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3 years ago

Find the union {6, 11, 15} U Ø

Mathematics

1 answer:

astra-53 [7]3 years ago

7 0

<h3>Answer: {6, 11, 15}</h3>

Explanation:

The Ø means "empty set". It's the set with nothing inside it, not even 0.

We can write Ø as { } which is a pair of curly braces with nothing between them.

The rule is that if we union any set A with Ø, then we'll get set A

A U Ø = A

Ø U A = A

In a sense, it's analogous to adding 0. So it's like saying A+0 = A and 0+A = A.

So that's why {6, 11, 15} U Ø = {6, 11, 15}

There's nothing to add onto the set {6, 11, 15}, so we just get the same thing back again.

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3 years ago

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3 years ago

Your part time job pays you $14/h. You work approximately 12 hours/week.

kirill115 [55]

Answer:

<u>BOX 1 </u>= 672.00

(14 * 12 = 168

168 * 4 = $ 672 )

2. <u>BOX 2</u> = 282.95

(Cell phone= $45 A MONTH

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<h2><em><u>please give brainliest </u></em></h2>

4 0

3 years ago

A^2(64a^2-3)/3=27-4a^2/4

frutty [35]

The answer is a = 3/4 = 0.75

First get rid of the paranthesis,
${a}^{2} (64 {a}^{2} - 3) = 64 {a}^{4} - 3 {a}^{2}$
Then set the denominators equal:
$\frac{ 256 {a}^{4} - 12 {a}^{2}}{12} = \frac{81 - 12 {a}^{2} }{12}$
Then remove the denominators and solve:
$256 {a}^{4} - 12 {a}^{2} = 81 - 12 {a}^{2}$
Eliminate -12a^2 by adding 12a^2 to both sides:
$256 {a}^{4} = 81$
Take the fourth root of them or take the square root twice:
$\sqrt[4]{256 {a}^{4} } = \sqrt[4]{81} \\ 4a = 3$
Divide both sides by 4:
$a = \frac{3}{4} = 0.75$

7 0

3 years ago

I will give BRAINLIEST to whoever is correct.

attashe74 [19]

28/11.2 = 2.5

since they are similar and openly stated as parrallelograms, you know that they have a scale of the longest edge of EFGH to the longest edge of JKLM

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4 years ago

Find the union {6, 11, 15} U Ø​

Find the union {6, 11, 15} U Ø