Answer:
The distance from A to B is 736.2 to the nearest tenth foot
Step-by-step explanation:
In ΔCAB
∵ m∠CAD = 30° ⇒ exterior angle of Δ at vertex A
∴ m∠CAD = m∠ACB + m∠ABC
∵ m∠ABC = 20°
∴ m∠ACB = 30° - 20° = 10°
We will use the sin rule to find the distance AB
∵ 
∴ 
≅ 736.2 to the nearest tenth foot
Answer:
2.29 ft of side length and 1.14 height
Step-by-step explanation:
a) Volume V = x2h, where x is side of square base and h is hite.
Then surface area S = x2 + 4xh because box is open.
b) From V = x2h = 6 we have h = 6/x2.
Substitude in formula for surface area: S = x2 + 4x·6/x2, S = x2 + 24/x.
We get S as function of one variable x. To get minimum we have to find derivative S' = 2x - 24/x2 = 0, from here 2x3 - 24 = 0, x3 = 12, x = (12)1/3 ≅ 2.29 ft.
Then h = 6/(12)2/3 = (12)1/3/2 ≅ 1.14 ft.
To prove that we have minimum let get second derivative: S'' = 2 + 48/x3, S''(121/3) = 2 + 48/12 = 6 > 0.
And because by second derivative test we have minimum: Smin = (12)2/3 + 4(12)1/3(12)1/3/2 = 3(12)2/3 ≅ 15.72 ft2
To find the surface area of the open net, you’ll have to find the area of the faces
So, since there are 6 boxes you’ll have to find the area of the faces and add it up to find the surface area
10x4=40x4=160
3x4=12x2=24
160+24=184
Hope this helps ;)
Answer: 35/3 or 11.6
Step-by-step explanation:
3x-7=28
Add 7 to each side
3x=35
Divide by 3
