Answer:

Step-by-step explanation:
For any angle
, we have that:

Quadrant:
means that
is in the first quadrant. This means that both the sine and the cosine have positive values.
Find the cosine:







Since the angle is in the first quadrant, the cosine is positive.

Answer:
14d + 12c -4
Step-by-step explanation:
Answer:
4485
Step-by-step explanation:
1679+2806= 4485
Hope this helps! :)
Answer:
The wedge cut from the first octant ⟹ z ≥ 0 and y ≥ 0 ⟹ 12−3y^2 ≥ 0 ⟹ 0 ≤ y ≤ 2
0 ≤ y ≤ 2 and x = 2-y ⟹ 0 ≤ x ≤ 2
V = ∫∫∫ dzdydx
dz has changed from zero to 12−3y^2
dy has changed from zero to 2-x
dx has changed from zero to 2
V = ∫∫∫ dzdydx = ∫∫ (12−3y^2) dydx = ∫ 12(2-x)-(2-x)^3 dx =
24(2)-6(2)^2+(2-2)^4/4 -(2-0)^4/4 = 20
Step-by-step explanation:
Do u mean d?
1.5d +3.25 = 1 +2.25d
3.25 = 1 + .75d -1.5d on both sides
2.25 = .75d subtraction 1 on both sides
3 = d divide both sides by .75