MTH209 Week 3 Chapter 5 Name: Score:
Show all work, as your work will be graded on both your final answer and on the process that you used. Points will be assigned as follows. Feel free to make the table bigger as needed to fit your work in.
Full credit – Solution correct and work shown where appropriate. Partial credit – Solution correct, but little work shown. Partial credit – Solution incorrect, all work shown where appropriate. No credit – No attempt at a solution or incorrect solution with little work shown.
|Text |Problem |Your Solution |Instructor |
| | | |Notes |
|5.1 #94 |[pic] |We have the area of a rectangle [pic], | |
| |[pic] |to find the length: | |
| | |[pic] Centimeters. | |
|5.1 #96 |[pic] |a) Rewrite A = P + Prt formula by factoring out the greatest | |
| | |common factor on the right-hand side. | |
| | |[pic] | |
| | |b) Find A if $8300 is invested for 3 years at a simple interest | |
| | |rate of 15%. | |
| | |[pic] | |
|5.2 #100 |[pic] |a) Factor out the price on the right-hand | |
| |[pic] |side of the formula. | |
| | |[pic] | |
| | |b) Write a formula [pic]for the monthly | |
| | |demand. | |
| | |The revenue R is the product of the | |
| | |price p and the demand, so the result is: | |
| | |[pic] | |
| | |c) Find [pic] | |
| | |[pic] | |
| | |d) Use the accompanying graph to | |
| | |estimate the price at which the | |
| | |revenue is maximized. Approximately | |
| | |how many pools will be sold monthly | |
| | |at this price? | |
| | |[pic] | |
| | |The revenue is maximized when | |
| | |[pic] | |
| | |When the price is 1875 dollars, the revenue will be maximized. | |
| | |Approximately 150 pools will be sold monthly. | |
| | |e) What is the approximate maximum | |
| | |revenue? | |
| | |[pic] | |
| | |The maximum revenue is approximately | |
| | |281,250 dollars | |
| | |f) Use the accompanying graph to | |
| | |estimate the price at which the | |
| | |revenue is zero. | |
| | |According to the graph, when the price is 0 dollars or 3700 | |
| | |dollars, the revenue will be zero. | |
|5.3 #110 |[pic] |Find A(5) | |
| | | | |
| | |Substitute 5 for x: | |
| | |[pic] | |
| | |Square meters. | |
| | |If the height of the sail is | |
| | |x + 3 meters, then what is the | |
| | |length of the base of the sail? | |
| | |We have the area of a triangle: | |
| | |[pic], the base then is: | |
| | |[pic] | |
| | |The length of the base is | |
| | |2x + 4 meters. | |
|5.4 #106 |[pic] |a) Rewrite the formula by factoring the right-hand side | |
| |[pic] |completely. | |
| | |[pic] | |
| | |b) Use the factored version of the formula to find N(3) | |
| | | | |
| | |[pic] | |
| | | | |
| | |c)Use the accompanying graph to estimate the time at which the | |
| | |workers are most efficient | |
| | |According to the graph, when | |
| | |the time = 5.5, the workers are most efficient. | |
| | | | |
| | |d)Use the accompanying graph to estimate the maximum number of | |
| | |components assembled per hour during an 8-hour shift. According | |
| | |to the graph, the maximum number of components assembled per | |
| | |hour during an 8 hours shift is approximately 245 components. | |
|5.5 #112 |[pic] |a)Find V(10) | |
| | |[pic] | |
| | |=168 cm3 | |
| | | | |
| | |b) If the new width is s - 6 centimeters, then what are the new | |
| | |length and height? | |
| | |[pic] | |
| | |The new length and height are | |
| | |[pic] cm and [pic] cm. | |
| | | | |
| | |c) Find the volume when s =10 by multiplying the length, width, | |
| | |and height. | |
| | |(10-6) = 4 cm | |
| | |(10-3) = 7 cm | |
| | |(10-4) = 6 cm | |
| | |(4)(7)(6) = [pic] | |
| | |Volume is [pic] | |
|5.6 #74 |[pic] |[pic] | |
| | |x= - 8 is out because is negative | |
| | |If [pic], then [pic] meters. | |
| | |Lengths of the sides are 8 and 6 meters. | |
|5.6 #92 |[pic] |a) What is the height of the wrench after 1 | |
| | |second? | |
| | |If [pic], then | |
| | |[pic] | |
| | |b) How long does it take for the wrench to reach the ground? | |
| | |[pic] | |
| | |[pic], is out because this is before the construction worker | |
| | |throws the wrench. Consequently, it takes 2 seconds for the | |
| | |wrench to reach the ground. | |
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Learning Team Log
Each problem should be completed by at least 2 members of your team and checked by everyone on the team. Once the team has decided an answer is correct, place it into the final version of the template.
