QuestionA cab charges \ $4 per mile, with an arithmetic fare. Find fares for 4, 5, and 6 miles:$ \mathrm{f}(4) $,$ \mathrm{f}(5) $,$ \mathrm{f}(6)$. Write a recursive rule.

Studdy Solution

STEP 1

Assumptions1. The fare for1 mile is $4. The fare for miles is$ 7.503. The fare for3 miles is $114. The fare increases in an arithmetic sequence

STEP 2

First, we need to find the common difference of the arithmetic sequence. We can do this by subtracting the fare for1 mile from the fare for2 miles.
$d = Fare\, for\,2\, miles - Fare\, for\,1\, mile$

STEP 3

Now, plug in the given values for the fares to calculate the common difference.
$d = \$7.50 - \$$

STEP 4

Calculate the common difference.
$d = \$7.50 - \$4 = \$3.50$

STEP 5

Now that we have the common difference, we can find the fare for4 miles. This can be done by adding the common difference to the fare for3 miles.
$are\, for\,4\, miles = Fare\, for\,3\, miles + d$

STEP 6

Plug in the values for the fare for3 miles and the common difference to calculate the fare for4 miles.
$are\, for\,4\, miles = \$11 + \$3.50$

STEP 7

Calculate the fare for4 miles.
$are\, for\,4\, miles = \$11 + \$3.50 = \$14.50$

STEP 8

Similarly, we can find the fare for5 miles by adding the common difference to the fare for4 miles.
$are\, for\,5\, miles = Fare\, for\,4\, miles + d$

STEP 9

Plug in the values for the fare for4 miles and the common difference to calculate the fare for5 miles.
$are\, for\,5\, miles = \$14.50 + \$3.50$

STEP 10

Calculate the fare for5 miles.
$are\, for\,5\, miles = \$14.50 + \$3.50 = \$18$

STEP 11

Finally, we can find the fare for6 miles by adding the common difference to the fare for5 miles.
$are\, for\,6\, miles = Fare\, for\,5\, miles + d$

STEP 12

Plug in the values for the fare for5 miles and the common difference to calculate the fare for6 miles.
$are\, for\,6\, miles = \$18 + \$.50$

STEP 13

Calculate the fare for6 miles.
$are\, for\,6\, miles = \$18 + \$3.50 = \$21.50$ The meter will display $.50 for miles,$ 18 for5 miles, and $21.50 for6 miles.
The recursive rule for the context is $f(n) = f(n-) + \$3.50$ where n is the number of miles and f(n) is the fare for n miles.

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QuestionA cab charges \ $4 per mile, with an arithmetic fare. Find fares for 4, 5, and 6 miles:$ \mathrm{f}(4) $,$ \mathrm{f}(5) $,$ \mathrm{f}(6)$. Write a recursive rule.

Studdy Solution

STEP 1

Assumptions1. The fare for1 mile is $4. The fare for miles is$ 7.503. The fare for3 miles is $114. The fare increases in an arithmetic sequence

STEP 2

First, we need to find the common difference of the arithmetic sequence. We can do this by subtracting the fare for1 mile from the fare for2 miles.
$d = Fare\, for\,2\, miles - Fare\, for\,1\, mile$

STEP 3

Now, plug in the given values for the fares to calculate the common difference.
$d = \$7.50 - \$$

STEP 4

Calculate the common difference.
$d = \$7.50 - \$4 = \$3.50$

STEP 5

Now that we have the common difference, we can find the fare for4 miles. This can be done by adding the common difference to the fare for3 miles.
$are\, for\,4\, miles = Fare\, for\,3\, miles + d$

STEP 6

Plug in the values for the fare for3 miles and the common difference to calculate the fare for4 miles.
$are\, for\,4\, miles = \$11 + \$3.50$

STEP 7

Calculate the fare for4 miles.
$are\, for\,4\, miles = \$11 + \$3.50 = \$14.50$

STEP 8

Similarly, we can find the fare for5 miles by adding the common difference to the fare for4 miles.
$are\, for\,5\, miles = Fare\, for\,4\, miles + d$

STEP 9

Plug in the values for the fare for4 miles and the common difference to calculate the fare for5 miles.
$are\, for\,5\, miles = \$14.50 + \$3.50$

STEP 10

Calculate the fare for5 miles.
$are\, for\,5\, miles = \$14.50 + \$3.50 = \$18$

STEP 11

Finally, we can find the fare for6 miles by adding the common difference to the fare for5 miles.
$are\, for\,6\, miles = Fare\, for\,5\, miles + d$

STEP 12

Plug in the values for the fare for5 miles and the common difference to calculate the fare for6 miles.
$are\, for\,6\, miles = \$18 + \$.50$

STEP 13

Calculate the fare for6 miles.
$are\, for\,6\, miles = \$18 + \$3.50 = \$21.50$ The meter will display $.50 for miles,$ 18 for5 miles, and $21.50 for6 miles.
The recursive rule for the context is $f(n) = f(n-) + \$3.50$ where n is the number of miles and f(n) is the fare for n miles.

