Answer:
uhh there are a lots of odd number there
probably 3 an including the other
Step-by-step explanation:
>3
Answer:
x1, x2 = 4.74 , -2.74
Step-by-step explanation:
To find the roots of a quadratic function we have to use the bhaskara formula
ax^2 + bx + c
x^2 - 2x - 13
a = 1 b = -2 c = -13
x1 = (-b + √ b^2 - 4ac)/2a
x2 =(-b - √ b^2 - 4ac)/2a
x1 = (2 + √ (2^2 - 4 * 1 * (-13)))/2 * 1
x1 = (2 + √ (4 + 52)) / 2
x1 = (2 + √ 56 ) / 2
x1 = (2 + 7.48) / 2
x1 = 9.48 / 2
x1 = 4.74
x2 = (2 - √ (2^2 - 4 * 1 * (-13)))/2 * 1
x2 = (2 - √ (4 + 52)) / 2
x2 = (2 - √ 56 ) / 2
x2 = (2 - 7.48) / 2
x2 = -5.48 / 2
x2 = -2.74
Answer:
what grade are u
Step-by-step explanation:
so that I c could definetly answer it
![\bf \cfrac{x}{4x+x^2}\implies \cfrac{\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}}{\begin{matrix} x \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~(4+x)}\implies \cfrac{1}{4+x}\qquad \{x|x\in \mathbb{R}, x\ne -4\}](https://tex.z-dn.net/?f=%5Cbf%20%5Ccfrac%7Bx%7D%7B4x%2Bx%5E2%7D%5Cimplies%20%5Ccfrac%7B%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D%7D%7B%5Cbegin%7Bmatrix%7D%20x%20%5C%5C%5B-0.7em%5D%5Ccline%7B1-1%7D%5C%5C%5B-5pt%5D%5Cend%7Bmatrix%7D~~%284%2Bx%29%7D%5Cimplies%20%5Ccfrac%7B1%7D%7B4%2Bx%7D%5Cqquad%20%5C%7Bx%7Cx%5Cin%20%5Cmathbb%7BR%7D%2C%20x%5Cne%20-4%5C%7D)
if you're wondering about the restriction of x ≠ -4, is due to that would make the fraction with a denominator of 0 and thus undefined.
