8q² + 4qm + 6q + 3m
4q(2q) + 4q(m) + 3(2q) + 3(m)
4q(2q + m) + 3(2q + m)
(4q + 3)(2q + m)
Answer:

Step-by-step explanation:
I will assume that the shaded region is the fire pit
Given


Required
Determine the area of the fire pit
First, calculate the area of the big circle.

Area of the small is:

The area of the pit is:







Answer:
-814°, -454°, -94°, 266°, 626°. Did you mean like that?
Answer:
The answer to the question is
The ratio of the two gas pressures
, that is Px to Py = 1/6
Step-by-step explanation:
Let the gases Volumes be V₁ and V₂
Where volume of X = V₁ and
volume of Y = V₂
The volume of Y is half the volume of X
∴ V₂ =
× V₁
Let the number of moles be n₁ and n₂ in X and Y respectively
therefore n₂ = 3 × n₁
The pressure of the gas in X is Pₓ and the pressure of the gas in Y is
then we have
P₁ × V₁ = n₁ × R × T₁ , and P₂ × V₂ = n₂ × R × T₂
(P₁ × V₁)/(n₁ × T₁) = (P₂ × V₂)/(n₂ × T₂)
but T₁ = T₂
Therefore
(P₁ × V₁)/n₁ = (P₂ × V₂)/n₂. However n₂ = 3 × n₁ and V₂ =
× V₁ therefore substituting in the equation we have
(P₁ × V₁)/n₁ = (P₂ ×
× V₁ )/(3 × n₁) from where
P₁ /P₂ = (
× V₁ × n₁)/(V₁×3 × n₁) =0.5/3 = 1/6
The ratio of
= 1/6