Yes, they can both be zero.
There are two simple strategies to compare two numbers, X and Y: you can calculate X - Y. If X - Y > 0 then X > Y, if X - Y = 0 then X = Y and if X - Y <0 then X < Y. Or you can calculate X/Y. Provided both are greater than 0, if X/Y > 0 then X > Y if X/Y = 1 then X = Y if X/Y <1 then X < Y
y = x - 1 y = -x + 1 (add both the equations) 2y = 0 y = 0 when y = 0, x = 1 So that the only solution is the intersection point of the lines, (1, 0).
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It is difficult to tell what expression you are trying to convey. 8 multiplied by x, and then take y away, is 8x-y x take away y, and then 8 multiplied by this number, is 8(x-y) 8 multiplied by x, and divided by y is 8x/y
Given an improper fraction of the from x/y where x and y are integers and x>=y>0,let x = q*y + r [q is the quotient and r the remainder when x is divided by y].Then x/y = (q*y + r)/y = q + r/yFor mathematically inclined people who wish to cover all bases for the assumptions that x>=y>0If y is negative, then the numerator and denominator of the fraction can be multiplied by -1 to give an equivalent fraction; except that now y > 0.If 0 < x < y then q = 0 so that x/y is a proper fraction and so there is no mixed number.If x < 0 then swap with -x (which will be >0), calculate the mixed number and put a negative sign in front of the answer.
if a column vector such as x y is multiplied by a raw vector such as ( 2 0), ( 2 o) x y = 2x so 2x is the image of x y
There are two simple strategies to compare two numbers, X and Y: you can calculate X - Y. If X - Y > 0 then X > Y, if X - Y = 0 then X = Y and if X - Y <0 then X < Y. Or you can calculate X/Y. Provided both are greater than 0, if X/Y > 0 then X > Y if X/Y = 1 then X = Y if X/Y <1 then X < Y
Given the linear equation 3x - 2y^6 = 0, the x and y intercepts are found by replacing the x and y with 0. This gives the intercepts of x and y where both = 0.
y = x - 1 y = -x + 1 (add both the equations) 2y = 0 y = 0 when y = 0, x = 1 So that the only solution is the intersection point of the lines, (1, 0).
-77
-575
It is difficult to tell what expression you are trying to convey. 8 multiplied by x, and then take y away, is 8x-y x take away y, and then 8 multiplied by this number, is 8(x-y) 8 multiplied by x, and divided by y is 8x/y
2 multiplied by x multiplied by y This problem cannot be solves because you do not know the values of x and y.
2y - 4x = -12 Algebraically rearrange to 2y = 4x - 12 Divide both sides by '2' y = 2x - 6 So at the 'y' intercept, x = 0 Substituting x = 0 We have y = 2(0) - 6 Anything multiplied '0' is equal; to '0' Hence y = 0 - 6 y = -6 is the value of the y-intercept.
x- and y-intercepts are point that lie on the axes, thus x-intercept point is (x, 0) and y-intercept point is (0, y). So let y = 0. replace 0 for y into the equation and solve for x. 2(0) - 12x = -6 -12x = -6 divide by -12 both sides x = 1/2 Thus, x-intercept point is (1/2, 0). Let x = 0. replace 0 for x into the equation and solve for y. 2y - 12(0) = -6 2y = -6 divide by 2 both sides x = -3 Thus, y-intercept point is (0, -3).
Given an improper fraction of the from x/y where x and y are integers and x>=y>0,let x = q*y + r [q is the quotient and r the remainder when x is divided by y].Then x/y = (q*y + r)/y = q + r/yFor mathematically inclined people who wish to cover all bases for the assumptions that x>=y>0If y is negative, then the numerator and denominator of the fraction can be multiplied by -1 to give an equivalent fraction; except that now y > 0.If 0 < x < y then q = 0 so that x/y is a proper fraction and so there is no mixed number.If x < 0 then swap with -x (which will be >0), calculate the mixed number and put a negative sign in front of the answer.
x-y=0 x=y so you will use that in the other equation by substituting every y with x 2x+y=0 2x+x=0 3x=0 x=0/3 x=0 then use that in the previous equation by substituting every x with 0 x=y 0=y; y=0 finally x=0 and y=0