[PLEASE HELP! WILL GIVE BRAINLIEST IF CORRECT!]
2 answers:
<h2>Answer:</h2>
Option: C is the correct answer.
C: x-axis: minutes in increments of 1; y-axis: temperature in increments of 1
<h2>Step-by-step explanation:</h2>As we know that x-axis is used to denote the independent variable and y-axis is used to represent the dependent variable.
In the given question the temperature is recorded over a period of time.
This means that the x-axis will represent the time in minutes and y-axis will represent the temperature.
Also, we are given a table as:
Minutes Temperature (°C)
0 -2
1 1
2 3
3 4
4 5
5 -4
6 2
7 -3
Hence, we see that the minutes are increasing by every 1 unit and the temperature also has a very small change and are integers, so we will give a increment of 1 in the y-axis as well.
By doing so the graph would be clear and easily understood.
Hence, the answer is:
C: x-axis: minutes in increments of 1; y-axis: temperature in increments of 1
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Remark
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.
If you don't start exactly the right way, you can get into all kinds of trouble. This is just one of those cases. I think the best way to start is to divide both terms by x^(1/2)
Step One
Divide both terms in the numerator by x^(1/2)
y= 6x^(1/2) + 3x^(5/2 - 1/2)
y =6x^(1/2) + 3x^(4/2)
y = 6x^(1/2) + 3x^2 Now differentiate that. It should be much easier.
Step Two
Differentiate the y in the last step.
y' = 6(1/2) x^(- 1/2) + 3*2 x^(2 - 1)
y' = 3x^(-1/2) + 6x I wonder if there's anything else you can do to this. If there is, I don't see it.
I suppose this is possible.
y' = 3/x^(1/2) + 6x
y' =
Frankly I like the first answer better, but you have a choice of both.