Answer:
=4x6−x3+x−3
Step-by-step explanation:
Answer:
D = L/k
Step-by-step explanation:
Since A represents the amount of litter present in grams per square meter as a function of time in years, the net rate of litter present is
dA/dt = in flow - out flow
Since litter falls at a constant rate of L grams per square meter per year, in flow = L
Since litter decays at a constant proportional rate of k per year, the total amount of litter decay per square meter per year is A × k = Ak = out flow
So,
dA/dt = in flow - out flow
dA/dt = L - Ak
Separating the variables, we have
dA/(L - Ak) = dt
Integrating, we have
∫-kdA/-k(L - Ak) = ∫dt
1/k∫-kdA/(L - Ak) = ∫dt
1/k㏑(L - Ak) = t + C
㏑(L - Ak) = kt + kC
㏑(L - Ak) = kt + C' (C' = kC)
taking exponents of both sides, we have

When t = 0, A(0) = 0 (since the forest floor is initially clear)


So, D = R - A =

when t = 0(at initial time), the initial value of D =

Step-by-step explanation:
12y-(-4-8y)
= 12y +4 + 8y. ['-' getting multiplied when brackets open]
= 12y + 8y +4. [ Adding the like terms ]
= <u>20 y + 4 (Ans)</u>
Answer:
Two angles are said to be supplementary if they add up to 180 degrees. So the supplement of an angle is obtained by subtracting it from 180. i.e., if x is an angle, its supplement is 180−x
Answer:
AB , AD , EF , EH all intersect AE ⇒ answer (c)
Step-by-step explanation:
∵ AE segment passing through A and E
∴ Each segment passing through A and E intersects AE
∴ AB intersects AE at A
∴ AD intersects AE at A
∴ EF intersects AE at E
∴ EH intersects AE at E
∴ The answer is (c)
Step-by-step explanation: