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Fittoniya [83]
3 years ago
11

Let A and B be subsets of R. (a) If x ∈ (A ∩ B)c, explain why x ∈ Ac ∪ Bc. This shows that (A ∩ B)c ⊆ Ac ∪ Bc. 12 Chapter 1. The

Real Numbers (b) Prove the reverse inclusion (A ∩ B)c ⊇ Ac ∪ Bc, and conclude that (A ∩ B)c = Ac ∪ Bc. (c) Show (A ∪ B)c = Ac ∩ Bc by demonstrating inclusion both ways.
Mathematics
1 answer:
cupoosta [38]3 years ago
7 0

Answer:

answer is -3 just subtract 4 from each side

Step-by-step explanation:

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Help please!!! I dont understand these questions<br><br><br>currently attaching photos dont delete
Katyanochek1 [597]

Answer:

  1. b/a
  2. 16a²b²
  3. n¹⁰/(16m⁶)
  4. y⁸/x¹⁰
  5. m⁷n³n/m

Step-by-step explanation:

These problems make use of three rules of exponents:

a^ba^c=a^{b+c}\\\\(a^b)^c=a^{bc}\\\\a^{-b}=\dfrac{1}{a^b} \quad\text{or} \quad a^b=\dfrac{1}{a^{-b}}

In general, you can work the problem by using these rules to compute the exponents of each of the variables (or constants), then arrange the expression so all exponents are positive. (The last problem is slightly different.)

__

1. There are no "a" variables in the numerator, and the denominator "a" has a positive exponent (1), so we can leave it alone. The exponent of "b" is the difference of numerator and denominator exponents, according to the above rules.

\dfrac{b^{-2}}{ab^{-3}}=\dfrac{b^{-2-(-3)}}{a}=\dfrac{b}{a}

__

2. 1 to any power is still 1. The outer exponent can be "distributed" to each of the terms inside parentheses, then exponents can be made positive by shifting from denominator to numerator.

\left(\dfrac{1}{4ab}\right)^{-2}=\dfrac{1}{4^{-2}a^{-2}b^{-2}}=16a^2b^2

__

3. One way to work this one is to simplify the inside of the parentheses before applying the outside exponent.

\left(\dfrac{4mn}{m^{-2}n^6}\right)^{-2}=\left(4m^{1-(-2)}n^{1-6}}\right)^{-2}=\left(4m^3n^{-5}}\right)^{-2}\\\\=4^{-2}m^{-6}n^{10}=\dfrac{n^{10}}{16m^6}

__

4. This works the same way the previous problem does.

\left(\dfrac{x^{-4}y}{x^{-9}y^5}\right)^{-2}=\left(x^{-4-(-9)}y^{1-5}\right)^{-2}=\left(x^{5}y^{-4}\right)^{-2}\\\\=x^{-10}y^{8}=\dfrac{y^8}{x^{10}}

__

5. In this problem, you're only asked to eliminate the one negative exponent. That is done by moving the factor to the numerator, changing the sign of the exponent.

\dfrac{m^7n^3}{mn^{-1}}=\dfrac{m^7n^3n}{m}

3 0
3 years ago
Please help me solve this
Afina-wow [57]

Answer:

y-27=8·(x-3)

Step-by-step explanation:

8 0
3 years ago
21/√7-5√7/2 solve it
umka21 [38]

Answer:

√72

Decimal Form:

1.32287565…

Step-by-step explanation:

To subtract fractions, find the LCD and then combine.

5 0
2 years ago
In a large school district, for every one male teacher, there are three female teachers. Four teachers are selected at random to
serg [7]

Answer:

It would be a female teacher 3 out of 4 times, and male 1 out of 4. You could make a circle, divide it into quarters, make a spinner and likely get male once of four times.

6 0
3 years ago
Refer to the picture above
vodomira [7]

Answer:

3.14

Step-by-step explanation:

First find the circumference of the circle:

2\pi r = Circumference.

2 * \pi * 6 = 12\pi

Find the ratio of the angle in relation to the entire circle:

30^o is what we have. So:

\frac{30^o}{360^o} = \frac{1}{12}

Use the ratio and multiply the circumference to find the length:

12\pi * \frac{1}{12} = \pi

Round answer to the hundredth:

\pi = 3.14

4 0
3 years ago
Read 2 more answers
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