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garik1379 [7]
3 years ago
13

lilli has $3.65 comprised of quarters and dimes floating on the bottom of her bag she has five more dimes than twice her quarter

s how many each type of coin does she have? lol pls help
Mathematics
1 answer:
marta [7]3 years ago
8 0

Answer:

= 3.65.

Step-by-step explanation:

You know that Tom has 23 coins in total. That means the number of dimes and the number of quarters equals 23.

D + Q = 23

You also know that the total value of the coins is $3.65. A dime is worth $0.10 and a quarter is worth $0.25, so:

0.10*D + 0.25*Q = 3.65

That says that 10 cents per dime times the number of dimes, plus 25 cents per quarter times the number of quarters is three dollars and sixty-five cents.

So you have a system of linear equations:

D + Q = 23

0.10D + 0.25Q = 3.65

I'm going to multiply the bottom row by 100 to clear the decimals.

D + Q = 23

10D + 25Q = 365

Now I can multiply the top row by -10 to eliminate the D terms

-10D - 10Q = -230

10D + 25Q = 365

________________

0D + 15Q = 135

15Q = 135

Q = 9

Tom has 9 quarters.

If D + Q = 23, and Q = 9, then:

D + 9 = 23

D = 14

Tom has 14 dimes.

Always good to double check:

9 quarters is worth $2.25. 14 dimes is worth $1.40. These should sum to $3.65.

2.25 + 1.40

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Solution:

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\begin{gathered} P^n_r=\frac{n!}{(n-r)!} \\  \end{gathered}

The combination formula is expressed as

\begin{gathered} C^n_r=\frac{n!}{(n-r)!r!} \\  \\  \end{gathered}

where

\begin{gathered} n\Rightarrow total\text{ number of objects} \\ r\Rightarrow number\text{ of object selected} \end{gathered}

Given that 6 objects are taken at a time from 8, this implies that

\begin{gathered} n=8 \\ r=6 \end{gathered}

Thus,

Number of permuations:

\begin{gathered} P^8_6=\frac{8!}{(8-6)!} \\ =\frac{8!}{2!}=\frac{8\times7\times6\times5\times4\times3\times2!}{2!} \\ 2!\text{ cancel out, thus we have} \\ \begin{equation*} 8\times7\times6\times5\times4\times3 \end{equation*} \\ \Rightarrow P_6^8=20160 \end{gathered}

Number of combinations:

\begin{gathered} C^8_6=\frac{8!}{(8-6)!6!} \\ =\frac{8!}{2!\times6!}=\frac{8\times7\times6!}{6!\times2\times1} \\ 6!\text{ cancel out, thus we have} \\ \frac{8\times7}{2} \\ \Rightarrow C_6^8=28 \end{gathered}

Hence, there are 28 combinations and 20160 permutations.

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Pressure is defined as the physical force exerted on an object.

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Given

The equation is

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