Eight dollars more than m algebraic expression
1 answer:
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Answer:
C. 1000 loss.
Step-by-step explanation:
Let x represent purchase price of each car.
We have been given that a used-car dealer sold one car at a profit of 25 percent of the dealer's purchase price for that car and sold another car at a loss of 20 percent of the dealer's purchase price for that car.
We can represent the car that dealer sold with 25% profit to find purchase price as:
Therefore, the purchase price of 1st car was $16,000.
We can represent the car that dealer sold with 20% loss to find purchase price as:
Therefore, the purchase price of 2nd car was $25,000.
The total purchase price of both cars would be
The total sale price of both cars .
We can see that the sale price of both car is less than purchase price by $1000, so the dealer got a loss of $1000.
Therefore, the dealer's total loss, in dollars, for the two transactions combined was 1000 and option C is the correct choice.
So we just do
cost/number of people=cost per 1 person
so
(6p)/(3p^2)
remember that you can split it up and make ones
=
times 
=
times times 
= times 1 times 1=2/p
easy way is
remember that
(x^m)/(x^n)=x^(m-n)
so
(6p)(3p^2)=6/(3p=2/p=2p^-1
each member pays 2/p
cost/number of people=cost per 1 person
so
(6p)/(3p^2)
remember that you can split it up and make ones
easy way is
remember that
(x^m)/(x^n)=x^(m-n)
so
(6p)(3p^2)=6/(3p=2/p=2p^-1
each member pays 2/p