Problem
|Problem |Team Member #1 |Team Member #2 |Answer |
| |Shannon |Ulises |agreed on by|
| | | |team? |
|5.1 #94 |XY= X^2+ 50X |Area of a painting. A rectangular painting with a width of x centimeters has an area of |#2 |
| |Then I would divide both sides by X |[pic] square centimeters. Find a binomial that represents the length. See the accompanying | |
| |Y=X+50 |figure. | |
| | |The area of a rectangle is given by[pic], so the length is: | |
| | |[pic] Centimeters. | |
|5.1 #96 |a) |Amount of an investment. The amount of an investment of P dollars for t years at simple |#2 |
| | |interest rate r is given by A = P + Prt. | |
| |A=P(1+RT) |a) Rewrite this formula by factoring out the greatest common factor on the right-hand side. | |
| | |[pic] | |
| | |b) Find A if $8300 is invested for 3 years at a simple interest rate of 15%. | |
| |b) |[pic] | |
| |A= 8300(1+0.15*3) | | |
| |=8673.50 | | |
|5.2 #100 |A) R= -.08p + 300p |Demand for pools. Tropical Pools sells an aboveground model for p dollars each. The monthly |#2 |
| |R= p(-.08p + 300) |revenue for this model is given by the formula | |
| | |[pic]. | |
| |B) p Would need to be multiplied by demand |Revenue is the product of the price p and the demand (quantity sold). | |
| |so that it can be R=p(-.08p + 300) because |a) Factor out the price on the right-hand | |
| |this is the same as multiplying by the |side of the formula. | |
| |demand as the revenue then it is going to be|[pic] | |
| |the same so the answer would be D=-.08P + 300|b) Write a formula [pic]for the monthly | |
| | |demand. | |
| |C) demand = -.08p + 300 |Since the revenue R is the product of the | |
| |Demand= -.08(3000) + 300 |price p and the demand, we can obtain: | |
| |Demand= -240 + 300 |[pic] | |
| |Demand = 60 |c) Find [pic] | |
| |D) D=-.08*P + 300 |[pic] | |
| | |d) Use the accompanying graph to | |
| |D= 150 |estimate the price at which the | |
| |E) -.0.08*1875^2+ 300* 1875=281250 |revenue is maximized. Approximately | |
| | |how many pools will be sold monthly | |
| | |at this price? | |
| | |[pic] | |
| | |The revenue is maximized when | |
| | |[pic] | |
| | |When the price is 1875 dollars, the revenue | |
| | |will be maximized. At this price, 150 pools | |
| | |will be sold monthly. | |
| | |e) What is the approximate maximum | |
| | |revenue? | |
| | |[pic] | |
| | |The maximum revenue is approximately | |
| | |281,250 dollars | |
| | |f) Use the accompanying graph to | |
| | |estimate the price at which the | |
| | |revenue is zero. | |
| | |By the graph, when the price is 0 dollars | |
| | |or 3700 dollars, the revenue will be zero. | |
|5.3 #110 |a) |Area of a sail. The area in square meters for a triangular sail is given by A(x) =x2 + 5x + |#2 |
| |A(x) =x^2+5x+6 |6. | |
| |A (5) =5^2+5*5+6 | | |
| |=25+25+6 |Find A(5) | |
| |=56 | | |
| | |Substitute 5 for x: | |
| |b) |[pic] | |
| |A=.5ab |Square meters. | |
| |.5(5+3)b=56 |If the height of the sail is | |
| |4b=56 |x + 3 meters, then what is the | |
| |b=14 m |length of the base of the sail? | |
| | |The area of a triangle is given | |
| | |by[pic]. So the base is: | |
| | |[pic] | |
| | |The length of the base is | |
| | |2x + 4 meters. | |
|5.4 #106 |Team Member #1 |Team Member #2 |#1 |
| |Jamie |Chelsey | |
| |a) Rewrite the formula by factoring the |a) Rewrite the formula by factoring the right-hand side completely. | |
| |right-hand side completely. |N(t) =-3t^3+23t^2+8t | |
| |N(t)= 3^3+23t^+8t |N(t) =-3t^3+23t^2+8t | |
| |N(t)= (-t)(3+^2-23t-8) |Factors of -3 are (-1, 3) and (1, -3) | |
| |N(t)=-t(3t+1)(t-8) |Factors of 8 are (1, 8), (-1, -8) (2, 4)((-2, -4) | |
| |b) Use the factored version of the formula to|Use: (-1, 3) and (1, 8) | |
| |find N(3). |N(t) = t(-t+8)(3t+1) | |
| |N(t)= 3^3+23t^+8t |b) Use the factored version of the formula to find N(3). | |
| |N(t)= (-t)(3+^2-23t-8) |c)Use the accompanying graph to estimate the time at which the workers are most efficient | |
