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.
I don't have the answer for the second one, but the answer to number 1 is 9.
6÷2(1 + 2) =
6÷2(3) =
3(3) =
9
Answer:
B
Step-by-step explanation:
The equation of a circle in standard form is
(x - h)² + (y - k)² = r²
where (h, k) are the coordinates of the centre and r is the radius
To obtain this form use the method of completing the square
Given
x² - 8x + y² - 2y - 8 = 0 ( add 8 to both sides )
x² - 8x + y² - 2y = 8
add ( half the coefficient of the x/y terms )² to both sides
x² + 2(- 4)x + 16 + y² + 2(- 1)y + 1 = 8 + 16 + 1
(x - 4)² + (y - 1)² = 25 → B
Answer:
Step-by-step explanation:
Given that you have eight cards. Five are green and three are yellow. The five green cards are numbered 1, 2, 3, 4, and 5. The three yellow cards are numbered 1, 2, and 3. The cards are well shuffled. You randomly draw one card.
G = card drawn is green
Y = card drawn is yellow
E = card drawn is even-numbered
List:
Sample space = {G1, G2, G3, G4, G5, Y1, Y2, Y3}
2) P(G) = 5/8
3) P(G/E) = P(GE)/P(E)
GE = {G2, G4}
Hence P(G/E) = 2/5
4) GE = {G2, G4}
P(GE) = 2/8 = 1/4
5) P(G or E) = P(G)+P(E)-P(GE)
= 5/8 + 3/8-2/8 = 3/5
6) No there is common element as G2 and G4
Cannot be mutually exclusive
