We are required to find an inequality which best represents the relationship between the number of hours gardening g and the total charge c
The inequality which best represents the relationship between the number of hours gardening g and the total charge c is c ≥ 15 + 12g
At least means greater than or equal to (≥)
fixed charge = $15
charges per hour = $12
Total charge = c
Number of hours = g
The inequality:
<em>Total charge ≥ fixed charge + (charges per hour × Number of hours</em>
c ≥ 15 + (12 × g)
c ≥ 15 + (12g)
c ≥ 15 + 12g
Therefore, the inequality which best represents the relationship between the number of hours gardening g and the total charge c is c ≥ 15 + 12g
Read more:
brainly.com/question/11067755
If you replace the x with -4 and multiple that by 0.5 then minus that from 14 you will get -16
Answer:
1. C) 3
2. D) -1
3. D) 7^2 - 8*2 - 16
4. B) 75
5. B) 6^2 + (2 - 8)*sqrt(81)
Step-by-step explanation:
1. (10 - (6-4)^2)/2
= (10 - 2^2)/2
= (10 - 4)/2
= 6/2
= 3
2. PEMDAS states that Multiplication is before Subtraction
8 - (5^2-7)/2
= 8 - (25-7)/2
= 8 - 18/2
= 8 - 9
= -1
3. D) 7^2 - (8*2) - 16
= 49 - 16 - 16
= 49 - 32
= 17
4. 3(2 + 3)^2
= 3(5)^2
= 3(25)
= 75
5. 6^2 + (2 - 8)*sqrt(81)
= 36 + (-6*9)
= 36 - 54
= -18
the answer is b
Step-by-step explanation:
hope this helps if not let me know have a blessed day
We are given a scenario and asked to choose which graph (described verbally) represents the scenario.
Let us break down the scenario piece by piece.
The first part of the scenario is that Kent walked to the bus station. His speed will be constant so that graph will show a line sloping upward.
Next, Kent waited for the bus. This will be represented by a horizontal line.
Then, he rode the bus. This will be represented by a sloping line but steeper than the first part of the graph since the speed of the bus is greater than Kent's speed.
Finally, Kent walked to work. The graph would still be a sloping line but the slope will be less than the previous part of the graph.
So, the answer is
<span>The line increases for 10 minutes, stays horizontal for 15 minutes, increases rapidly for another 25 minutes, then increases slowly for 5 minutes.</span>