| |N(t)=-t(3t+1)(t-8) |Chelsey = Worker Efficiency is 8 hours | |
| |=N(t)=-3t^3+23t^2+8 |Ulises = According to graph, when the time =5.5, | |
| |N(3)=-3(3*3+1)*(3-8) |the workers are most efficient. | |
| |N(3)=3(10)(-5) |d) Use the accompanying graph to estimate the maximum number of components assembled per | |
| |N(3)=(15)(10)=150 |hour during an 8-hour shift. | |
| |Answer: N(3)=150 |Ulises = According to the graph, the maximum number of | |
| |c)Use the accompanying graph to estimate the |components assembled per hour during an 8 hours shift is approximately 245 components. | |
| |time at which the workers are most efficient | | |
| |The peak would be after approximately 5.27 | | |
| |hours. | | |
| | | | |
| |d) Use the accompanying graph to estimate the| | |
| |maximum number of components assembled per | | |
| |hour during an 8-hour shift. | | |
| |Replacing t's with 3 the new is : | | |
| |n(3)=3(3)^3+23(3)^2+8(3) | | |
| | | | |
| |n(3)=3(3)^3+23(3)^2+8(3) | | |
| |n(3)=3*3=9*3=27 *3=81 3*3=9*23=207 | | |
| |207+81=288 8*3=24 | | |
| |288+24= 312 | | |
| |Answer: n(3)=312 | | |
|5.5 #112 | |Decreasing cube. Each of the three dimensions of a cube with sides of length s centimeters |#2 |
| | |is decreased by a whole number of centimeters. The new volume in cubic centimeters is given | |
| | |by | |
| | |[pic] | |
| | |a) Find V(10). | |
| | |V(s) = s^3-13s^2+54s-72 | |
| | |1000-1300+540-72 = 168 | |
| | |168 cm^3 | |
| | |b) If the new width is s - 6 centimeters, then what are the new length and height? | |
| | |Used factors of 168 in the place of (s-#) to find L and W | |
| | |(s-6) (s-3) (s-4) | |
| | |Length is s-3 | |
| | |Width is s-4 | |
| | |c) Find the volume when s =10 by multiplying the length, width, and height. | |
| | |(10-6)(10-3)(10-4) | |
| | |(4)(7)(6) = 168cm^3 | |
| | |Volume is 168cm^3 | |
|5.6 #74 |Rectangular stage. One side of a rectangular |Rectangular stage. One side of a rectangular stage is 2 meters longer than the other. If the|#2 |
| |stage is 2 meters longer than the other. If |diagonal is 10 meters, then what are the lengths of the sides? | |
| |the diagonal is 10 meters, then what are the |a^2+b^2=c | |
| |lengths of the sides? |(x+2)^2 + (x)^2 = 10^2 | |
| |Problem L=W+2 |x^2 +4x +4 (x)^2 = 100 | |
| | |2x^2+4x+4 = 100 | |
| | 10^2=L^2+W^2 | 2 2 | |
| | |x^2 +2x + 2 = 50 | |
| |Simplify: 100= |x^2+2x-48 = 0 | |
| |(w^2+4w+4)+w^2 |(x+8)(x+6) | |
| |Reorder terms: 100= | | |
| |(4+4w+w^2)+w^2 | | |
| |combine like terms: w^2+w^2 =2w^2 | | |
| | | | |
| | 100=4+4w+2w^2 | | |
| | | | |
| |Divide by 2: 100=2w^2+4w+4| | |
| | | | |
| | 50=w^2+2w+2 | | |
| | | | |
| |Subtract 50 from each | | |
| |side: w^2+2w-48=0 | | |
| |Solve for w : | | |
| |(-8+-1w)(6+-1w)=0 | | |
| | w=6| | |
| |(meters) and L = w+2 or 6+2 =8 (length) | | |
| | | | |
| |10^2=L^2+W^2 replace and check: 10^2=8^2+6^2 | | |
| |or 100= 64+36 | | |
|5.6 #92 |Throwing a wrench. An angry construction |Throwing a wrench. An angry construction worker throws his wrench downward from a height of |#2 |
| |worker throws his wrench downward from a |128 feet with an initial velocity of 32 feet per second. The height of the wrench above the | |
| |height of 128 feet with an initial velocity |ground after t seconds is given by[pic]. | |
| |of 32 feet per second. The height of the |a) What is the height of the wrench after 1 second? | |
| |wrench above the ground after t seconds is |5(t)=-16t^2-32t+128 | |
| |given by[pic]. |t=1 | |
| |a) What is the height of the wrench after 1 |-16-32+128 = 80 feet | |
| |second? |b) How long does it take for the wrench to reach the ground? | |
| |S(t)=-16t^2-32t+128 |0 = -16t^2-32t+128 | |
| |t=1 so rewrite problem: |0 = -t^2-2t+8 | |
| |s(1)=-16(1^2)-32(1)+128 |0 = (t+4)(-t+2) | |
| |S=80 feet or s=0 t+128 |t = 2 | |
| |b) How long does it take for the wrench to |2 seconds | |
| |reach the ground? | | |
| |rewrite problem: | | |
| |0=-16t^2-32 | | |
| |0=t^2+2t-8 | | |
| |0=(t+4)(t-2) | | |
| |t= 2 seconds | | | 