5x times the square root of y
The square root of anything to an even power is that thing to half the power. √(5x^13) = √(x^12 5x) = √(x^12) √(5x) = x^6 √(5x) But is that really any simpler?
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "divided by", "equals". Please use "brackets" (or parentheses) because it is impossible to work out whether sqrt 5x plus 1 is sqrt (5x + 1) or sqrt(5x) + 1.
(5x - i)(5x + i) where i is the imaginary unit equivalent to the square root of negative one.
(5x - i)(5x + i) where i is the imaginary unit equivalent to the square root of negative one.
a+ square root of b has a conjugate a- square root of b and this is used rationalize the denominator when it contains a square root. If we want to multiply 5 x square root of 10 by something to get rid of the radical you can multiply it by square root of 10. But if we look at 5x( square root of 10 as ) 0+ 5x square root of 10 then the conjugate would be -5x square root of 10
5x times the square root of y
The square root of anything to an even power is that thing to half the power. √(5x^13) = √(x^12 5x) = √(x^12) √(5x) = x^6 √(5x) But is that really any simpler?
5x
Derivative with respect to 'x' of (5x)1/2 = (1/2) (5x)-1/2 (5) = 2.5/sqrt(5x)
Unfortunately, limitations of the browser used by Answers.com means that we cannot see most symbols. It is therefore impossible to give a proper answer to your question. Please resubmit your question spelling out the symbols as "plus", "minus", "times", "divided by", "equals". Please use "brackets" (or parentheses) because it is impossible to work out whether sqrt 5x plus 1 is sqrt (5x + 1) or sqrt(5x) + 1.
(5x - i)(5x + i) where i is the imaginary unit equivalent to the square root of negative one.
(5x - i)(5x + i) where i is the imaginary unit equivalent to the square root of negative one.
sqrt(20*9) = sqrt(2*2*5*9) =2*sqrt(45)
That doesn't factor neatly. Applying the quadratic equation, we find two imaginary solutions: (5 plus or minus the square root of -15) divided by 2.x = 2.5 + 1.9364916731037085ix = 2.5 - 1.9364916731037085iwhere i is the square root of negative one.
10
That doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: (-5 plus or minus the square root of -15) divided by 2.x = -2.5 + 1.9364916731037085ix = -2.5 - 1.9364916731037085iwhere i is the square root of negative one.