Answer: (A - B)' = {1, 2, 3, 5, 6, 7, 8, 9, 10,12, 13}
Explanation:
The roster form of a set is the elements listed inside brackets { }.
In this case it is easy to visualize A - B directly of the Venn diagram: when you subtract from A those elements that are also in B, the result is A - B = {4, 11, 14}.
Now take into account tha (A - B)' is the complement of (A - B), this is all the elements that are not in A - B.
Being A - B = {4, 11, 14} then to show (A - B)' you just have to list all the elements that are on the figure except 4, 11 and 14.
The result, in roster form, is (A - B)' = {1, 2, 3, 5, 6, 7, 8, 9, 10, 12, 13}
Answer:
7/60
Step-by-step explanation:
Because the question is phrased such that it asks "the ratio of time with father <em>to...."</em>, we know to put the time with father on top, or first. Similarly, since it is asked to put the ratio of time with the father to the time in daycare, we can put the daycare amount second/on the bottom. Therefore, we can write the ratio as
(42 minutes)/(6 hours)
To make both of these in terms of minutes to simplify it, we know that 60 minutes = 1 hour, so 1 hour / 60 minutes =1 . Since multiplying something by 1 keeps it the same, we can multiply our ratio by this amount. We want to multiply with the hour on top so the hours cancel out, as so:

To simplify this even further, we can note that 6 is a factor of both 42 and 360*, so we can multiply this fraction by (1/6) / (1/6) to get
(42/360) * (1/6)/(1/6) = 7/60 as our final answer
If this doesn't come up immediately, you can always find the greatest common factor by listing the factors of each number and finding the highest one that fits both, e.g.
for 42: 42, 21, 14, 7 , 6...
for 360: 360,180, ... 6,5....
The value of 3 in 6300<span> is </span>10<span> times the value of 3 in </span>530.
<span>It's because 3 in 530 is tens and 3 in 6300 is hundreds.</span>
Step-by-step explanation:
Consider a function
f
(
x
)
which is twice differentiable. The graph of such a function will be concave upwards in the intervals where the second derivative is positive and the graph will be concave downwards in the intervals where the second derivative is negative. To find these intervals we need to find the inflection points i.e. the x-values where the second derivative is 0.