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vagabundo [1.1K]
3 years ago
12

Jonathon has a collection of coins worth a total of $10.75. If he has one-fourth as many dimes as pennies, and two-thirds as man

y dimes as nickels, how many of each coin does Jonathon have?
Mathematics
2 answers:
Mrrafil [7]3 years ago
4 0

Step-by-step explanation:

d = number of dimes

n = number of nickels

p = number of pennies

d×$0.10 + n×$0.05 + p×$0.01 = $10.75

d = p/4

d = n × 2/3

now we need to transform these equation and substitute the variable equations in the main equation, so that we have actually only one variable left. and that we can solve.

so, it seems to me transforming p and n into expressions of d might be the easiest one.

d = p/4

=>

p = 4d

d = n × 2/3

=>

3d = 2n

n = d × 3/2

so, now putting that into the main equation

d×$0.10 + (d × 3/2)×$0.05 + 4d×$0.01 = $10.75

since we have a fraction in there, let's multiply everything by that denominator (2) to be fraction free :

2d×$0.10 + 3d×$0.05 + 8d×$0.01 = $21.50

now we can simplify the terms by multiplying the integer factors of d into the value decimals

d×0.20 + d×0.15 + d×0.08 = 21.50

now we can simply add up ask the factors of d

d×(0.2 + 0.15 + 0.08) = 21.50

d×0.43 = 21.50

d = 21.5 / 0.43 = 50

p = 4d = 4×50 = 200

n = d × 3/2 = 50 × 3/2 = 25 × 3 = 75

so he has 50 dimes, 75 nickels and 200 pennies

bulgar [2K]3 years ago
4 0

Answer:

  • 200 pennies, 75 nickels, 50 dimes

Step-by-step explanation:

  • Pennies = p, penny = 1¢
  • Dimes = d, dime = 5¢
  • Nickels = n, nickel = 10¢
  • Total = $10.75 = 1075¢

<u>Equations:</u>

  • d = 1/4p ⇒ p = 4d
  • d = 2/3n ⇒ n = 1.5d
  • 1p + 5n + 10d = 1075

<u>Substitute p and n into third equation and solve for d:</u>

  • 4d + 5*1.5d + 10d = 1075
  • 21.5d = 1075
  • d = 1075/21.5
  • d = 50

<u>Find p and n:</u>

  • p = 4*50 = 200
  • n = 1.5*50 = 75

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Answers are in bold.

22. b/c

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=====================================================

Explanation:

Problem 22

To get the cosine ratio, we divide the adjacent over hypotenuse.

cos(angle) = adjacent/hypotenuse

cos(x) = b/c

----------------------------------

Problem 24

To get the sine ratio, we divide opposite over hypotenuse.

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In the fourth step, I multiplied top and bottom by sqrt(17) to rationalize the denominator.

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Problem 26

tan(angle) = opposite/adjacent

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----------------------------------

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Problem 34

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----------------------------------

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Let x be the length of side AC.

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tan(angle) = opposite/adjacent

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