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If the factors of a polynomial are x-2 and x-5 what values of x make that polynomial 0?
The values that make each of the factors zero. In other words, you need to solve:x - 2 = 0 and: x - 5 = 0
What is the slope of the line x -5?
x - 5 is not an equation, it is an expression. It cannot represent a line and so there is no question of any slope. Since x is a variable, it can take on infinitely many values, thus the value of x - 5 depends on x. If we impose a variable y which will represents all the values produced by x - 5, then we form a linear equation with 2 variables, y = x - 5. As we know, this is the slope-intercept form of an equation, y = mx + b. Thus, the slope is 1.
What values of x are solutions of 3x2 11x 20?
If you mean: 3x squared -11x -20 = 0 Then: x = -5 or x = 4/3
How do you find all values of x for which the function f is increasing?
You need to find all values of x for which the first derivative of the function is greater than zero. Example: f(x)=x^2+5x+6 f'(x)=2x+5 = 0 x=-5/2 Consider intervals (minus infinity, -5/2) and (-5/2, infinity). Choose any number in each interval and examine the sign of f'(x), that is 2x+5. If it is positive, the function increases on that interval. If it is negative, the function decreases on that interval. In this case, the function increases on (minus infinity, -5/2) and decreases on (-5/2, infinity).
Which two values of x are roots of the polynomial x2 plus 3x - 5?
To find the roots of the polynomial (x^2 + 3x - 5), we need to set the polynomial equal to zero and solve for x. So, (x^2 + 3x - 5 = 0). To solve this quadratic equation, we can use the quadratic formula: (x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}), where a = 1, b = 3, and c = -5. Plugging these values into the formula, we get (x = \frac{-3 \pm \sqrt{3^2 - 41(-5)}}{2*1}), which simplifies to (x = \frac{-3 \pm \sqrt{29}}{2}). Therefore, the two values of x that are roots of the polynomial are (x = \frac{-3 + \sqrt{29}}{2}) and (x = \frac{-3 - \sqrt{29}}{2}).
