Craig has every 13th night and Edie has every 5th night off.
You have to find LCM - the Least Common Multiple that is the smallest ("least") number that both 13 and 5 will divide into.
Since numbers 13 and 5 are both prime, then LCM(13,5)=13·5=65.
This means, they will have the same every 65th night off.
The length of the GH segment is 13
Step-by-step explanation:
For solving this problem we need to remember some of the circle corollaries-
When two-chord intersects each other, the product of the chord segments are equal
The above corollary can be easily understood by looking at a diagram attached below-
In the figure, EF and GH are two chords intersecting at K
Thus, EK*KF= GK*KH
Values of the EK, KF, GK are given as 5, 6 and 3 respectively
Substituting the values we get
5*6=3*KH
KH= 10
We know that GH= GK+KH
Thus GH= 3+10= 13
We have that
using a graph tool--------------> graph the <span> cosecant function
</span>see the attached figure
the answer is the option B
Answer:
Subtract One-half from both sides of the equation.
Divide both sides by 6/7
Multiply both sides by 7/6
Step-by-step explanation:
just took the question on ed
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.