MATH SOLVE

3 months ago

Q:
# hi,i need help with figuring out how to solve real world word problems with polynomial and integer expressions.Thankyou

Accepted Solution

A:

Part a.

Fixed charge for the month: $30

Charge per hour: $0.50 daytime

Charge per hour: $0.25 nights and weekends

Let's say in one month, a person parks for d hours of daytime and n hours of nights and weekends.

The total monthly charge would be

cost = 30 + 0.50d + 0.25n

Now let's see what Trent did.

He parked for 47 hours in one month.

h of the 47 hours are nights and weekends.

Let x = number of daytime hours.

x + h = 47

x = 47 - h

He parked h hours of night and weekends, and he parked 47 - h hours of daytime.

Now we use h for night and weekend hours and 47 - h for daytime hours in the expression above.

cost = 30 + 0.50d + 0.25n

cost = 30 + 0.50(47 - h) + 0.25h

Answer to part a.: 30 + 0.50(47 - h) + 0.25h

Part b.

We are told the actual number of night and weekend hours, which we called h above, is 12. h = 12.

Now we use the cost expression we found in part a. with 12 in for h.

cost = 30 + 0.50(47 - h) + 0.25h

cost = 30 + 0.50(47 - 12) + 0.25(12)

cost = 30 + 0.50(35) + 0.25(12)

cost = 30 + 17.50 + 3

cost = 50.5

Answer to part b.: $50.50

Fixed charge for the month: $30

Charge per hour: $0.50 daytime

Charge per hour: $0.25 nights and weekends

Let's say in one month, a person parks for d hours of daytime and n hours of nights and weekends.

The total monthly charge would be

cost = 30 + 0.50d + 0.25n

Now let's see what Trent did.

He parked for 47 hours in one month.

h of the 47 hours are nights and weekends.

Let x = number of daytime hours.

x + h = 47

x = 47 - h

He parked h hours of night and weekends, and he parked 47 - h hours of daytime.

Now we use h for night and weekend hours and 47 - h for daytime hours in the expression above.

cost = 30 + 0.50d + 0.25n

cost = 30 + 0.50(47 - h) + 0.25h

Answer to part a.: 30 + 0.50(47 - h) + 0.25h

Part b.

We are told the actual number of night and weekend hours, which we called h above, is 12. h = 12.

Now we use the cost expression we found in part a. with 12 in for h.

cost = 30 + 0.50(47 - h) + 0.25h

cost = 30 + 0.50(47 - 12) + 0.25(12)

cost = 30 + 0.50(35) + 0.25(12)

cost = 30 + 17.50 + 3

cost = 50.5

Answer to part b.: $50.50