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yuradex [85]
3 years ago
9

Choose all the fractions that are in simplest form. 9/14 7/21 6/15 18/23 8/15

Mathematics
2 answers:
IceJOKER [234]3 years ago
8 0

Answer:

9/14, 18/23 and 8/15

Step-by-step explanation:

The fractions in their simplest forms are:

1) 9/14

2) 18/23

3) 8/15

Rufina [12.5K]3 years ago
7 0

Answer:

9/14 ; 18/23 ; 8/15

Step-by-step explanation:

7/21 = 1/3

6/15 = 2/3

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A student raises her grade from a 75 to a 90 what is the percent of increas in the students grade average
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4 years ago
The cost of a car rental is ​$25 per day plus 22 per mile. You are on a daily budget of ​$90. Write and solve an inequality to f
lara31 [8.8K]

Answer: You can drive, at most, 2.95 miles per day.

x ≤ 2.95

Step-by-step explanation:

In a single day, you can spend at most $90

Then if C represents the cost of renting the car, then we will have the inequality:

C ≤ $90

Now let's find the equation for C.

We know that we have a fixed cost of $25 plus $22 per mile, then if you drive x miles, the total cost will be $25 plus x times $22, or:

C = $25 + x*$22.

We can now replace that in the inequality:

$25 + x*$22 ≤ $90

Now let's isolate the variable x

x*$22 ≤ $90 - $25

x*$22 ≤ $65

x ≤ $65/$22 = 2.95

x ≤ 2.95

You can drive at most, 2.95 miles per day.

To find other inequalities with the same solution we can start with the solution:

x ≤ 2.95

Now let's multiply both sides by a number (the units of the number can be dollars, in that way we can make a similar problem)

Let's multiply both sides by $10:

x*$10 ≤ 2.95*$10 = $29.5

x*$10 ≤ $29.5

Now let's add the same number in both sides, for example, $5.

x*$10 + $5 ≤ $29.5 + $5 = $34.5

x*$10 + $5 ≤ $34.5

We could write this problem as:

"To rent a cab in your city, you have an initial cost of $5, plus $10 for each mile driven. How many miles could you drive if at most you can spend $34.50?"

You could be more creative with the problem, but this is the way in which you can craft problems of this type when you already know the solution.

5 0
3 years ago
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