Answer:
x<±1 and x< -3
Step-by-step explanation:
Given the inequality x^2-1/3x+9<0, we are to find the values of x that satisfies the inequality
x^2-1/3x+9<0
x^2-1/3x+9 (3x+9)²<0* (3x+9)²
(x^2-1)(3x+9) < 0
x²-1 < 0 and 3x + 9 <0
x²-1 < 0
x²<1
x<±√1
x<±1
Also 3x + 9 <0
3x < -9
x < -9/3
x < -3
Hence the required values of x are x<±1 and x< -3
Answer:
B
Step-by-step explanation:
4*60=240
240/13.7*25=437
<span>the sum of the first and three times the second is 54.
x + 3y = 54
3y = 54 - x
y= 54 /3 - x / 3 = 18 -x /3
</span>xy = x (18 -x /3) = 18x - x^2 / 3 = - 1/3 x ^ 2 + 18 x
<span>
</span><span>since a<0, it is concave downward, and the vertex is the maximum value. That vertex occurs at x= - b /2a = (-18) / (-2/3) = 12
</span>
<span>The first number is x=12.
The second is y = </span><span>18 -x /3 = 18- 12/ 3 = 18 - 4 = 14 </span>
