[tex]x+(x+1)+(x+2)\leq 12\\
3x+3\leq 12\\
3x\leq 9\\
x\leq 3\\
\boxed{\text{3,4,5\ or\ lesser}}
Answer: The only one of these that's an integer is 9.
Explanation: Let X be the integer. Then
X - 3/X = 26/3. Multiplying by 3X on both sides:
3X^2 - 9 = 26X
3X^2 - 26X - 9 = 0
X = (26 +/- SQRT(26^2-(4*(-27))))/6 = (26 +/- 28) / 6,
so the two solutions for X are 1/3 and 54/6 = 9.
The answer is #1. The mean of Team A is less than the mean of Team B, and the standard deviation of Team A is greater than the standard deviation of Team B.
Least common multiple=lcm
lcm(9,7)=63
126(8/9 + (x-4)/7 )=
126(7*8+9(x-4)/63=
126(56+9x-36)/63=
126(20+9x)/63= (126/63=2)
2(20+9x)=
40+18x
Answer: 126(8/9 + (x-4)/7 )= 40+18x
Hallo ^_^
Quadratic Equation :
ax² + bx + c = 0
From the task :
x² + 1/20 x = 1/20
Switch sides :
x² + 1/20x - 1/20 = 0
Multiple each sides by 2020(x² + 1/20x - 1/20) = 0(20)
⇒ 20x² + x - 1 = 0
From that :
a = 20, b = 1 and c = - 1
Solve x using quadratic formula :
Substitution using the formula :

Sorry for my bad english :)