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Shalnov [3]
3 years ago
10

Logan has $1.95 in dimes and quarters in his pocket. He has 2 more dimes than quarters. (Keep in mind that dimes are worth 10c a

nd quarters 25c.)
1. Write an equation that will help you determine the number of quarters.
2. Solve the equation showing all your steps.
3. Tell how many quarters and dimes he has.
Mathematics
1 answer:
Katarina [22]3 years ago
5 0

Answer:

17 dimes 1 quarter

Step-by-step explanation:

logan has 1.95

step by step explanation:

a dime can not be more than 10 c

and less than 10c

a quarter can not be more than 25c

and less than 25 c

so logan must have 17 dimes and 1 quarter

key = 1 dime = 10c, 1 quarter = 25c

$1.70 + 25c = $1.95

hope this helps!!

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Answer:

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are\ in\ AP

Step-by-step explanation:

Given that (b-c)^2, (c-a)^2 , (a-b)^2 are in AP

To prove: \dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)} are in AP

From given as we know if p , q, r are in AP then 2q= p+r.

2(c-a)^2= (b-c)^2+(a-b)^2\\\\\Rightarrow 2(c^2+a^2-2ac)=b^2+c^2-2bc+a^2+b^2-2ab\\\\\Rightarrow 2c^2+2a^2-4ac= 2b^2+c^2+a^2 -2bc-2ab\\\\\Rightarrow a^2+c^2-2b^2-4ac= -2bc-2ab\\\\\Rightarrow a^2-2b^2+c^2= 4ac-2bc-2ab

Now

\dfrac{1}{(b-c)},\dfrac{1}{(c-a)} ,\dfrac{1}{(a-b)}2\dfrac{1}{(c-a)} =\dfrac{1}{(b-c)}+\dfrac{1}{(a-b)}\\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-b+b-c}{(b-c)(a-b)} \\\\\Rightarrow \dfrac{2}{(c-a)}= \dfrac{a-c}{(b-c)(a-b)} \\\\\Rightarrow2(b-c)(a-b) = (c-a)(a-c) \\\\\Rightarrow 2(ab-b^2-ac+bc)= -(a-c)^2\\\\\Rightarrow 2ab- 2b^2-2ac+2bc = -a^2-c^2+2ac\\\\\Rightarrow a^2-2b^2+c^2=4ac-2ab-2bc

Which is the result of AP

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Hence proved

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