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Airida [17]
3 years ago
10

i have 17 coins. N of them are nickels and the rest are dimes. write an expression in two different ways for the amount of money

that i have.(hint: one is the other simplified)
Mathematics
1 answer:
sergiy2304 [10]3 years ago
4 0

Answer:

0.05N - 0.1N + 1.7 (in dollars

1.7 - 0.05N (in dollars)

Step-by-step explanation:

Given :

Total number of coins = 17

Number of nickels = N

Number of dimes = 17 - N

1 nickel = 5 cent = $0.05

1 dime = 10 cent = $0.1

The amount of money :

(Value of nickel * number of nickels) + (value of dime * number of dime)

(0.05 * N) + (0.1 * (17-N))

0.05N + 1.7 - 0.1N

0.05N - 0.1N + 1.7

-0.05N + 1.7 = 1.7 - 0.05N

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9 boards

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so to figure out how much one third of the porch needs you find half of what it takes to fill in 2/3 of the porch and you get 3.

so 3=1/3 of the porch

then you simply do 3times 3 or 3+3+3 and you get 9

so it takes 9 boards to fill the whole porch

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3 years ago
Blake has a glass tank. First, he wants to put some marbles in it, all of the same volume. Then, he wants to fill the tank with
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32 liters is the volume of the glass tank

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2 years ago
If your friend leaves your house at a speed of 8mph and two hours later you follow in your car at 40mph, how long will It be bef
dusya [7]

Answer:

  • 1/2 hour
  • 20 miles

Step-by-step explanation:

Your friend has a head start of (2 h)×(8 mi/h) = 16 mi.

When you leave, you gain on your friend at the rate of 40 mph - 8 mph = 32 mph. So, it will take ...

  (16 mi)/(32 mi/h) = 1/2 h

for you to reduce the distance between you to zero.

In that 1/2 hour, you have traveled ...

  (1/2 h)×(40 mi/h) = 20 mi

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3 years ago
What is the probability that a five-card poker hand contains the ace of hearts?
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<span>What is the probability that a five-card poker hand contains the ace of hearts?
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6 0
3 years ago
Which expression is equivalent to [(3xy^-5)^3/(x^-2y^2)^-4]^-2?
Mademuasel [1]

Answer:- a.The given expression is equivalent to  \frac{x^{10}y^{14}}{729}



Given expression:- [\frac{(3xy^{-5})^3}{(x^{-2}y^2)^{-4}}]^{-2}

=[\frac{(3)^3x^3y^{-5\times3}}{x^{-2\times-4}y^{2\times-4}}]^{-2}.........(a^m)^n=a^{mn}

=[\frac{27x^3y^{-15}}{x^8y^{-8}}]^{-2}

=[27x^{3-8}y^{-15-(-8)}]^{-2}............\frac{a^m}{a^n}=a^{m-n}

=[27x^{-5}y^{-7}]^{-2}=(27)^{-2}(x^{-5})^{-2}(y^{-7})^{-2}.........(a^m)^n=a^{mn}

=\frac{1}{(27)^2}(x^{10}y^{14})=\frac{x^{10}y^{14}}{729}

Thus a. is the right answer.


6 0
3 years ago
Read 2 more answers
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