Question

course mth 158

I was just checking to see how long it usually takes to get grades back on our tests, and also where we can view those grades(blackboard?). Thank you.

Blackboard's gradebook has been made unusable in recent versions. You need to submit the Test Taken form and/or email me requesting results.

I believe I have your tests and they are graded. I can let you know the results tomorrow if you email me.

Please specify which test or tests are you asking about.

assignment #027

027. `query 27

College Algebra

07-16-2008

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10:48:35

3.6.6. x = -20 p + 500, 0<=p<=25

What is the revenue function and what is the revenue if 20 units are sold?

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RESPONSE -->

To find p...

x = -20p + 500

20p = 500 - x

p = (500 - x) / 20

R(x) = x ((500 - x) / 20) =

x((500/20) - (x/20)) =

-(1/20)x^2 + 25x

R(20) = -(1/20)(20)^2 + 25(20) =

-20 + 500 =

$480

confidence assessment: 2

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10:50:34

** revenue = demand * price = x * p = (-20 p + 500) * p = -20 p^2 + 500 p

If price = 24 then we get

R = -20 * 24^2 + 500 * 24 = 480. **

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RESPONSE -->

The end results are the same, but I dont understand what else happened... my problem said to express the revenue if 20 units were sold...

self critique assessment: 2

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11:07:13

3.6.10. P = (x, y) on y = x^2 - 8.

Give your expression for the distance d from P to (0, -1)

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RESPONSE -->

d = sqrt[(x-0)^2 + (y-(-1)^2] =

sqrt[x^2 + (x^2 - 8 + 1)^2] =

sqrt(x^2 + x^4 -7x^2 - 7x^2 + 49)

sqrt(x^4 - 12x^2 + 49)

d(x) = sqrt(x^4 - 12x^2 + 49)

confidence assessment: 2

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11:08:05

** P = (x, y) is of the form (x, x^2 - 8).

So the distance from P to (0, -1) is

sqrt( (0 - x)^2 + (-1 - (x^2-8))^2) =

sqrt(x^2 + (-7-x^2)^2) =

sqrt( x^2 + 49 - 14 x^2 + x^4) =

sqrt( x^4 - 13 x^2 + 49). **

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RESPONSE -->

I see I subtracted 2x^2 instead of x^2 from 14x^2. Oops.

self critique assessment: 3

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11:10:42

What are the values of d for x=0 and x = -1?

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RESPONSE -->

d(0) = sqrt(0^4 - 12(0)^2 + 49) =

sqrt(49) =

7

d(-1) = sqrt((-1)^4 - 12(-1)^2 + 49)

sqrt(38) =

5 sqrt(13)

confidence assessment: 2

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11:12:32

If x = 0 we have

sqrt( x^4 - 13 x^2 + 49) = sqrt(0^4 - 13 * 0 + 49) = sqrt(49) = 7.

If x = -1 we have

sqrt( x^4 - 13 x^2 + 49) = sqrt((-1)^4 - 13 * (-1) + 49) = sqrt( 64) = 8.

Note that these results are the distances from the x = 0 and x = 1 points of the graph of y = x^2 - 8 to the point (0, -1). You should have a sketch of the function and you should vertify that these distances make sense. **

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RESPONSE -->

Ah, again, I forgot to change the 12 to 13. I'm sorry

self critique assessment: 2

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11:27:18

3.6. 18 (was and remains 3.6.18). Circle inscribed in square.

What is the expression for area A as a function of the radius r of the circle?

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RESPONSE -->

A(r) = (2r)(2r) = 4r

confidence assessment: 2

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11:28:45

A circle inscribed in a square touches the square at the midpoint of each of the square's edges; the circle is inside the square and its center coincides with the center of the square. A diameter of the circle is equal in length to the side of the square.

If the circle has radius r then the square has sides of length 2 r and its area is (2r)^2 = 4 r^2.

The area of the circle is pi r^2.

So the area of the square which is not covered by the circle is 4 r^2 - pi r^2 = (4 - pi) r^2. **

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RESPONSE -->

Was I supposed to put the area of the square not covered by the circle? that wasn't one of the questions

self critique assessment: 2

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11:29:19

What is the expression for perimeter p as a function of the radius r of the circle?

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RESPONSE -->

P(r) = (2r)(4) = 8r

confidence assessment: 2

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11:29:24

The perimeter of the square is 4 times the length of a side which is 4 * 2r = 8r. **

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RESPONSE -->

ok

self critique assessment: 3

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11:47:15

3.6.27 (was 3.6.30). one car 2 miles south of intersection at 30 mph, other 3 miles east at 40 mph

Give your expression for the distance d between the cars as a function of time.

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RESPONSE -->

The distance formula is h^2 = L^2 + L^2 for a triangle, and that is what this situation creates.

d^2 = dc^2 + dc^2, d meaning the distance between the cars, and dc meaning the cars.

d^2 = (2 - 30t)^2 + (3 - 40t)^2

d(t) = sqrt[(2 - 30t)^2 + (3 - 40t)^2]

sqrt(4 - 120t + 900t^2 + 9 - 240t + 160t^2)

sqrt(2,500t^2 - 360t + 5)

confidence assessment: 2

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11:48:18

At time t the position of one car is 2 miles south, increasing at 30 mph, so its position function is 2 + 30 t.

The position function of the other is 3 + 40 t.

If these are the x and the y coordinates of the position then the distance between the cars is

distance = sqrt(x^2 + y^2) = sqrt( (2 + 30 t)^2 + (3 + 40t)^2 ) = sqrt( 4 + 120 t + 900 t^2 + 9 + 240 t + 1600 t^2) = sqrt( 2500 t^2 + 360 t + 13). **

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RESPONSE -->

I was thinking it was 4 -9 instead of -4 -9 for some reason.

self critique assessment: 3

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Good responses. See my notes and let me know if you have questions. &#