x = -1
y = 3
3x2 - y2 = 3 (-1)2 - (3)2 = 3 (1) - 9 = -6
Set the equation equal to zero. 3x2 - x = -1 3x2 - x + 1 = 0 The equation is quadratic, but can not be factored. Use the quadratic equation.
3x2 + 27x = 30 ∴ x2 + 9x - 10 = 0 ∴ (x + 10)(x - 1) = 0 ∴ x ∈ {-10, 1}
3x2 + 22x + 7 = 3x2 +21x + x + 7 = 3x(x + 7) + (x + 7) = (x + 7)(3x + 1)
Proper form first. X2 + Y2 = 1 Y2 = 1 - X2 Y = (+/-) sqrt(1 - X2) -------------------------- zero out the X Y = (+/-) sqrt(1 - 02) Y = 1 ----------------the radius of this circle
The only answer I couldn't think of is 1*7
y2-3x2+6x+6y= 18 is in standard form. The vertex form would be (y+3)2/24 - (x-1)2/8 = 1
Set the equation equal to zero. 3x2 - x = -1 3x2 - x + 1 = 0 The equation is quadratic, but can not be factored. Use the quadratic equation.
3x2 + 12x = - 1 ie 3x2 + 12x + 1 = 0 has no rational roots. The irrational roots are [-12 +/- sqrt(132)]/6 = -3.915 and -0.085
What do you want to convert it to? x2 + y2 = 2x If you want to solve for y: x2 + y2 = 2x ∴ y2 = 2x - x2 ∴ y = (2x - x2)1/2 If you want to solve for x: x2 + y2 = 2x ∴ x2 - 2x = -y2 ∴ x2 - 2x + 1 = 1 - y2 ∴ (x - 1)2 = 1 - y2 ∴ x - 1 = ±(1 - y2)1/2 ∴ x = 1 ± (1 - y2)1/2
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.
Well 3x2 is 6 and 6 to the 2nd power is 36 and if you subtract one it is 35. you would write like this (3x2)x6-1
1 x 157, 157 x 1
3x2 + 27x = 30 ∴ x2 + 9x - 10 = 0 ∴ (x + 10)(x - 1) = 0 ∴ x ∈ {-10, 1}
1 and 3
3x2 + 22x + 7 = 3x2 +21x + x + 7 = 3x(x + 7) + (x + 7) = (x + 7)(3x + 1)
x-y = 1 => x = y+1 x2+y2 = 5 => (y+1)(x+1)+y2 = 5 2y2+2y-4 = 0 y = -2 or y = 1 So the points of intersection are: (-1, -2) and (2, 1)
Proper form first. X2 + Y2 = 1 Y2 = 1 - X2 Y = (+/-) sqrt(1 - X2) -------------------------- zero out the X Y = (+/-) sqrt(1 - 02) Y = 1 ----------------the radius of this circle