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QuestionA cab charges \4permile,withanarithmeticfare.Findfaresfor4,5,and6miles:4 per mile, with an arithmetic fare. Find fares for 4, 5, and 6 miles: 4permile,withanarithmeticfare.Findfaresfor4,5,and6miles:\mathrm{f}(4),, ,\mathrm{f}(5),, ,\mathrm{f}(6)$. Write a recursive rule.

Studdy Solution

STEP 1

Assumptions1. The fare for1 mile is 4.Thefareformilesis4. The fare for miles is 4.Thefareformilesis7.503. The fare for3 miles is $114. The fare increases in an arithmetic sequence

STEP 2

First, we need to find the common difference of the arithmetic sequence. We can do this by subtracting the fare for1 mile from the fare for2 miles.d=Fare for 2 miles−Fare for 1 miled = Fare\, for\,2\, miles - Fare\, for\,1\, miled=Farefor2miles−Farefor1mile

STEP 3

Now, plug in the given values for the fares to calculate the common difference.d=$7.50−$d = \$7.50 - \$d=$7.50−$

STEP 4

Calculate the common difference.d=$7.50−$4=$3.50d = \$7.50 - \$4 = \$3.50d=$7.50−$4=$3.50

STEP 5

Now that we have the common difference, we can find the fare for4 miles. This can be done by adding the common difference to the fare for3 miles.are for 4 miles=Fare for 3 miles+dare\, for\,4\, miles = Fare\, for\,3\, miles + darefor4miles=Farefor3miles+d

STEP 6

Plug in the values for the fare for3 miles and the common difference to calculate the fare for4 miles.are for 4 miles=$11+$3.50are\, for\,4\, miles = \$11 + \$3.50arefor4miles=$11+$3.50

STEP 7

Calculate the fare for4 miles.are for 4 miles=$11+$3.50=$14.50are\, for\,4\, miles = \$11 + \$3.50 = \$14.50arefor4miles=$11+$3.50=$14.50

STEP 8

Similarly, we can find the fare for5 miles by adding the common difference to the fare for4 miles.are for 5 miles=Fare for 4 miles+dare\, for\,5\, miles = Fare\, for\,4\, miles + darefor5miles=Farefor4miles+d

STEP 9

Plug in the values for the fare for4 miles and the common difference to calculate the fare for5 miles.are for 5 miles=$14.50+$3.50are\, for\,5\, miles = \$14.50 + \$3.50arefor5miles=$14.50+$3.50

STEP 10

Calculate the fare for5 miles.are for 5 miles=$14.50+$3.50=$18are\, for\,5\, miles = \$14.50 + \$3.50 = \$18arefor5miles=$14.50+$3.50=$18

STEP 11

Finally, we can find the fare for6 miles by adding the common difference to the fare for5 miles.are for 6 miles=Fare for 5 miles+dare\, for\,6\, miles = Fare\, for\,5\, miles + darefor6miles=Farefor5miles+d

STEP 12

Plug in the values for the fare for5 miles and the common difference to calculate the fare for6 miles.are for 6 miles=$18+$.50are\, for\,6\, miles = \$18 + \$.50arefor6miles=$18+$.50

STEP 13

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QuestionA cab charges \ $4 per mile, with an arithmetic fare. Find fares for 4, 5, and 6 miles:$ \mathrm{f}(4) $,$ \mathrm{f}(5) $,$ \mathrm{f}(6)$. Write a recursive rule.

Assumptions1. The fare for1 mile is $4. The fare for miles is$ 7.503. The fare for3 miles is $114. The fare increases in an arithmetic sequence

First, we need to find the common difference of the arithmetic sequence. We can do this by subtracting the fare for1 mile from the fare for2 miles.
$d = Fare\, for\,2\, miles - Fare\, for\,1\, mile$

Now, plug in the given values for the fares to calculate the common difference.
$d = \$7.50 - \$$

Calculate the common difference.
$d = \$7.50 - \$4 = \$3.50$

Now that we have the common difference, we can find the fare for4 miles. This can be done by adding the common difference to the fare for3 miles.
$are\, for\,4\, miles = Fare\, for\,3\, miles + d$

Plug in the values for the fare for3 miles and the common difference to calculate the fare for4 miles.
$are\, for\,4\, miles = \$11 + \$3.50$

Calculate the fare for4 miles.
$are\, for\,4\, miles = \$11 + \$3.50 = \$14.50$

Similarly, we can find the fare for5 miles by adding the common difference to the fare for4 miles.
$are\, for\,5\, miles = Fare\, for\,4\, miles + d$

Plug in the values for the fare for4 miles and the common difference to calculate the fare for5 miles.
$are\, for\,5\, miles = \$14.50 + \$3.50$

Calculate the fare for5 miles.
$are\, for\,5\, miles = \$14.50 + \$3.50 = \$18$

Finally, we can find the fare for6 miles by adding the common difference to the fare for5 miles.
$are\, for\,6\, miles = Fare\, for\,5\, miles + d$

Plug in the values for the fare for5 miles and the common difference to calculate the fare for6 miles.
$are\, for\,6\, miles = \$18 + \$.50$