Answer:
6
Step-by-step explanation:
To solve this problem, you simply plug 2 in for x in the equation of f(x):
f(2) = 2 + 4 = 6
Thus, the answer is 6.
Hope this helps!
Answer:
All of the above
Step-by-step explanation:
dy/dt = y/3 (18 − y)
0 = y/3 (18 − y)
y = 0 or 18
d²y/dt² = y/3 (-dy/dt) + (1/3 dy/dt) (18 − y)
d²y/dt² = dy/dt (-y/3 + 6 − y/3)
d²y/dt² = dy/dt (6 − 2y/3)
d²y/dt² = y/3 (18 − y) (6 − 2y/3)
0 = y/3 (18 − y) (6 − 2y/3)
y = 0, 9, 18
y" = 0 at y = 9 and changes signs from + to -, so y' is a maximum at y = 9.
y' and y" = 0 at y = 0 and y = 18, so those are both asymptotes / limiting values.
9 hope that this helps use a online calculator it’s easier !!
14/25 = 56/100
29/50 = 58/100
53/100
13/20 = 65/100
3/5 = 60/100
order least to greatest:
53/100, 14/25, 29/50, 3/5, 13/20
Answer:
i)
Find the attached
ii)
The mathematical model that best fits the data is;
y = 7.19 + 12.8 ln x
Step-by-step explanation:
i)
A scatter-plot can easily be constructed using applications such as Ms. Excel and Stat-Crunch.
In Ms. Excel we first enter the data in any two adjacent columns. Next, highlight the data, then click the insert ribbon and select the scatter-plot option.
Excel returns a scatter-plot chart as shown in the attachment below.
ii)
After obtaining the scatter-plot, we shall need to add a trend line in order to determine the mathematical model that best fits the data given.
Click anywhere inside the chart, then select the design tab under chart tools. Click on the Add Chart element in the upper left corner of the excel workbook and select more trend-line options. This feature will enable us to fit any trend-line to our data.
Select any trend line option ensuring you check the boxes; Display Equation on chart and Display R-squared value on chart.
Find the attached for the various trend-lines.
The mathematical model that best fits the data is;
y = 7.19 + 12.8 ln x
Since it has the largest R-squared value of 0.9905