Answer:
1) Between 0 and 1 minute the rate of change is 25 ft./min
2) Between 1 and 2 minute the rate of change is 0 ft./min
3) Between 2 and 4 minute the rate of change is -12.5 ft./min
Step-by-step explanation:
1) Between 0, and 1 feet, we have;
The rate of change = (Final height - Initial height)/(Final time - Initial time)
The rate of change = (40 - 15)/(1 - 0) = 25 ft/min
Between 0 and 1 minute the rate of change = 25 ft./min
2) Between 1, and 2 feet, we have;
The rate of change = (40 - 40)/(2 - 1) = 0 ft/min
Between 1 and 2 minute the rate of change = 0 ft./min
3) Between 2, and 4 feet, we have;
The rate of change = (15 - 40)/(4 - 2) = -12.5 ft/min
Between 2 and 4 minute the rate of change = -12.5 ft./min
Answer:
y=6
Step-by-step explanation:
x=20
y=10
x=15 > 20x=15
x=15/20
x=3/5
y=10(3/5)
=6
Answer: ax=10 cause u multiply 4x10=40=40
Step-by-step explanation:
I'll proceed by evaluating total length and surface area by scanned images. For measurements, I suggest to use winrhizo or any similar software. In this case the most important thing to think about is to use the appropiate acquisition parameters, I mean, to set a good resolution during scanning. Pixels of the aquired image should be smaller than root hairs diameter. Therefore, let me try to suggest you a good acquisition resolution. Making a quick research, it seems that hair root diameters range between 0,012 and 0,017 mm, so I suggest a resolution that will give a pixel dimension of 0,002 mm. Root hair diameters should be composed by 6-8 pixels, that is enough for winrhizo measurements. Resolution is measured in dpi, so in "dots/pixel per inch". A inch is 25,4 mm. We need pixels with a diameter of 0,002 mm, so in 25,4 mm we found 12700 pixels. That means 12700 dpi. It's a very high resolution, you need a good scanner and much space in your computer.
what would you like me to do with these numbers????
