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Lilit [14]
2 years ago
5

A boy had 20 cents. He bought x pencils for 3 cents each. If y equals the number of pennies left, write an equation showing the

dependence of y on x. What is the domain of the function?
Mathematics
1 answer:
Vesna [10]2 years ago
6 0

Hey there!!

The total number of cents - 20

Cost for each pencil - 3 cents

Number of pencils bought - ' x '

y = number of pennies left

What is the domain ?

Show how y is dependent on x ..

Let's get this into an equation :

... The total cost for x pencils bought = 3x

... Number of pennies left = 20 - 3x

... y = number of pennies left

... y = 20 - 3x

Notice : If the x value changes, the y value changes too

... If x = 1 , then , y = 17; If x = 2 , then , y = 14

Hence, we could say y is dependent upon x

Domain = ?

... Remember - The total number of pennies = 20 ; hence, the total cost cannot go above 20 cents.

Hence, we will have to work with inequalities

The equation  :

... 20 ≥ 3x

Divide 3 on both sides

20 / 3 ≥ x

x ≥ 6.667

Let's take this as 6

Hence , the domain will be :

D : { 1 , 2 , 3 , 4 , 5 , 6 }

Hope my answer helps!

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klio [65]
Since the Venus orbits round the sun, the sun is the center of the circular path of the revolution of the planet, Venus.

Thus, the distance of the planet, Venus fron the sun is given by the distance between the points (0, 0) and (41, 53).

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Thus, the distance between the points (0, 0) and (41, 53) is given by:

d= \sqrt{(41-0)^2+(53-0)^2}  \\  \\ = \sqrt{41^2+53^2} = \sqrt{1,681+2,809}  \\  \\ = \sqrt{4,490} =67 \ units

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8 0
1 year ago
6 tractors take 10days to collect the harvest. How long would it take 18 tractors to do the same amount of work?
Nutka1998 [239]
This is a question of proportionality. However, it is not direct proportionality as it is expected that as the number of tractors increases, the work is finished faster as opposed to fewer number of tractors. This is referred to as inverse proportionality.

Therefore;
6 tractors ----- 10 days
18 tractors --- x days

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x = (6*10)/18 = 10/3 days = 3 days, and 8 hours.

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6 0
2 years ago
For the line that passes through Y(3,0), parallel to ↔DJ with D(-3,1) and J(3,3) I need to find the slope and to write a point s
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Answer:

Step-by-step explanation:

2/5

do the run over rise

3 0
2 years ago
A craft vendor must sell at least $300 worth of merchandise to make a profit. Scarves sell for $10 each and hats sell for $20 ea
valina [46]
There is a missing graph in the problem given. However, we can simply solve the equation using the given data.

Items to be sold: scarves and hats. Minimum of 20 items sold in all.
Scarves sell for 10 each and hats sell for 20 each. Must sell at least 300 worth of merchandise to make profit. 

Let s represent scarves and h represent hats.

10s + 20h <u>></u> 300
s + h <u>></u> 20

We use inequality because the problem states "at least". 

s + h = 20
10s + 20h = 300

s = 20 - h
10(20-h) + 20h = 300
200 - 10h + 20h = 300
10h = 300 - 200
10h = 100
h = 100/10
h = 10

s = 20 - h
s = 20 - 10
s = 10

s + h <u>></u> 20
10 + 10 <u>></u> 20

10s + 20h <u>></u> 300
10(10) + 20(10) <u>></u> 300
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7 0
1 year ago
Read 2 more answers
Every day, Jorge buys a lottery ticket. Each ticket has a probability of of winning a prize. After six days, what is the probabi
Romashka [77]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Every day, Jorge buys a lottery ticket. Each ticket has a 0.16 probability of winning a prize. After six days, what is the probability that Jorge has won at least one prize? Round your answer to four decimal places.

Answer:

The probability that Jorge has won at least one prize after six days is

P(at least 1 win) = 0.6487

Step-by-step explanation:

Every day, Jorge buys a lottery ticket which has a 0.16 chance of winning a prize.

We want to find out the probability that Jorge has won at least one prize after six days.

P(at least 1 win) = 1 - P(not winning for 6 days)

We know that the probability of winning is 0.16 then the probability of not winning is

P(not winning) = 1 - 0.16 = 0.84

For 6 days,

P(not winning for 6 days) = 0.84×0.84×0.84×0.84×0.84×0.84

P(not winning for 6 days) = 0.84⁶

P(not winning for 6 days) = 0.3513

Finally,

P(at least 1 win) = 1 - P(not winning for 6 days)

P(at least 1 win) = 1 - 0.3513

P(at least 1 win) = 0.6487

6 0
2 years ago
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