9514 1404 393
Answer:
(0, π/4] ∪ (π, 5π/4]
Step-by-step explanation:
Multiplying by 2 gives ...
cot(x) ≥ 1
The cotangent function decreases from ∞ to 1 in the domain (0, π/4], and again in the domain (π, 5π/4]. The solution is the union of these two intervals.
x ∈ (0, π/4] ∪ (π, 5π/4]
_____
(a, b] is interval notation for a < x ≤ b
Answer:
D
Step-by-step explanation:
Answer:
The parenthesis need to be kept intact while applying the DeMorgan's theorem on the original equation to find the compliment because otherwise it will introduce an error in the answer.
Step-by-step explanation:
According to DeMorgan's Theorem:
(W.X + Y.Z)'
(W.X)' . (Y.Z)'
(W'+X') . (Y' + Z')
Note that it is important to keep the parenthesis intact while applying the DeMorgan's theorem.
For the original function:
(W . X + Y . Z)'
= (1 . 1 + 1 . 0)
= (1 + 0) = 1
For the compliment:
(W' + X') . (Y' + Z')
=(1' + 1') . (1' + 0')
=(0 + 0) . (0 + 1)
=0 . 1 = 0
Both functions are not 1 for the same input if we solve while keeping the parenthesis intact because that allows us to solve the operation inside the parenthesis first and then move on to the operator outside it.
Without the parenthesis the compliment equation looks like this:
W' + X' . Y' + Z'
1' + 1' . 1' + 0'
0 + 0 . 0 + 1
Here, the 'AND' operation will be considered first before the 'OR', resulting in 1 as the final answer.
Therefore, it is important to keep the parenthesis intact while applying DeMorgan's Theorem on the original equation or else it would produce an erroneous result.
Answer:
Step-by-step explanation:
1) a very poor question indeed. It has at least two possible solutions
c = 3x + 2 or c = 3(x + 2)
when x = 2
c = 3(2) + 2 = 8 or c = 3(2 + 2) = 12
when x = 5
c = 3(5) + 2 = 17 or c = 3(5 + 2) = 21
2)
y = x + 2
when x = 2
y = 2 + 2 = 4
when x = 5
y = 5 + 2 = 7
Answer:
-1.5
Step-by-step explanation:
