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aksik [14]
3 years ago
6

Suppose there is a pile of quarters dimes and pennies with a total value of $1.07 how much of each coin can be present without b

eing able to make change for a dollar? Of there are multiple selections of the coins that will work choose the selection with the largest total numbers of coins.
Mathematics
1 answer:
Korolek [52]3 years ago
4 0
Hello,

Very nice as problem.

2 solutions:
1 quater,8 dimes, 2 pennies
and
3 quaters,3 dimes, 2 pennies

since
107=( 0, 0, 107) but : 100= 0*25+ 0*10+ 100
107=( 0, 1, 97) but : 100= 0*25+ 1*10+ 90
107=( 0, 2, 87) but : 100= 0*25+ 2*10+ 80
107=( 0, 3, 77) but : 100= 0*25+ 3*10+ 70
107=( 0, 4, 67) but : 100= 0*25+ 4*10+ 60
107=( 0, 5, 57) but : 100= 0*25+ 5*10+ 50
107=( 0, 6, 47) but : 100= 0*25+ 6*10+ 40
107=( 0, 7, 37) but : 100= 0*25+ 7*10+ 30
107=( 0, 8, 27) but : 100= 0*25+ 8*10+ 20
107=( 0, 9, 17) but : 100= 0*25+ 9*10+ 10
107=( 0, 10, 7) but : 100= 0*25+ 10*10+ 0
107=( 1, 0, 82) but : 100= 1*25+ 0*10+ 75
107=( 1, 1, 72) but : 100= 1*25+ 1*10+ 65
107=( 1, 2, 62) but : 100= 1*25+ 2*10+ 55
107=( 1, 3, 52) but : 100= 1*25+ 3*10+ 45
107=( 1, 4, 42) but : 100= 1*25+ 4*10+ 35
107=( 1, 5, 32) but : 100= 1*25+ 5*10+ 25
107=( 1, 6, 22) but : 100= 1*25+ 6*10+ 15
107=( 1, 7, 12) but : 100= 1*25+ 7*10+ 5
107=( 1, 8, 2) is good
107=( 2, 0, 57) but : 100= 2*25+ 0*10+ 50
107=( 2, 1, 47) but : 100= 2*25+ 1*10+ 40
107=( 2, 2, 37) but : 100= 2*25+ 2*10+ 30
107=( 2, 3, 27) but : 100= 2*25+ 3*10+ 20
107=( 2, 4, 17) but : 100= 2*25+ 4*10+ 10
107=( 2, 5, 7) but : 100= 2*25+ 5*10+ 0
107=( 3, 0, 32) but : 100= 3*25+ 0*10+ 25
107=( 3, 1, 22) but : 100= 3*25+ 1*10+ 15
107=( 3, 2, 12) but : 100= 3*25+ 2*10+ 5
107=( 3, 3, 2) is good
107=( 4, 0, 7) but : 100= 4*25+ 0*10+ 0



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1.60/20 =0.08

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Please Please Please help with this math problem
katovenus [111]
  1. The revenue as a function of x is equal to -x²/20 + 920x.
  2. The profit as a function of x is equal to -x²/20 + 840x - 6000.
  3. The value of x which maximizes profit is 8,400 and the maximum profit is $3,522,000.
  4. The price to be charged to maximize profit is $500.

<h3>How to express the revenue as a function of x?</h3>

Based on the information provided, the cost function, C(x) is given by 80x + 6000 while the demand function, P(x) is given by -1/20(x) + 920.

Mathematically, the revenue can be calculated by using the following expression:

R(x) = x × P(x)

Revenue, R(x) = x(-1/20(x) + 920)

Revenue, R(x) = x(-x/20 + 920)

Revenue, R(x) = -x²/20 + 920x.

Expressing the profit as a function of x, we have:

Profit = Revenue - Cost

P(x) = R(x) - C(x)

P(x) = -x²/20 + 920x - (80x + 6000)

P(x) = -x²/20 + 840x - 6000.

For the value of x which maximizes profit, we would differentiate the profit function with respect to x:

P(x) = -x²/20 + 840x - 6000

P'(x) = -x/10 + 840

x/10 = 840

x = 840 × 10

x = 8,400.

For the maximum profit, we have:

P(x) = -x²/20 + 840x - 6000

P(8400) = -(8400)²/20 + 840(8400) - 6000

P(8400) = -3,528,000 + 7,056,000 - 6000

P(8400) = $3,522,000.

Lastly, we would calculate the price to be charged in order to maximize profit is given by:

P(x) = -1/20(x) + 920

P(x) = -1/20(8400) + 920

P(x) = -420 + 920

P(x) = $500.

Read more on maximized profit here: brainly.com/question/13800671

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Martin had 24
inna [77]

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Answer:

x = -9

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Hey there :)

- tan²x + sec²x = 1    or    1 + tan²x = sec²x

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\frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}

\frac{sinx}{cosx} = tanx so \frac{sin^2x}{cos^2x} = tan^2x
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Therefore,
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1 = - tan²x = sec²x or sec²x - tanx = 1

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