There are an infinite number of combinations for x, y, and z in this case. Just assign any number for y and for z, then calculate the value for x.
There are an infinite number of combinations for x, y, and z in this case. Just assign any number for y and for z, then calculate the value for x.
There are an infinite number of combinations for x, y, and z in this case. Just assign any number for y and for z, then calculate the value for x.
There are an infinite number of combinations for x, y, and z in this case. Just assign any number for y and for z, then calculate the value for x.
x equals -12 and y equals 1/4 of -12, so y = -3.
y = x/k so k = 1/25; when x = 1, x/k, hence y, = 25
x+y=1 x=1-y or y=1-x or if you fill that in 1-y+y=1 1 = 1 or 1-x+x = 1 1 = 1
X = 2 and Y = 3.
y equals x plus 4 when y equals 0 then x equals 2i i is the square root of negative 1
x - y = xydifferentiating wrt x1 - (dy/dx) = x(dy/dx) + y(x + 1)(dy/dx) + y + 1 = 0
It is linear. The highest power is 1 (x = x1, y = y1) so it is linear.
If [ y = x + 2 ], then x is not -1 when y = 5.If [ y = x + 2 ],then when x = -1, y = 1,and when y = 5, x = 3.
In the equations Y=X-1 and Y=-X+1, the solution is (1,0)
x1:y1 = x2:y2 4:-2 = x2:5 x2 = (4*5)/-2 x2 = -10
solution: y = 0 x = -1
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
The solution is: x = 1 and y = -1
y = 6*x1/2 - 3x dy/dx = 6*(1/2)*x-1/2 - 3 = 3x-1/2 - 3 = 3/sqrt(x) - 3 or 3[1/sqrt(x) - 1]
Yes, y=x-4 is linear, since the highest power on x is 1 (y = x1-4)
xy = x ÷x y = 1
When x = 1 then y = 4