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What is the derivative of y equals square root of x when x is positive?
y = x^(1/2) (NB power of '1/2' mean the 'square root'. Hence dy/dx = (1/2)x^(-1/2) or dy/dx = 1/ [2x^(1/2)]
X2-6x plus 1 equals 0?
x2 - 6x + 1 = 0 x = [-(-6) +&- square root of [(-6)2 - 4(1)(1)]]/2(1) x = (6 +&- square root of 32)/2 x = [6 +&- 4(square root of 2)]/2 x = 3 +&- 2(square root of 2) x = 3 + 2(square root of 2) or x = 3 - 2(square root of 2) Check:
What is x2 plus x-5 equals 0?
x=1/2(-1-square root of 21) and x=1/2(square root of 21 -1)
How do you differentiate square root x?
The derivative of ( x1/2 ) with respect to 'x' is [ 1/2 x-1/2 ], or 1/[2sqrt(x)] .
What number do you add to the polynomial x2 plus x to complete the square?
X^2 + X = 0 halve the linear term (1) and square it then add to both sides X^2 + X + 1/4 = 1/4 factor left; gather terms right (X + 1/2)^2 = 1/4 (X + 1/2)^2 - 1/4 = 0 (-1/2,-1/4) vector and the number 1/4 was added to both sides completing the square
