If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!
If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!
If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!
If you use n terms from the Taylor expansion, the absolute value of the error is less than [|x|^(2n+1)]/(2n+1)!
No.
The square root of a polynomial is another polynomial that, when multiplied by itself, yields the original polynomial. Not all polynomials have a square root that is also a polynomial; for example, the polynomial (x^2 + 1) does not have a polynomial square root in the real number system. However, some polynomials, like (x^2 - 4), have polynomial square roots, which in this case would be (x - 2) and (x + 2). Finding the square root of a polynomial can involve techniques such as factoring or using the quadratic formula for quadratic polynomials.
Try the quadratic formula. X = -b ± (sqrt(b^2-4ac)/2a)
false
false
The remainder theorem states that if you divide a polynomial function by one of it's linier factors it's degree will be decreased by one. This theorem is often used to find the imaginary zeros of polynomial functions by reducing them to quadratics at which point they can be solved by using the quadratic formula.
No.
The square root of a polynomial is another polynomial that, when multiplied by itself, yields the original polynomial. Not all polynomials have a square root that is also a polynomial; for example, the polynomial (x^2 + 1) does not have a polynomial square root in the real number system. However, some polynomials, like (x^2 - 4), have polynomial square roots, which in this case would be (x - 2) and (x + 2). Finding the square root of a polynomial can involve techniques such as factoring or using the quadratic formula for quadratic polynomials.
Try the quadratic formula. X = -b ± (sqrt(b^2-4ac)/2a)
From the Division Algorithm for Polynomials theorem,f(x) = q(x)g(x) + r(x) or we say:dividend = (quotient)(divisor) + (remainder)In our case,quotient = x^2 - 5x - 6; divisor = x - 3; and remainder = 5.Substitute what you know into the formula, and you will have:f(x) = (x^2 - 5x - 6)(x - 3) + 5f(x) = x^3 - 5x^2 - 6x - 3x^2 + 15x + 18 + 5f(x) = x^3 - 5x^2 - 3x^2 - 6x + 15x + 18 + 5f(x) = x^3 - 8x^2 + 9x + 23 (this is the required polynomial)
false
false
True
There is no element that corresponds to the abbreviation Ln. The closest approximation to this chemical formula is ZnCl2, or zinc chloride.
diameter = circumference/pi (pi = 3.14159265 or 3.14 for rough approximation).
It is 1/3, not 0.33 which is an approximation. The derivation of this formula requires knowledge of integration. For basic mathematical details follow the link below.
There is a formula for the difference of squares. In this case, the answer is (C + D)(C - D)