Assignment 15

course MTH 158

F󷠟́assignment #015

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015. `query 15

College Algebra

06-27-2008

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10:38:55

1.7.20 (was 1.2.30). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) million to lend at 19% or 16%, max lent at 16% to average 18%.

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RESPONSE -->

x = amount lent at 16%

.16x + .19(1,000,000 - x) = .18 (1,000,000)

.16x + 190,000 - .19x = 180,000

-.03x = -10

(-.03x) / -.03 = (-10) / -.03

x = 333.33

confidence assessment: 1

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10:39:08

** Good. The details:

If x is the amount lent at 16%, then the amount lent at 19% is 1,000,000 - x.

Interest on x at 16% is .16 x, and interest on 1,000,000 - x at 19% is .19 (1,000,000 - x).

This is to be equivalent to a single rate of 18%. 18% of 1,000,000 is 180,000 so the total interest is 1,000,000.

So the total interest is .16 x + .19(1,000,000 - x), and also 180,000. Setting the two equal gives us the equation

.16 x + .19(1,000,000 - x) = 180,000. Multiplying both sides by 100 to avoid decimal-place errors we have

16 x + 19 ( 1,000,000 - x) = 18,000,000. Using the distributive law on the right-hand side we get

16 x + 19,000,000 - 19 x = 18,000,000. Combining the x terms and subtracting 19,000,000 from both sides we have

-3 x = 18,000,00 - 19,000,000 so that

-3 x = -1,000,000 and

x = -1,000,000 / (-3) = 333,333 1/3. **

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RESPONSE -->

ok

self critique assessment: 2

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10:39:33

1.7.36 (was 1.2.36). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 3 mph current, upstream takes 5 hr, downstream 2.5 hr. Speed of boat?

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RESPONSE -->

The number of this problem in the book is actually 1.7 26, not 36.

Using the distance = rate * time formula...

Upstream | D | x-3 | 5

Downstream | D | x+3 | 2.5

Since the distance is the same in both columns, I can set them equal to each other.

(x-3) * 5 = (x+3) * 2.5

5x - 15 = 2.5x + 7.5

2.5x = 22.5

x = 9

confidence assessment: 2

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10:39:46

STUDENT SOLUTION:

Speed of the boat is 9 mph, I used the equation 5(x - 3) = 2.5(x + 3)

Reasoning is that it took 5 hours for the boat to travel against the 3mph current, and then traveled the same distance with the 3mph current in 2.5 hours.

INSTRUCTOR COMMENT:

Good. The details:

If we let x be the water speed of the boat then its actual speed upstream is x - 3, and downstream is x + 3.

Traveling for 5 hours upstream, at speed x - 3, we travel distance 5 ( x - 3).

Traveling for 2.5 hours downstream, at speed x + 3, we travel distance 2.5 ( x + 3).

The two distance must be the same so we get

5 ( x - 3) = 2.5 ( x + 3) or

5 x - 15 = 2.5 x + 7.5. Adding -2.5 x + 15 to both sides we get

2.5 x = 22.5 so that

x = 22.5 / 2.5 = 9.

So the water speed is 9 mph. **

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RESPONSE -->

ok

self critique assessment: 3

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10:40:09

1.7.32 (was 1.2.42). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) pool enclosed by deck 3 ft wide; fence around deck 100 ft. Pond dimensions if pond square, if rectangular 3/1 ratio l/w, circular; which pond has most area?

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RESPONSE -->

a. If the perimeter is 100 ft, then 100/4 = 25 ft

Then there are 3 feet of deck on each side, so

25 - 6 = 19

Dimensions are 19ft x 19ft

b. the width will be x and the length will be 3x

2 (x+6) +2 (3x + 6) = 100

2x + 12 + 6x + 12 = 100

8x + 24 = 100

8x = 76

x = 9.5

Dimensions are 9.5ft x 28.5ft

c. Let x = diameter

Circumference = 2 pi r, or diameter * pi

100 = x(pi) + 6

94 = x(pi)

x = 29.92

d. the square pond's area is 361 ft^2, the rectangle pond's area is 270.75 ft^2 and the circular pond's area is approx 703.09 ft^2.

