Answer:
Shannon run 1.5 km more on Friday than on Saturday.
Step-by-step explanation:
From the given table
- Distance run on Friday = 4/2 = 2 km
- Distance run on Saturday = 1/2 = 0.5 km
- In order to run how many more kilometers Shannon run on Friday than on Saturday, we need to subtract the distance run on Saturday from the distance run on Friday.
i.e.
Friday run - Saturday run = 2 - 0.5
= 1.5 km
Thus, Shannon run 1.5 km more on Friday than on Saturday.
Answer:
6.25n + 3.50
$34.75
Step-by-step explanation:
Break down the important information given in the problem.
The one-time delivery fee is 3.50. This is only paid one time an never again, making it the <u>constant</u>, a number that does not change.
Each lunch costs 6.25. This amount will increase depending on how many times Mr. Jackson orders lunches, "n" times. This number is the <u>rate</u> because it changes. The rate is attached to the variable.
If you add the amounts together, that is the total cost of ordering lunches.
6.25n + 3.50
(Reember expressions do not have the equal sign).
To find the cost of ordering 5 lunches, use the expression. Substitute "n" for 5 because "n" represents the number of lunches ordered.
6.25n + 3.50
= 6.25(5) + 3.50 Simplify by multiplying 5 and 6.25
= 31.25 + 3.50 Add the two values
= 34.75 Total cost of 5 lunches
Therefore the cost of ordering 5 lunches is $34.75.
Answer:
Option D) F
Step-by-step explanation:
we have
-----> inequality A
The solution of the inequality A is the shaded area below the dashed line 
The y-intercept of the dashed line is (0,10)
The x-intercept of the dashed line is (5,0)
----> inequality B
The solution of the inequality B is the shaded area below the dashed line 
The y-intercept of the dashed line is (0,-2)
The x-intercept of the dashed line is (4,0)
The solution of the system of inequalities is the shaded area between the two dashed lines
see that attached figure
Remember that
If a ordered pair is a solution of the system of inequalities, then the ordered pair must lie on the shaded area of the solution
therefore
The solution are the points
E, F and G
