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course MTH 158

????z???????????assignment #021

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021. `query 21

College Algebra

03-03-2007

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14:49:58

2.7.8 (was 2.6.6). y inv with sqrt(x), y = 4 when x = 9.

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RESPONSE -->

y=k/sqrt(x)

4=k/sqrt(9) simplify

4 = k / 3 multiplt by 3

k = 12

confidence assessment: 2

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14:57:35

** The inverse proportionality to the square root gives us y = k / sqrt(x).

y = 4 when x = 9 gives us

4 = k / sqrt(9) or

4 = k / 3 so that

k = 4 * 3 = 12.

The equation is therefore

y = 12 / sqrt(x). **

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RESPONSE -->

I forgot the end but, i do under stand

self critique assessment: 2

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15:20:09

2.7.12 (was 2.6.10). z directly with sum of cube of x and square of y; z=1 and x=2 and y=3.

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RESPONSE -->

z=k*x^3+y^2 Fill in

1=k*2^3+3^2 simplify

1=k*8+9 Subtract 9

-8=k*8 devide

-8/8=k So k=-1

1=k*2^3+3^2 = 1=-1*2^3+3^2

1=-1*2^3+3^2 = 1=-1*8+9 Subtract 9 and multiply

-8 =-8

{z=-1*x^3+y^2}

confidence assessment: 2

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15:24:37

** The proportionality is

z = k (x^3 + y^2).

If x = 2, y = 3 and z = 1 we have

1 = k ( 2^3 + 3^2) or

17 k = 1 so that

k = 1/17.

The proportionality is therefore

z = (x^3 + y^2) / 17. **

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RESPONSE -->

z=k*x^3+y^2 Fill in

1=k*2^3+3^2 simplify

1=k*8+9 Subtract 9

-8=k*8 devide

-8/8=k So k=-1

I missed it from the start but i was close k should multiply by both (x^3+y^2)

I agree, and that is exactly the point you missed. Everything else was good.

self critique assessment: 2

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15:42:51

2.7.20 (was 2.6.20). Period varies directly with sqrt(length), const 2 pi / sqrt(32)

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RESPONSE -->

I dont fully understand the problum. I can only move things around.

t=k(2pi/sqrt(32)) multiply sqrt(32)

t*sqrt(32)=k*2pi 2pi =6.28

t*sqrt(32)=k*6.28 divide 6.28

(t*sqrt(32))/6.28=k

I dont think this is correct

confidence assessment: 0

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15:45:16

** The equation is

T = k sqrt(L), with k = 2 pi / sqrt(32). So we have

T = 2 pi / sqrt(32) * sqrt(L). **

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RESPONSE -->

That is what i had on my paper but i didnt think that is what you wonted.it was to simple.

self critique assessment: 2

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15:46:24

**** What equation relates period and length? ****

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RESPONSE -->

t=k*sqrtl

confidence assessment: 2

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16:20:43

2.7.34 (was 2.6.30). Resistance dir with lgth inversely with sq of diam. 432 ft, 4 mm diam has res 1.24 ohms. **** What is the length of a wire with resistance 1.44 ohms and diameter 3 mm? Give the details of your solution.

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RESPONSE -->

r = k*l/d^2

1.24=k*432/4^2

1.24=k*432/8 multiply by 8

4^2 is 16; otherwise you're on the right track

9.92 =k*432

k = 9.92/432 or k =m .023

1.44= .023*l/9

12.96=.023*l

l = 12.96/.023 or l = 563.5

wire is 563ft 6in long

confidence assessment: 2

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16:26:39

** We have

R = k * L / D^2. Substituting we obtain

1.24 = k * 432 / 4^2 so that

k = 1.24 * 4^2 / 432 = .046 approx.

Thus

R = .046 * L / D^2.

Now if R = 1.44 and d = 3 we find L as follows:

First solve the equation for L to get

L = R * D^2 / (.046). Then substitute to get

L = 1.44 * 3^2 / .046 = 280 approx.

The wire should be about 280 ft long. **

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RESPONSE -->

1.24 = k * 432 / 4^2 so that

k = 1.24 * 4^2 / 432 = .046 approx.

that is a neet trick

self critique assessment: 2

you would have gotten .046, except for the errror on 4^2.

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K???F?????V?????a??assignment #022

022. `query 22

College Algebra

03-03-2007

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16:51:32

3.1.20 (was 3.1.6) (-2,5),(-1,3),(3,7),(4,12)}Is the given relation a function? Why or why not? If so what are its domain and range?

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RESPONSE -->

Yes. passes the verticol test.

domain = -2,-1,0,1,2,3,4

range = 3-12

confidence assessment: 2

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16:55:38

This relation is a function because every first element is paired with just one second element--there are no distinct ordered pairs with the same first element.

the domain is ( -2,-2,3,4)

the range is ( 5,3,7,12)

Another way of saying that this is a function is that every element of the domain appears only once in the relation.

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RESPONSE -->

I understand this but you put that the domain is ( -2,-2,3,4) why wouldnt that be ( -2,-1,3,4)

self critique assessment: 2

That's a typo; the sets should also have been set off by brackets

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17:42:10

3.1.34 (was 3.1.20) f(0), f(1), f(-1), f(-x), -f(x), f(x+1), f(2x), f(x+h) for 1 - 1 / (x+1)^2What are you expressions for f(0), f(1), f(-1), f(-x), -f(x), f(x+1), f(2x), f(x+h)?query 3.1.30. y = (3x-1)/(x+2)

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RESPONSE -->

1 - 1 / (x+1)^2

a = 1 - 1 / (0+1)^2 = 0

b = 1 - 1 / (1+1)^2 = 3/4

c = 1 - 1 / (-1+1)^2 = 1

-1 + 1 = 0, which would give you 0 in your denominator, making the expression undefined

d = 1 - 1 / (-x+1)^2 = 1 - 1 / x^2-2x+1

e =-1( 1 - 1 / (x+1)^2)= 1 - 1 / x^2+2x+1

f = 1 - 1 / (x+1+1)^2 = 1 - 1 / x^2+4x+4

g= 1 - 1 / (2x+1)^2 = 1 - 1 / 4x^2+4x+1

h = 1 - 1 / (x+h+1)^2 = 1 - 1 / x^2+2hx+h^2+1

confidence assessment:

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17:48:15

STUDENT SOLUTION WITH INSTRUCTOR COMMENTS: f(x) = 1- 1/(x+2)^2

f(0) = 1- 1/ (0+2)^2

f(0) = 1-1/4

f(0) = 3/4

f(1) = 1- 1/ (3)^2

f(1) = 1- 1/9

f(1) = 8/9

f(-1) = 1- 1/(-1+2)^2

f(-1)= 1-1

f(-1)= 0

f(-x)= 1- 1/(-x+2)^2

f(-x)= 1 -1/ (x^2-4x+4)

-f(x) = -(1- 1/(x+2)^2)

-f(x)= -(1 - 1/ (x^2+4x+4))

-f(x) = (1/(x^2 + 4x + 4)) - 1

** Your answer is right but you can leave it in factored form:

f(-x) = -(1 - 1/(x+2)^2)

= -1 + 1 / (x+2)^2. **

f(x+1) = 1- 1/((x+1) + 2)^2. This can be expanded as follows, but the expansion is not necessary at this point:

= 1- 1/ ((x+1)^2 +2(x+1) + 2(x+1)+4)

= 1- 1/ ((x+1)^2 +8x+8)

= 1- 1/ (x^2+2x+1+8x+8)

= 1- 1/(x^2 + 10x +9)

** Good algebra, and correct, but again no need to expand the square, though it is perfectly OK to do so. **

f(2x)= 1-1/(2x+2)^2

= 1- 1/(4x^2+8x+4)

** same comment **

f(x+h)= 1- 1/((x+h)+2)^2

= 1- 1/((x+h)^2 + 4(x+h) + 4)

= 1- 1/ (x^2 + 2xh + h^2+4x+4h+4)

** same comment **

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RESPONSE -->

I forgot the -f(x) and hit cancle on the previce screen to go back it didn't work.

Also I think the problem was f(x) = 1- 1/(x+1)^2 not

f(x) = 1- 1/(x+2)^2 but i might be mistaken.

self critique assessment: 0

Your answers were good; see my one note

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