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2 years ago

If set A contains seven distinct numbers and set B contains three distinct letters, how many elements are in (A U B)?

Mathematics

1 answer:

Ira Lisetskai [31]2 years ago

4 0

<h3>Answer: 10 elements</h3>

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Explanation:

Let's use an example to see why this is

A = {1,2,3,4,5,6,7}

B = {x,y,z}

A U B = {1,2,3,4,5,6,7, x,y,z}

We simply glue the two sets together to form one single big set. Because sets A and B have no overlap, this means we dont have to worry about tossing out duplicates (since there aren't any).

There are 7 items in set A, and 3 items in set B.

Therefore, we have 7+3 = <u>10 items in set A U B</u>

The symbol U refers to the set union operator.

If you wanted to make a Venn Diagram, then you'd have two circles overlapping. In circle A will be the numbers 1 through 7. These values are not in the overlapped region. For circle B, the letters x,y,z are inside but not in the overlapped region. The overlapped region remains empty.

The set A U B talks about the collection of everything mentioned. It's the set of stuff in set A or in set B or in both sets.

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If you want a formula to write down on a notecard or reference sheet, then it would be

n(A U B) = n(A)+n(B)

where A and B are mutually exclusive, aka disjoint. This means they have nothing in the overlapped region

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a) $\sqrt{144-288t+208t^2}$ b.) -12knots, 8 knots c) No e) $4\sqrt{13}$

Step-by-step explanation:

We know that the initial distance between ships A and B was 12 nautical miles. Ship A moves at 12 knots(nautical miles per hour) south. Ship B moves at 8 knots east.

We know that at time t , the ship A has moved $12\dot t (n.m)$ and ship B has moved $8\dot t (n.m)$ . We also know that the ship A moves closer to the line of the movement of B and that ship B moves further on its line.

Using Pythagorean theorem, we can write the distance s as:

$\sqrt{(12-12\dot t)^2 + (8\dot t)^2}\\s=\sqrt{144-288t+144t^2+64t^2}\\s=\sqrt{144-288t+208t^2}$

We want to find $\frac{ds}{dt}$ for t=0 and t=1

$\sqrt{144-288t+208t^2}|\frac{d}{dt}\\\\\frac{ds}{dt}=\frac{1}{2\sqrt{144-288t+208t^2}}\dot (-288+416t)\\\\\frac{ds}{dt}=\frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\\frac{ds}{dt}(0)=\frac{208\dot 0-144}{\sqrt{144-288\dot 0 + 209\dot 0^2}}=-12knots\\\\\frac{ds}{dt}(1)=\frac{208\dot 1-144}{\sqrt{144-288\dot 1 + 209\dot 1^2}}=8knots$

We know that the visibility was 5n.m. We want to see whether the distance s was under 5 miles at any point.

Ships have seen each other = $s\leq 5\\\\\sqrt{144-288t+208t^2}\leq 5\\\\144-288t+208t^2\leq 25\\\\199-288t+208t^2\leq 0$

Since function $f(x)=199-288x+208x^2$ is quadratic, concave up and has no real roots, we know that $199-288x+208x^2>0$ for every t. So, the ships haven't seen each other.

Attachedis the graph of s(red) and ds/dt(blue). We can see that our results from parts b and c were correct.

Function ds/dt has a horizontal asympote in the first quadrant if

$\lim_{t \to \infty} \frac{ds}{dt}$

So, lets check this limit:

$\lim_{t \to \infty} \frac{ds}{dt}=\lim_{t \to \infty} \frac{208t-144}{\sqrt{144-288t+208t^2}}\\\\=\lim_{t \to \infty} \frac{208-\frac{144}{t}}{\sqrt{\frac{144}{t^2}-\frac{288}{t}+208}}\\\\=\frac{208-0}{\sqrt{0-0+208}}\\\\=\frac{208}{\sqrt{208}}\\\\=4\sqrt{13}$