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Ghella [55]
3 years ago
14

Oscar charges $25 for the first hour of tutoring and $15 for each additional hour or fraction of an hour. The rates that Latoya

charges for x hours of tutoring are modeled with the function shown.
Who will charge more for 5.5 hours, Oscar or Latoya? How much more?


Who?

How much more?
Mathematics
1 answer:
Hatshy [7]3 years ago
8 0

Answer:

Layota

$47.5

Step-by-step explanation:

Let number of hours = x

Given that:

Oscar's charges :

First hour = $25

Additional hours = $15

Thus :

Total charge : 25 + 15(x - 1)

The piece wise equation given for Layota's charge rate :

First 3 hours = $30

Next 3 hours = $20

Hours greater Than 6 = $10

If total hours ; x = 5.5 hours

Oscar :

$25 + $15(5.5 - 1)

$25 + $15(4.5)

$25 + $67.5

= $92.5

Layota:

First 3 hours :

$30 * 3 = $90

Hours left :

$20 * (5.5 - 3)

$20 * 2.5 = 50

Total = $90 + $50 = $140

Layota charges more : 140 > 92.5

Difference :

$140 - $92.5

= $47.5

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Sally bought some candy bars for $1.25 each and some sodas for $1.49 each. If she bought a total of 38 items and spent $50.38 ho
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Answer:

26 candy bars and 12 sodas

Step-by-step explanation:

This problem can be solved by a simple system of equations

I am going to call x the number of candy bars that Sally bought and y the number of sodas that Sally bought.

The fact that she bought 38 itens means the x + y = 38.

She spent $50.38. The value she spent is the sum of the number of candy bars multplied by the value of the candy bar and the number of sodas multiplied by value of a soda. So 1.25x + 1.49y = 50.38

So we have to solve the following system:

1)x + y = 38

2)1.25x + 1.49y = 50.38

From 1), we have that x = 38 - y. I am going to replace it in 2)

1.25(38-y) + 1.49y = 50.38

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6 0
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-5/6x = -10/3<br> solve for x
ololo11 [35]

Answer:

x = 4

Step-by-step explanation:

First you need to multiply both sides of the equation by -6/5

-6/5 × (-5/6x) = -6/5 × -10/3

Then you need to calculate and reduce. First, you'll reduce the numbers with the greatest common divisor, 6.

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