A/Wall if you have 20 and 30 that box is missing # so you will need to times and you will get something like this 20×30=60 and that's the missing #
B/Its going to be about 768 so you need to times and if you don't get 768
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.
Step-by-step explanation:
(a) ∫₋ₒₒ°° f(x) dx
We can split this into three integrals:
= ∫₋ₒₒ⁻¹ f(x) dx + ∫₋₁¹ f(x) dx + ∫₁°° f(x) dx
Since the function is even (symmetrical about the y-axis), we can further simplify this as:
= ∫₋₁¹ f(x) dx + 2 ∫₁°° f(x) dx
The first integral is finite, so it converges.
For the second integral, we can use comparison test.
g(x) = e^(-½ x) is greater than f(x) = e^(-½ x²) for all x greater than 1.
We can show that g(x) converges:
∫₁°° e^(-½ x) dx = -2 e^(-½ x) |₁°° = -2 e^(-∞) − -2 e^(-½) = 0 + 2e^(-½).
Therefore, the smaller function f(x) also converges.
(b) The width of the intervals is:
Δx = (3 − -3) / 6 = 1
Evaluating the function at the beginning and end of each interval:
f(-3) = e^(-9/2)
f(-2) = e^(-2)
f(-1) = e^(-1/2)
f(0) = 1
f(1) = e^(-1/2)
f(2) = e^(-2)
f(3) = e^(-9/2)
Apply Simpson's rule:
S = Δx/3 [f(-3) + 4f(-2) + 2f(-1) + 4f(0) + 2f(1) + 4f(2) + f(3)]
S ≈ 2.5103
Answer:
Width = 9
Step-by-step explanation:
(P=2L+2W) = 72
L = 3W<- length is triple the width
(P = 2(3W) + 2W) = 72
P = 8W =72
W = 9
