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3 years ago

Basic Set Notation. Help Pls.

Mathematics

1 answer:

balandron [24]3 years ago

4 0

Hello!

The set X ∪ Y, can be found by combining the given sets of the sets into one set.

You are given set X = {5, 10, 15} and set Y = {3, 6, 9, 12, 15}.

∪ represents union, so you put these sets together.

Also, since the element 15 is in both sets, it should be in the X ∪ Y set, once.

Therefore the answer is A, X ∪ Y = {3, 5, 6, 9, 10, 12, 15}.

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There were 490 people at play. The admission price was $3.00 for adults and $1.00 for children. The admission receipts were $990

dusya [7]

Hope this helps! This is system of equations.

8 0

4 years ago

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A 48 in, by 36 in. poster is reduced to an image that is 8 in. by 6 in. in size to be used for a flyer. What is the scale factor

Aliun [14]

Divide the original size by the reduced size:

48 / 8 = 6
36 / 6 = 6

Both sides were reduced by the same amount, and because the image was made smaller the scale factor would be a fraction less than 1.

The scale factor would be 1/6

The answer is B.

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4 years ago

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Determine the time necessary for P dollars to double when it is invested at interest rate r compounded annually, monthly, daily,

Rudik [331]

Answer:

Part 1) 8.17 years

Part 2) 4.98 years

Part 3) 4.95 years

Part 4) 4.95 years

Step-by-step explanation:

we know that

The compound interest formula is equal to

$A=P(1+\frac{r}{n})^{nt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

n is the number of times interest is compounded per year

Part 1) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded annually

in this problem we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=1$

substitute in the formula above

$2p=p(1+\frac{0.14}{1})^{t}$

$2=(1.14)^{t}$

Apply log both sides

$log(2)=log[(1.14)^{t}]$

$log(2)=(t)log(1.14)$

$t=log(2)/log(1.14)$

$t=8.17\ years$

Part 2) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded monthly

in this problem we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=12$

substitute in the formula above

$2p=p(1+\frac{0.14}{12})^{12t}$

$2=(\frac{12.14}{12})^{12t}$

Apply log both sides

$log(2)=log[(\frac{12.14}{12})^{12t}]$

$log(2)=(12t)log(\frac{12.14}{12})$

$t=log(2)/12log(\frac{12.14}{12})$

$t=4.98\ years$

Part 3) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% compounded daily

in this problem we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14\\n=365$

substitute in the formula above

$2p=p(1+\frac{0.14}{365})^{365t}$

$2=(\frac{365.14}{365})^{365t}$

Apply log both sides

$log(2)=log[(\frac{365.14}{365})^{365t}]$

$log(2)=(365t)log(\frac{365.14}{365})$

$t=log(2)/365log(\frac{365.14}{365})$

$t=4.95\ years$

Part 4) Determine the time necessary for P dollars to double when it is invested at interest rate r=14% continuously

we know that

The formula to calculate continuously compounded interest is equal to

$A=P(e)^{rt}$

where

A is the Final Investment Value

P is the Principal amount of money to be invested

r is the rate of interest in decimal

t is Number of Time Periods

e is the mathematical constant number

we have

$t=?\ years\\ P=\$p\\A=\$2p\\r=14\%=14/100=0.14$

substitute in the formula above

$2p=p(e)^{0.14t}$

Simplify

$2=(e)^{0.14t}$

Apply ln both sides

$ln(2)=ln[(e)^{0.14t}]$

$ln(2)=(0.14t)ln(e)$

Remember that ln(e)=1

$ln(2)=(0.14t)$

$t=ln(2)/(0.14)$

$t=4.95\ years$

4 0

3 years ago

mr. Patel travelled from Delhi to Mumbai by road in his car. a). If he travelled in his car 770 km over 22 days, then how many k

Bumek [7]

Answer:

a. 35 Kilometres.

b. 60 litres.

c. 8150 sweets.

Step-by-step explanation:

Given the following data;

Distance traveled = 770 km
Number of days = 22 days

a. To find how many kilometres he traveled in one day;

770 km = 22 days

X km = 1 day

Cross-multiplying, we have;

22X = 770

X = 770/22

X = 35 Kilometres.

b. To find how many litres he would need to go 840 km;

1 litre = 14 kilometres

X litres = 840

Cross-multiplying, we have;

14X = 840

X = 840/14

X = 60 litres.

c. To find how many sweets are there in his car;

Number of sweet box, Ns = 326 boxes

Number of sweets in each box, Sw = 25 sweets

Total number of sweets = Ns * Sw

Total number of sweets = 326 * 25

Total number of sweets = 8150 sweets.

4 0

3 years ago

1) what is the value of x ?

Vikentia [17]

Answer:

a). x = 9

b). m∠ABC = 27°

c). m∠CBD = 153°

Step-by-step explanation:

1). Property of linear pair of angles states, "sum of linear pair of angles is 180°"

3x + (15x + 18) = 180°

18x + 18 = 180

18x = 180 - 18

18x = 162

x = 9

2). m∠ABC = 3x°

= 3×9

= 27°

3). m∠CBD = (15x + 18)°

= (15×9 + 18)°

= (135 + 18)°

= 153°

6 0

3 years ago