Assume (a,b) has a minimum element m.
m is in the interval so a < m < b.
a < m
Adding a to both sides,
2a < a + m
Adding m to both sides of the first inequality,
a + m < 2m
So
2a < a+m < 2m
a < (a+m)/2 < m < b
Since the average (a+m)/2 is in the range (a,b) and less than m, that contradicts our assumption that m is the minimum. So we conclude there is no minimum since given any purported minimum we can always compute something smaller in the range.
Answer:
Step-by-step explanation:
<u>Given</u>
- Q(s) = -4s^3 + 7s^2 - 24;
- s = -4 and 1
<u>Verifying zeroes</u>
Q(-4) =
- -4(-4)^3 + 7(-4)^2 - 24 =
- 256 + 112 - 24 =
- 344
- Incorrect as 344 ≠ 0
Q(1) =
- -4(1)^3 + 7(1)^2 - 24 =
- -12 + 7 - 24 =
- -29
- Incorrect as -29 ≠ 0
Answer:
Y = -5/7x - 6/7
Step-by-step explanation:
m = y2-y1 / x2-x1
Hope this helps. Pls give brainliest.
F(x) = 4 [cos (x)]^2 - 3 = 0
4[cos(x)]^2 = 3
cos(x) = √3 / 2
That happens in the first and fourth quadrants, for the angles 30 degrees and 330 degrees.
Answer: x = 30 degrees and x = 330 degrees
