Please help me with 11th grade math // Geometry

Please I’ve attached the question can you help me

Andreyy89

Answer:

1. n[(A U B) - C] = {23, 24, 27, 29, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 61, 63}

2. n[(A - B) U C] = n[A U C] = {6, 10, 12, 15, 20, 23, 29, 30, 31, 37, 41, 43, 47, 53, 59, 60, 61}

3. D. I, II and III.

Step-by-step explanation:

U = {21, 22, 23, ..., 64}

A prime number is a number that can be divided only by 1 and itself.

A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

B = {24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63}

C = {6, 10, 12, 15, 20, 30, 60}

1. Find n[(A U B) - C]

n(A U B) = {23, 24, 27, 29, 30, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 60, 61, 63}

n[(A U B) - C] = {23, 24, 27, 29, 30, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 60, 61, 63} - {6, 10, 12, 15, 20, 30, 60}

Since only 30 and 60 are common to n(A U B) and C, we therefore remove them it and have:

n[(A U B) - C] = {23, 24, 27, 29, 33, 36, 37, 39, 41, 42, 43, 45, 47, 48, 51, 53, 54, 57, 59, 61, 63}

2. Find n[(A - B) U C]

To get n(A - B) we remove all the elements of B in A. Since there are no common elements between A and B, we therefore have:

A - B = A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

C = {6, 10, 12, 15, 20, 30, 60}

Therefore, we have:

n[(A - B) U C] = n[A U C] = {6, 10, 12, 15, 20, 23, 29, 30, 31, 37, 41, 43, 47, 53, 59, 60, 61}

3. Which of the following is/are true?

I. A ∩ B = A ∩ C

A ∩ B = ∅

A ∩ C = ∅

Therefore, A ∩ B = A ∩ C is true.

II. A - B = A - C

A - B = A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

A - C = A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

Therefore, A - B = A - C is true.

III. A ∩ (B ∪ C) = ∅

A = {23, 29, 31, 37, 41, 43, 47, 53, 59, 61}

B U C = {6, 10, 12, 15, 20, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63}

Therefore, A ∩ (B ∪ C) = ∅ is true.

Therefore, the correct option is D i.e. I, II and III are true.

4 0

3 years ago

How to put one hundred forty one million six hundred twenty thousand in standard form

Brums [2.3K]

Answer:

140,620,000

Step-by-step explanation:

There are two types of standard form. The one above (given to you) is the normal standard form. However, there is another one. This one would look like:

140,620,000 = 1.4062 x 10^8

Remember that in this standard form, you must move the decimal to just after the first place value, and the amount of moves you made for the decimal is the amount in the power over 10.

~

6 0

3 years ago

Read 2 more answers

Please help i need it for tomorrow With explanation please

Gemiola [76]

Answer:

pangalang ng pair angles figure

5 0

2 years ago

Read 2 more answers

Thomas is in charge of selling roses for the Graduation dance. The roses sell for $3.50 each. He estimates that the expenses of

KonstantinChe [14]

Answer:

P= $3.5x -$30

Step-by-step explanation:

Let the number of roses Thomas bought and sold be x.

Hence the total selling price would be;

$3.5 × x= $3.5x

The profit = selling price-expenses

P= $3.5x -$30

6 0

3 years ago

Let f(x) = x+2/x+6 f^-1(-6) =

GREYUIT [131]

We can explicitly find the inverse. If $f^{-1}(x)$ is the inverse of $f(x)$ , then

$f\left(f^{-1}(x)\right) = \dfrac{f^{-1}(x)+2}{f^{-1}(x)+6} = x$

Solve for the inverse :

$\dfrac{f^{-1}(x) + 2}{f^{-1}(x) + 6} = x$

$\dfrac{f^{-1}(x) + 6 - 4}{f^{-1}(x) + 6} = x$

$1 - \dfrac4{f^{-1}(x) + 6} = x$

$1 - x = \dfrac4{f^{-1}(x) + 6}$

$f^{-1}(x) + 6 = \dfrac4{1-x}$

$\implies f^{-1}(x) = \dfrac4{1-x} - 6$

Then when x = -6, we have

$f^{-1}(-6) = \dfrac4{1-(-6)} - 6 = \dfrac47-6 = \boxed{-\dfrac{38}7}$

Alternatively, we can first solve for x such that $f(x) = -6$ . Then taking the inverse of both sides, $x = f^{-1}(-6)$ . (The difference in this method is that we don't compute the inverse for all x.)