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MAVERICK [17]
3 years ago
5

every evening jenna empties her pockets and puts her change in a jar. At the end of the week she counts her money. One week she

had 40 coins, all of them dimes and quarters. when she added them up, she had a total of $7.75. how many dimes did jenna have
Mathematics
1 answer:
Darya [45]3 years ago
8 0
Let Q= the number of quarters and D= the number of dimes.
Q+D= 40 coins
0.25Q+0.1D= $7.75
Rewrite equation 1 as: D = 40-Q and substitute into equation 2
0.25Q+0.1(40-Q)= 7.75 Simplify the left side
0.25Q+4-0.1Q= 7.75 Subtract 4 from both sides
0.25Q-0.1Q= 3.75 Convime like terms on left side
0.15Q= 7.75 Divide both sides by 0.15
Q=25 and D= 40-Q
D= 40-25 which is 15
So Jenna had 25 quarters and 15 dimes.
<span>To check multiply 25 x 25 and 15x10 which equals 625 and 150 this equals 775 so this is 7.75</span>
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<h2>Answer:</h2>

<em><u>Recursive equation for the pattern followed is given by,</u></em>

a_{n}=a_{n-1}+(n-1)^{2}

<h2>Step-by-step explanation:</h2>

In the question,

The number of interaction for 1 child = 0

Number of interactions for 2 children = 1

Number of interactions for 3 children = 5

Number of interaction for 4 children = 14

So,

We need to find out the pattern for the recursive equation for the given conditions.

So,

We see that,

a_{1}=0\\a_{2}=1\\a_{3}=5\\a_{4}=14\\

Therefore, on checking, we observe that,

a_{n}=a_{n-1}+(n-1)^{2}

On checking the equation at the given values of 'n' of, 1, 2, 3 and 4.

<u>At, </u>

<u>n = 1</u>

a_{n}=a_{n-1}+(n-1)^{2}\\a_{1}=a_{1-1}+(1-1)^{2}\\a_{1}=0+0=0\\a_{1}=0

which is true.

<u>At, </u>

<u>n = 2</u>

a_{n}=a_{n-1}+(n-1)^{2}\\a_{2}=a_{2-1}+(2-1)^{2}\\a_{2}=a_{1}+1\\a_{2}=1

Which is also true.

<u>At, </u>

<u>n = 3</u>

a_{n}=a_{n-1}+(n-1)^{2}\\a_{3}=a_{3-1}+(3-1)^{2}\\a_{3}=a_{2}+4\\a_{3}=5

Which is true.

<u>At, </u>

<u>n = 4</u>

a_{n}=a_{n-1}+(n-1)^{2}\\a_{4}=a_{4-1}+(4-1)^{2}\\a_{4}=a_{3}+9\\a_{4}=14

This is also true at the given value of 'n'.

<em><u>Therefore, the recursive equation for the pattern followed is given by,</u></em>

a_{n}=a_{n-1}+(n-1)^{2}

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