[Students: Please create yourself a little Excel sheet for calculating AFS and HFS. These are sort of tedious calculations that you will have to do many times]. Just underneath the fresh litter-layer on the forest floor of a Spodosol, is an O horizon that is 2.2 inches thick. This O horizon is dominated by well-decomposed soil organic matter (SOM), and the SOM content of this O horizon is 92.2 % by weight. Further data indicate this soil organic matter contains 5.3 % nitrogen (N) by weight. What is the approximate weight (in pounds N) of nitrogen in the organic matter of the AFS for this O horizon that is 2.2 inches thick, with a bulk density of 0.52 grams per cm3 (43,560 square feet per acre, 62.4 pounds per cubic foot of water)

it has changed from the width and the height.

step-by-step explanation:

18.75 gallons

step-by-step explanation:

answer: first option i.e. 'a' is correct.

step-by-step explanation:

since we have given that

number of minutes to wrap a small gift box = 4 minutes

we need to find the number of gifts to be wrapped in 5 minutes

since there is a direct variation, as the number of minutes will increase with   number of gifts to be wrapped.

so, let the number of gifts to be wrapped in 5 minutes be x.

so, it becomes,

hence, first option 'a'   is correct.

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

the factor is 2

step-by-step explanation:

let f(x)=8(3^(x−2)+2)

to stretch a graph by 3, it would be 3f(x)

g(x) =3f(x)

g(x) = 3*(8(3^(x−2))+2)

g(x) = 24 (3^(x-2) )+6

x     −2     0     2     4     6

f(x) 1/64 1/16   1/4     1     4

1/64 * z * z = 1/16

i multiplies by z twice because it its 2 steps from -2 to 0

1/64 z^2 = 1/16

multiply by 64 to clear the fraction on the left side

1/4 * 64 z^2 =1/16 *64

z^2 = 4

take the square root of each side

sqrt(z^2) = ±sqrt(4)

z = ±2  

we are multiplying by 2 each time.