it may contains one or two or any number of literals in each term.Ex: F(x,y,z) = x + yz + x|y|Ex: F(x,y) = (x+y)x|
The standard form is y = mx + c In this case it will be y = 3x/6 + 4/6 so that y = x/2 + 2/3
Standard form would be x and y on one side of the equal sign with no coefficients. x-y+6=0 in standard form would be: x-y=-6.
If you mean y = -x+3 then it is x+y-3 = 0 in standard form
Yes, y = mx +b is standard form and y = 4/3x - 10, is in standard form y=y m = 4/3 x=x b = - 10
Not entirely sure what standard form is, but... probably y = x + 5
The y-intercept is c in the standard form. The x-intercept is -c/m.
x + y = 3
y = x + 7
X-y=1
Standard form is 5x - 2y = -23
If you mean: y = -1-x then the straight line equation can be expressed as x+y+1 = 0
CANONICAL FORM: In canonical form each term (minterm or maxterm) must and should contains all literals corresponding to each variable of the function. It is obtained from truth table.Ex: F(x,y,z) = xy|z + xyz + x|y|z|Ex: F(x,y) = (x+y)(x|+y)STANDARD FORM: In standard form each term doesn't contains literals corresponding to each variable of the function. that is it may contains one or two or any number of literals in each term.Ex: F(x,y,z) = x + yz + x|y|Ex: F(x,y) = (x+y)x|
In the equation y x-5 2 plus 16 the standard form of the equation is 13. You find the answer to this by finding the value of X.