Answer:
B and C
Step-by-step explanation:
We need to find which one of these numbers is a perfect square. We know 7 and 13 can be eliminated since they're prime numbers so they can't be perfect squares. B and C are both perfect squares so that's the answer.
Answer:
56 units²
Step-by-step explanation:
Each triangle has an area that is ...
... A = 1/2bh = 1/2·7·4
There are 4 such triangles, so the total area is ...
... 4A = 4(1/2)·7·4 = 2·7·4 = 56 . . . . units²
_____
An area formula customarily used when the diagonals are pependicular to each other is that the area is half the product of their lengths.
... A = (1/2)d1·d2 = (1/2)·14·8 = 56
Answer:
the y -intercpt is the 3
the slope is -3/4
Step-by-step explanation:
this is because the 3 is the starting point on the y-axis and the slope is -3/4 because you are going down 3 to the right 4 ---hope this helps
Answer:
Following are the answer to this question:
Step-by-step explanation:
For Option a:
Its optimal mixed approach for Player A's to Player A
A1 with a chance of .05 utilizing technique
Using the .60 chance strategy for A2
Use the .35 possibility strategy for A3
For Option b:
Optimal level mixed approach for team B:
Use the strategy for B1 with a probability of 50
Using the chance strategy for B2 at .50
no use strategy for B3
For Option c:
The estimated gain of Player A will be= 3.500
For Option d:
The estimated loss of Player B will be 3.500
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.