The Circular Pond has the most area

confidence assessment: 2

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10:40:29

** If the deck is circular then its circumference is C = 2 pi R and its radius is r = C / (2 pi). C is the 100 ft length of the fence so we have

R = 100 ft / ( 2 pi ) = 50 ft / pi.

The radius of the circle is 3 ft less, due to the width of the deck. So the pool radius is

r = 50 ft / pi - 3 ft.

This gives us pool area

A = pi r^2 = pi ( 50 / pi - 3)^2 = pi ( 2500 / pi^2 - 300 / pi + 9) = 524, approx..

If the pool is square then the dimensions around the deck are 25 x 25. The dimensions of the pool will be 6 ft less on each edge, since each edge spans two widths of the deck. So the area would be

A = 19 * 19 = 361.

The perimeter of the rectangular pool spans four deck widths, or 12 ft. The perimeter of a rectangular pool is therefore 12 ft less than that of the fence, or 100 ft - 12 ft = 88 ft.

If the pool is rectangular with length 3 times width then we first have for the

2 l + 2 w = 88 or

2 (3 w) + 2 w = 88 or

8 w = 88, giving us

w = 11.

The width of the pool will be 11 and the length 3 times this, or 33.

The area of the pool is therefore 11 * 33 = 363.

The circular pool has the greatest area, the rectangular pool the least. **

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RESPONSE -->

Where are the rest of the answers?

self critique assessment: 1

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10:54:45

1.7.44 (was 1.2.54). Explain how you set up and solved an equation for the problem. Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text) 20 lb bag 25% cement 75% sand; how much cement to produce 40% concentration?

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RESPONSE -->

x = pounds of resulting mix

I'm going to set up and use cross-multiplication for this because it's simpler and easier.

25/20 * 40/x

25x = 800

x = 32

The difference between the original mix (20) and the resulting mix (32) is 12, thus

12lbs of cement must be added.

confidence assessment: 2

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10:56:59

** If x stands for the amount of cement added then we have the following:

Original amount of cement in bag is 25% of 20 lb, or 5 lb.

Original amount of sand in bag is 75% of 20 lb, or 15 lb.

The final amount of cement will therefore be 5 lb + x, the final amount of sand will be 15 lb and the final weight of the mixture will be 20 lb + x.

The mix has to be 40%, so (amt of cement) / (total amt of mixture) = .40.

This gives us the equation

(5 + x) / (20 + x) = .40. Multiplying both sides by 20 + x we have

5 + x = .40 ( 20 + x ). After the distributive law we have

5 + x = 80 + .40 x. Multiplying by 100 we get

500 + 100 x = 800 + 40 x. Adding -40 x - 500 to both sides we have

60 x = 300 so that

x = 300 / 60 = 5.

We should add 5 lbs of cement to the bag. **

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RESPONSE -->

My mistake, I thought I could do it differently. Understood now.

self critique assessment: 2

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11:05:56

1.7.52 (was 1.2.60). Include your equation and the reasoning you used to develop the equation. Problem (note that this statement is for instructor reference; the full statement was in your text): without solving what's wrong with prob how many liters 48% soln added to 20 liters of 25% soln to get 58% soln?

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RESPONSE -->

Well, the litres of 25% ethanol that you would add to the 48% would actually decrease the concentration, not increase to 58% no matter how much you added.

What happens?

x = amount of liters of 25% to be added

25x + 20(48) = 58

25x + 960 = 58

- 960 - 960 =

25x = -902

x = -36.08

A negative answer doesn't work with the problem.

confidence assessment: 2

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11:06:23

** Solution from Previous Student and Instructor Comment:

It's not possible, adding a 25% solution to a 48% solution is only going to dilute it, I don't really know how to prove that algebraically, but logically that's what I think.

(This is much like the last problem, that I don't really understand).

INSTRUCTOR COMMENT: Right but the 48% solution is being added to the 25% solution.

Correct statement, mostly in your words

Adding a 48% solution to a 25% solution will never give you a 58% solution. Both concentrations are less than the desired concentration. **

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RESPONSE -->

ok

self critique assessment: 2

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11:06:30

Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

confidence assessment:

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11:06:59"

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