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choli [55]
3 years ago
12

A chef makes a fruit salad using pears and mangos. Pears and mangos each cost $2.25 per fruit. The chef buys x pears and y mango

s. Which two expressions represent the chef's cost of making the fruit salad?
Mathematics
1 answer:
GrogVix [38]3 years ago
3 0

Answer:

P=2.25x-----------1

M=2.25y----------2

Step-by-step explanation:

Step one:

given data

we are told that the cost of pears= $2.25 each

       and cost of mango= $2.25 each

Step two:

the chef bought x pears and y mangoes

let the total cost of pears be P

and for mango be M

P=2.25x-----------1

M=2.25y----------2

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Jamie has a collection of 78 nickels, dimes, and quarters worth $12.40. If the number of quarters is
fgiga [73]

Jamie has 25 nickels, 14 dimes and 39 quarters in his collection

Step-by-step explanation:

The given is:

1. Jamie has a collection of 78 nickels, dimes, and quarters

2. They worth $12.40

3. If the number of quarters is  doubled, the value becomes $22.15

Assume that there are n nickels, d dimes and q quarters

∵ The collection of Jamie has 78 coins

∴ n + d + q = 78 ⇒ (1)

∵ 1 nickels = 5 cents

∵ 1 dimes = 10 cents

∵ 1 quarter = 25 cents

∵ The collection of Jamie worth $12.40

- Change the value of the collection to cents

∵ $1 = 100 cents

∴ The collection of Jamie worth = 12.40 × 100 = 1240 cents

∴ 5n + 10d + 25q = 1240

- Divide all terms by 5 to simplify it

∴ n + 2d + 5q = 248 ⇒ (2)

∵ When the number of quarters is  doubled, the value becomes

   $22.15

∵ The number of quarters is q

∴ The new number of quarters is 2q

∵ The value of coins is $22.15

∵ $22.15 = 22.15 × 100 = 2215 cents

∴ 5n + 10d + 25(2q) = 2215

∴ 5n + 10d + 50q = 2215

- Divide each term by 5 to simplify it

∴ n + 2 d + 10q = 443 ⇒ (3)

Subtract equation (2) from equation (3)

∵ n + 2d + 5q = 248 ⇒ (2)

∵ n + 2 d + 10q = 443 ⇒ (3)

∴ 5q = 195

- Divide both sides by 5

∴ q = 39

Substitute the value of q in equations (1) and (2)

∵ n + d + q = 78 ⇒ (1)

∴ n + d + 39 = 78

- Subtract 39 from both sides

∴ n + d = 39 ⇒ (4)

∵ n + 2d + 5q = 248 ⇒ (2)

∴ n + 2d + 5(39) = 248

∴ n + 2d + 195 = 248

- Subtract 195 from both sides

∴ n + 2d = 53 ⇒ (5)

Subtract equation (4) from equation (5)

∵ n + d = 39 ⇒ (4)

∵ n + 2d = 53 ⇒ (5)

∴ d = 14

Substitute the value of d in equation (4)

∵ n + d = 39 ⇒ (4)

∴ n + 14 = 39

- Subtract 14 from both sides

∴ n = 25

Jamie has 25 nickels, 14 dimes and 39 quarters in his collection

Learn more:

You can learn more about solving system of equations in

brainly.com/question/13168205

brainly.com/question/2115716

#LearnwithBrainly

5 0
3 years ago
What is the graph for the above relationship?
Vesna [10]

Second graph.

<h2>Explanation:</h2>

Hello! You haven't provided any relationship, but I found it on the internet and it is the table attached below. The point-slope form of the equation of a line is given by:

y=mx+b

So:

y-Intercept: \\ \\ (0,b) \\ \\ \\ From \ the \ table: \\ \\ (0,b)=(0,46) \\ \\ So: \\ \\ b=46

Finding the slope:

m=\frac{39-46}{1-0} \\ \\ m=-7

So the equation of the line is:

y=-7x+46

From the graphs, only option 2 and 4 have negative slope and only option 2 passes through (0,46)

Conclusion: Correct option is second graph.

<h2>Learn more:</h2>

Parallel lines: brainly.com/question/12169569

#LearnWithBrainly

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3 years ago
Which expression is equivalent to (2g^5)^3
lianna [129]
Answer is 8g^15
Pls mark me brainliest
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3 years ago
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Create a data set that has six values, a mean of 7, and a range of 15
Nezavi [6.7K]
12,13,14,15,16,1 is the six values
5 0
4 years ago
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a parking meter contains $8.25 in dimes and quarters. after emptying the meter, the worker counts 51 coins. how many quarters an
Marianna [84]
<h3>30 dimes and 21 quarters are present</h3>

<em><u>Solution:</u></em>

Let "d" be the number of dimes

Let "q" be the number of quarters

1 dime = $ 0.10

1 quarter = $ 0.25

The worker counts 51 coins

Therefore,

d + q = 51

d = 51 - q ---------- eqn 1

<em><u>A parking meter contains $8.25 in dimes and quarters</u></em>

Therefore,

0.10d + 0.25q = 8.25 -------- eqn 2

<em><u>Substitute eqn 1 in eqn 2</u></em>

0.10(51 - q) + 0.25q = 8.25

5.1 - 0.10q + 0.25q = 8.25

0.15q = 3.15

Divide both sides by 0.15

<h3>q = 21</h3>

Substitute q = 21 in eqn 1

d = 51 - 21

<h3>d = 30</h3>

Thus 30 dimes and 21 quarters are present

8 0
3 years ago
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