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shtirl [24]
3 years ago
5

if you’re good with set theory in math 30 please help with questions 31 and 32!! real answers only !!

Mathematics
1 answer:
GREYUIT [131]3 years ago
7 0

Answer:  31) d     32) c

<u>Step-by-step explanation:</u>

31)

A = {1, 3, 5, 15}

B = {2, 3, 5, 7}

A ∪ B = {1, 2, 3, 5, 7, 15}

C = {2, 4, 6, 8}

(A ∪ B) ∩ C = {1, 2, 3, 5, 7, 15} ∩ {2, 4, 6, 8} = {2}

<em>2 is the only number in both sets</em>

32)

The pattern is: reflection, add symbol, ..., reflection, add symbol.

The last image shown is: add symbol

So the next image will be: reflection (flipped to the left)

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Need help with questions B, C and D. If you would graphing with functions help? :)
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The answer is attached here.
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Sung Lee invests $4,000 at age 18. He hopes the investment will be worth $16,000 when he turns 25. If the interest compounds con
irinina [24]

Answer:

The growth rate he needs to achieve his goal is approximatelly 19.8%

Step-by-step explanation:

Since the sum will be compounded continuously we have to use the appropriate formula given below:

M = C*e^(r*t)

Where "M" is the final amount, C is the initial amount, r is the interest rate and t is the time elapsed. Since Sung Lee will invest that sum at 18 years old and he wants to recieve the return at 25, then the time elapsed is given by 25 -18 = 7 years. We can now apply the data to the formula:

16000 = 4000*e^(r*7)

4000*e^(7*r) = 16000

e^(7*r) = 16000/4000 = 4

ln[e^(7*r)] = ln(4)

7*r = ln(4)

r = ln(4)/7 = 0.198

The rate of interest is given by (r)*100%, so we have (0.198)*100% = 19.8%.

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3 years ago
Determine if the set of ordered pairs is a relation or a function.
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Which of the following points would be a solution to this system of linear inequalities? ​
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Let's plug in both x and y values in the equation and check if the inequality is true...

\red{ \rule{35pt}{2pt}} \orange{ \rule{35pt}{2pt}} \color{yellow}{ \rule{35pt} {2pt}} \green{ \rule{35pt} {2pt}} \blue{ \rule{35pt} {2pt}} \purple{ \rule{35pt} {2pt}}

(4,1) \\ y \leqslant x + 1 \\ y <  -  \frac{x}{2}  - 1

Substitute x as 4 and y as 1

1 \leqslant 4 + 1 \\ 1 <  -  \frac{4}{2}  - 1

0  \leqslant  4 + 1 \\ 1 <  - 2 - 1

0 \leqslant 5 \\ 1 <  - 3

  • <em>This is not a solution because both are not true, Only one equation is giving true as answer which is not correct!~</em>

\purple{ \rule{300pt}{3pt}}

(0, - 3) \\ y \leqslant x + 1 \\ y <  -  \frac{x}{2}  - 1

Substitute x as 0 and y as -3

- 3 \leqslant 0 + 1 \\  - 3 <  -  \frac{0}{2}  - 1

- 3 \leqslant 1 \\  - 3 <  - 0 - 1

- 3 \leqslant 1 \\  - 3 <  - 1

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3 years ago
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