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iren [92.7K]
3 years ago
11

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:
aleksandrvk [35]3 years ago
8 0
10.35 collections for the coin s worth
Anastasy [175]3 years ago
7 0

The number of dimes, pennies and nickels are required.

The number of dimes is 50, pennies is 200 and nickels is 75.

Let the number of dimes be x

From the question the number of dimes is one-fourth the number of pennies and two-thirds the number of nickels.

So,

The number of pennies is 4x

and number of nickels is \dfrac{3}{2}x

1 nickel = 0.05$

1 penny = 0.01$

1 dime = 0.01$

Total value of the coins is $10.75

0.1x+0.05\times \dfrac{3}{2}x+0.01\times 4x=10.75\\\Rightarrow 0.215x=10.75\\\Rightarrow x=\dfrac{10.75}{0.215}\\\Rightarrow x=50

The number of dimes is 50.

The number of pennies is 4x=4\times50=200

The number of nickels is \dfrac{3}{2}x=\dfrac{3}{2}\times50=75.

Learn more:

brainly.com/question/24875263

brainly.com/question/20842274

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The sum of two terms of gp is 6 and that of first four terms is 15/2.Find the sum of first six terms.​
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Given:

The sum of two terms of GP is 6 and that of first four terms is \dfrac{15}{2}.

To find:

The sum of first six terms.​

Solution:

We have,

S_2=6

S_4=\dfrac{15}{2}

Sum of first n terms of a GP is

S_n=\dfrac{a(1-r^n)}{1-r}              ...(i)

Putting n=2, we get

S_2=\dfrac{a(1-r^2)}{1-r}

6=\dfrac{a(1-r)(1+r)}{1-r}

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Putting n=4, we get

S_4=\dfrac{a(1-r^4)}{1-r}

\dfrac{15}{2}=\dfrac{a(1-r^2)(1+r^2)}{1-r}

\dfrac{15}{2}=\dfrac{a(1+r)(1-r)(1+r^2)}{1-r}

\dfrac{15}{2}=6(1+r^2)            (Using (ii))

Divide both sides by 6.

\dfrac{15}{12}=(1+r^2)

\dfrac{5}{4}-1=r^2

\dfrac{5-4}{4}=r^2

\dfrac{1}{4}=r^2

Taking square root on both sides, we get

\pm \sqrt{\dfrac{1}{4}}=r

\pm \dfrac{1}{2}=r

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Case 1: If r is positive, then using (ii) we get

6=a(1+0.5)  

6=a(1.5)  

\dfrac{6}{1.5}=a  

4=a

The sum of first 6 terms is

S_6=\dfrac{4(1-(0.5)^6)}{(1-0.5)}

S_6=\dfrac{4(1-0.015625)}{0.5}

S_6=8(0.984375)

S_6=7.875

Case 2: If r is negative, then using (ii) we get

6=a(1-0.5)  

6=a(0.5)  

\dfrac{6}{0.5}=a  

12=a  

The sum of first 6 terms is

S_6=\dfrac{12(1-(-0.5)^6)}{(1+0.5)}

S_6=\dfrac{12(1-0.015625)}{1.5}

S_6=8(0.984375)

S_6=7.875

Therefore, the sum of the first six terms is 7.875.

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