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.
For this case we have the following multiplication:
We can rewrite this multiplication in the following way:
From here, we apply the distributive property.
We have then:
Therefore, Jack's strategy is:
Write the number in an expanded form and then apply the distributive property.
Answer:
Jack's strategy is:
Write the number in an expanded form and then apply the distributive property.
Hello.
C) x=4π/3
The variable x in the cotangent argument has a unit coefficient, so the period is π, just as it is in the parent function cot(x).
Can you graph y = cot(x)? By subtracting the constant π/6 from the argument, that graph is translated to the right by π/6. Just as with cot(x), it is decreasing everywhere.
Have a nice day
The answer is A
there is not enough hours to build everything and A has the right equation
Answer:
log9
Step-by-step explanation:
hello :
note : a>0 and b>0
1) log(ab) = loga+logb
2) loga^n = nloga.....n in N
log(9x^5) + 5 log (1/x) = log(9x^5) + log (1/x)^5 = log((9x^5) (1/x)^5)
= log(9x^5)/(x^5)) = log9 because : (1/x)^5 = 1/x^5
