I think it would be B but I'm not sure
Answer: You would put the 35° and the 4x+7° equal to each other.
Step-by-step explanation:
1. 4x+7=35
2. 4x+7=35 -> Subtract 7 from the positive 7 and the 35
-7 -7
3. 4x=28 -> Divide 4x and 28
4. x=7
Answer:
m∠BAC = 105°
m∠FAB = 75°
Step-by-step explanation:
By using the property of an exterior angle of a triangle,
Measure of an exterior angle is equal to the sum of opposite two angles of a triangle.
From the triangle given in the picture,
m∠ABC + m∠BCA + m∠CAB = 180°
(13x - 3)° = (3x + 2)° + 55°
13x - 3 = 3x + 57
13x - 3x = 57 + 3
10x = 60
x = 6
m∠FAB = (13x - 3)° = 75°
m∠ABC = (3x + 2)° = 20°
Since, ∠BAC and ∠FAB are the linear pair of angles,
m∠BAC + m∠FAB = 180°
m∠BAC + 75° = 180°
m∠BAC = 180° - 75° = 105°
That would be 1/4 * 28 = 7 ins^2
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
