Answer:
The second one:
L∪J={i,j,k,l,m,n,o}
Step-by-step explanation:
The union is the elements listed in either set.
So since l,m,n, and o are elements of set L, they will also be elements of whatever it is "unioned" with.
Since i,j,k,l and m are elements of set J, they will also be elements of whatever it is "unioned" with.
When you write the union, just be sure to include each element that occurs in either set once.
So the union of L and J is {i,j,k,l,m,n,o}.
The answer is the second one.
The intersection would actually be that upside down U thing, the ∩ symbol. The intersection of two sets is a list of elements that both sets include. So here the intersection would just consist of the elements l amd m.
Answer:
a2 - 3ab + b2
Step-by-step explanation:
just saying the 2s are squared :)
Step-by-step explanation:
12 + 16 ÷ 4 × 5 - 8
12 + 4 × 5 - 8
12 + 20 - 8
32 - 8
24
<span>Benchmark fractions are created when we make two different fractions have the same numerator or denominator. In thisi case, we will produce the same denominator. The LCM of 10 and 12 is 60. So the fractions become 42/60 and 25/60. Since the fraction with the larger numerator has a larger value when the denominators are the same, 7/10 is larger than 5/12.</span>
Answer:
C (the third table, from the second picture).
Step-by-step explanation:
First, we need to find the slope of the graph.
Two conspicuous points are (0, -3) and (2, 1).
The slope is: (1 - -3) / (2 - 0) = (1 + 3) / 2 = 4 / 2 = 2.
A: In the table, the y-values increase by 2, while the x-values increase by 4. 2 / 4 = 1 / 2 = 0.5. The slope is not the same as the graph.
B: In the table, the y-values decrease by 2, while the x-values increase by 4. -2 / 4 = -1 / 2 = -0.5. The slope is not the same as the graph.
C: In the table, the y-values increase by 4, while the x-values increase by 2. 4 / 2 = 2 / 1 = 2. The slope is the same as the graph, so C is your answer.
D: In the table, the y-values decrease by 4, while the x-values increase by 2. -4 / 2 = -2 / 1 = -2. The slope is not the same as the graph.
Hope this helps!
