A) Perimeter of Trapezium = 8+5+6+15.5
= 34.5in
Area= 1/2*(a+b)*h
=1/2*(8+15.5)*4
=47in^2
B)Perimeter= 12+6+8.5
=26.5mm
Area=1/2*b*h
=1/2*12*4
=24mm^2
Answer:
Step-by-step explanation:
She must write an email to the new employees explaining how to access documents that need editing and what to do with new versions of the documents.
Answer:
$6.8
Step-by-step explanation:
suppose Molly's account balance = X
25 + 2.2 = X * 4
27.2 = X*4
27.2/4 = X = 6.8
Answer: 1. X+2
2. 6x-3
3. -2x-1
4. 9x5
Step-by-step explanation:
1. X+4-3+1 There is only one variable, so you don’t need to worry about that. Just do all of the numerical operations. 4-3+1= 2. So it would be x+2
2. X-5+5x+2. Again, do the numerical operations first. -5+2=-3. X+5x=6x. So it would be 6x-3.
3. 2x+7-4x-8. 7-8=-1. 2x-4x=-2x. So it would be -2x-1.
4. (4+3)x+2x-5. Do the operation in the parenthesis first. Then it would be (7)x or just 7x+2x-5. Simplify to 9x-5.
Answer:
P_max = 9.032 KN
Step-by-step explanation:
Given:
- Bar width and each side of bracket w = 70 mm
- Bar thickness and each side of bracket t = 20 mm
- Pin diameter d = 10 mm
- Average allowable bearing stress of (Bar and Bracket) T = 120 MPa
- Average allowable shear stress of pin S = 115 MPa
Find:
The maximum force P that the structure can support.
Solution:
- Bearing Stress in bar:
T = P / A
P = T*A
P = (120) * (0.07*0.02)
P = 168 KN
- Shear stress in pin:
S = P / A
P = S*A
P = (115)*pi*(0.01)^2 / 4
P = 9.032 KN
- Bearing Stress in each bracket:
T = P / 2*A
P = T*A*2
P = 2*(120) * (0.07*0.02)
P = 336 KN
- The maximum force P that this structure can support:
P_max = min (168 , 9.032 , 336)
P_max = 9.032 KN