5x’2-4x+7-4x-2
We add the two 4
5x’2-8x+7-2
5x’2-8x+5
The answer is
5x’2 -8x +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
Answer:
the value of (2/3*0)-3 is -3
Step-by-step explanation:
This is a problem in direct variation. For every five minutes that goes by, Liz utters 225 words. The general form of an equation of direct variation is
w = k x, where w is the # of words, k is the constant of proportionality and x is the number of minutes that have elapsed.
Find k by dividing 225 words by 5 minutes, to find the number of words per minute.
Next: How many minutes are there between 10:30 a.m. and 11:15 a.m.? Calculate w=kx, using your constant of proportionality, k, and that number of minutes.
