Set of equations that can be used to calculate rate for each plumber:
2A+8B+8C = 1,400 --- (1)
4A+7B+10C = 1,660 --- (2)
3A+9B+9C = 1,660 --- (3)
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2*(1) - (2)
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4A+16B+16C = 2,800
4A+7B+10C = 1,660 -
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9B+6C = 1,140 --- (4)
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3(2) -4(3)
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12A+21B+30C = 4,980
12A+36B+36C = 6,600 -
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-15B-6C = -1,620 --- (5)
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(4) + (5)
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9B+6C = 1140
-15B-6C = -1620 +
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-6B = -480 => 6B = 480 => B = 480/6 = 80
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Using (4), 9(80)+6C = 1140
720+6C = 1140 => 6C = 1140-720 = 420 => C = 420/6 = 70
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Using (1), 2A+8(80)+8(70) = 1400
2A+640+560 =1400 => 2A = 1400-640-560 = 200 => A = 200/2 = 100
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The rates are:
A = $100
B = $80
C = $70
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On Thursday, number of calls: A = 4 hrs, B = 6 hrs, C = 3 hrs
Money earned = 4*100+6*80+3*70 = $1,090
Answer:
a) 38"
b) L = 5.7'
Step-by-step explanation:
a) Given
P = 9,500 lb/ft
Thick = 18"
We assume an allowable bearing pressure of σ = 3000 lb/sf
If Stress = Force / Area
σ = P/A
Solving for area
A = P/σ ⇒ A = 9,500 lb/ft/3000 lb/sf
⇒ A = 3.167 sf, or 3.167 ft wide, per foot of length.
⇒ A = 38" wide
We can see the pic 1 in order to understand the answer.
b) Given
P = 65,000 pounds = 65,000 lb
Thick = 18" = 1.5'
We assume an allowable bearing pressure of σ = 2000 lb/sf
If Stress = Force / Area
σ = P/A
Solving for area
A = P/σ ⇒ A = 65,000 lb/2000 lb/sf
⇒ A = 32.5 sf
then A = L² ⇒ L = √A = √(32.5 sf) = 5.7 ft
Finally L = 5.7'
We can see the pic 2 in order to understand the answer.
Answer:
Step-by-step explanation:
Let the length is x, then the width is 1/4x.
<u>Find the perimeter of rectangle:</u>
- P = 2(l + w)
- P = 2(x + 1/4x) = 2x + 1/2x = 2 1/2x = 2.5x
Step-by-step explanation:
According to this trigonometric function, −C gives you the OPPOSITE terms of what they really are, so be EXTREMELY CAREFUL:
![\displaystyle Phase\:[Horisontal]\:Shift → \frac{\frac{π}{3}}{3} = \frac{π}{9} \\ Period → \frac{2}{3}π](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7B%5Cfrac%7B%CF%80%7D%7B3%7D%7D%7B3%7D%20%3D%20%5Cfrac%7B%CF%80%7D%7B9%7D%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7B3%7D%CF%80)
Therefore we have our answer.
Extended Information on the trigonometric function
![\displaystyle Vertical\:Shift → D \\ Phase\:[Horisontal]\:Shift → \frac{C}{B} \\ Period → \frac{2}{B}π \\ Amplitude → |A|](https://tex.z-dn.net/?f=%5Cdisplaystyle%20Vertical%5C%3AShift%20%E2%86%92%20D%20%5C%5C%20Phase%5C%3A%5BHorisontal%5D%5C%3AShift%20%E2%86%92%20%5Cfrac%7BC%7D%7BB%7D%20%5C%5C%20Period%20%E2%86%92%20%5Cfrac%7B2%7D%7BB%7D%CF%80%20%5C%5C%20Amplitude%20%E2%86%92%20%7CA%7C)
NOTE: Sometimes, your vertical shift might tell you to shift your graph below or above the <em>midline</em> where the amplitude is.
I am joyous to assist you anytime.
