what is the domain of g(x) equals square root of x plus 1? √(x+1) ≥ 0 x+1≥0 x≥-1 Domain: [-1,∞)
The center of the circle is at (0, 0) and its radius is the square root of 1 which is 1
sqrt(a)+sqrt(b) is different from sqrt(a+b) unless a=0 and/or b=0. *sqrt=square root of
Domian is x>-6 Range is y> or equal to 0
x²+2x-15=0 x1=-2/2 - Square root of ((2/2)²+15) x1=-1-4 x1=-5 x2=-2/2 + Square root of ((2/2)²+15) x2=-1+4 x2=3
quadratics have the form ax2+bx+c=0 the discriminant is the square root of (b2-4ac) = square root of (16-16) =square root of 0 = 0
what is the domain of g(x) equals square root of x plus 1? √(x+1) ≥ 0 x+1≥0 x≥-1 Domain: [-1,∞)
The square root of both 0 and 1 equals the square of 0 and 1
Using the quadratic equation formula: x = -3 - the square root of 3 or x = -3 + the square root of 3
y equals x plus 4 when y equals 0 then x equals 2i i is the square root of negative 1
The answer is 0.
x=1/2(-1-square root of 21) and x=1/2(square root of 21 -1)
If 2x2 + 3x + 9 = 0 then x = -0.75 - 1.984i or -0.75 + 1.984i where i is the imaginary square root of -1.
x^2 plus 4 = 0x^2 = -4square root both sidesx = the square root of -4x = 2i
It is sqrt(0), which equals 0.
The center of the circle is at (0, 0) and its radius is the square root of 1 which is 1
sqrt(a)+sqrt(b) is different from sqrt(a+b) unless a=0 and/or b=0. *sqrt=square root of