18- 19- 20

course MTH 158

yݶOLh~l~assignment #018

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018. `query 18

College Algebra

02-25-2007

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13:27:23

2.4.30 (was 2.3.24). Slope 4/3, point (-3,2)

Give the three points you found on this line and explain how you obtained them.

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RESPONSE -->

3 points of Slope 4/3, point (-3,2) are (0,6) (3,10) and (6,14)

Ijust graphed the point then went up 4 and over three

confidence assessment: 2

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13:31:04

STUDENT SOLUTION:

(-3,2) slope 4/3. Move 3 units in the x direction, 4 in the y direction to get

((-3+3), (2+4)), which simplifies to

(0,6)

(-3,2) slope 4/3 = -4/-3 so move -3 units in the x direction and -4 in the y direction to get

((-3-3), (2-4)) which simplifies to

(-6,-2)

From (0,6) with slope 4/3 we move 4 units in the y direction and 3 in the x direction to get

((0+3), (6+4)), which simplifies to

(3,10). The three points I obtained are

(-6,-2), (0,6), (3,10).

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RESPONSE -->

Ok i miss understood.

self critique assessment: 2

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14:52:45

query 2.4.36 (was 2.3.30). Line thru (-1,1) and (2,2) **** Give the equation of the line and explain how you found the equation.

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RESPONSE -->

Find slope= 1/3

Then y-y1=m(x-x1)

So y-2 = 1/3(x-2) Or y-1 = 1/3(x +1)

= y = 1/3x Or y = 1/3(x +2)

y = 1/3x + 6/3 =y = 7

I was under the impresion that thees should be =

confidence assessment: 1

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15:08:18

STUDENT SOLUTION: The slope is m = (y2 - y1) / (x2 - x1) = (2-1)/(2- -1) = 1/3.

Point-slope form gives us

y - y1 = m (x - x1); using m = 1/3 and (x1, y1) = (-1, 1) we get

y-1=1/3(x+1), which can be solved for y to obtain

y = 1/3 x + 4/3.

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RESPONSE -->

You put that y-1=1/3(x+1) can be solved for y to obtain y = 1/3 x + 4/3. I dont understand this

I Add the the 1 to get y =1/3(x+2) then I gt 1/3x+ 2/3

Good explanation of your work.

You only missed one step. Let me know if you don't understand the following correction:

When you add 1 to both sides you get

y = 1/3 (x + 1) + 1.

You can't add the 1 to the expression inside the parentheses, because everything in the parentheses is multiplied by 1/3.

When you multiply things out you get

y = 1/3 x + 1/3 + 1. Since 1/3 + 1 = 4/3, this is

y = 4/3 x + 1.

self critique assessment: 2

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15:25:35

2.4.46 (was 2.3.40). x-int -4, y-int 4 **** What is the equation of the line through the given points and how did you find the equation?

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RESPONSE -->

Find slope = 4/4

then y-0=4/4(x-4)

This should read y - 0 = 4/4 ( x - (-4) ), which is the same as

y - 0 = 4/4 ( x + 4).

=y = 1x - 4

= y= -3x

1x - 4 is not -3. 1x and -4 are not like terms and cannot be combined.

1x is generally expressed just as x, but 1x is not incorrect.

confidence assessment: 2

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15:29:44

STUDENT SOLUTION: The two points are (0, 4) and (4, 0). The slope is therefore m=rise / run = (4-0)/(0+4) = 1.

The slope-intercept form is therefore y = m x + b = 1 x + 4, simplifying to

y=x+4.

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RESPONSE -->

I am having dificulty chosing witch formula to use

You are being methodical and showing all your steps, which is very good. This allows me to explain the steps you miss.

Point slope , slope intercept or equation slope intercept

self critique assessment: 2

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15:35:49

2.4.56 (was 2.4.48). y = 2x + 1/2. **** What are the slope and the y-intercept of your line and how did you find them?

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RESPONSE -->

y = 2x + 1/2

Slope = 2

y int = 1/2

because slope intercept is Y = mx+b

confidence assessment: 2

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15:36:46

** the y intercept occurs where x = 0, which happens when y = 2 (0) + 1/2 or y = 1/2. So the y-intercept is (0, 1/2).

The slope is m = 2.**

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RESPONSE -->

thats good

self critique assessment: 2

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޴|ꤛ{ӵF

assignment #019

019. `query 19

College Algebra

02-25-2007

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15:42:13

2.5.22 (was 2.4.18) Parallel to x - 2 y = -5 containing (0,0) **** Give your equation for the requested line and explain how you obtained it.

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RESPONSE -->

find slope = 4/2

then y-0 = 4/2 (x-0)

So y = 4/4x

confidence assessment: 2

Good, except I'm not sure how you got your slope, which is incorrect.

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16:03:00

** The equation x - 2y = -5 can be solved for y to give us

y = 1/2 x + 5/2.

A line parallel to this will therefore have slope 1/2.

Point-slope form gives us

y - 0 = 1/2 * (x - 0) or just

y = 1/2 x. **

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RESPONSE -->

ok

self critique assessment: 2

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16:38:39

2.5.28 (was 2.4.24) Perpendicular to x - 2 y = -5 containing (0,4) **** Give your equation for the requested line and explain how you obtained it.

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RESPONSE -->

x - 2 y = -5 ; (0,4) Subtract x

- 2 y =-x -5 Divide by -2

y =(-x -5)/-2 = y = 1/2 x -5/-2

=y = 1/2 x 2.5 M = 1/2 So Perpendicular M=-2

Vry good, but note that y = 1/2 x 2.5 should have been written y = 1/2 x + 2.5

y-4 = -2 (x-0) y = -2(x + 4) { y = -2x - 8 }

y-4 = -2 (x-0) gives you

y = -2(x - 0) + 4. The 4 doesn't go inside the parentheses, where everything is being multiplied by -2. The expression then simplifies to

y = -2 x + 4.

confidence assessment: 2

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16:42:11

** The equation x - 2y = -5 can be solved for y to give us

y = 1/2 x + 5/2.

A line perpendicular to this will therefore have slope -2/1 = -2.

Point-slope form gives us

y - 4 = -2 * (x - 0) or

y = -2 x + 4. **

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RESPONSE -->

in the distributive part I my +4 by the -2 i am still not sure how this can be skiped.

self critique assessment: 2

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ݞOgƮӅHh

assignment #019

019. `query 19

College Algebra

02-26-2007

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17:09:56

2.3.10 (was 2.4.30). (0,1) and (2,3) on diameter **** What are the center, radius and equation of the indicated circle?

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RESPONSE -->

center, = (1,2)

radius = 1.2/5 using (x - h)^2 + (y - k)^2 = R^2

equation = x^2 + y^2 - 1x - 4y - 3.6 = 0

that should be right

confidence assessment: 2

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17:17:03

** The distance from (0,1) to (2,3) is sqrt( (2-0)^2 + (3-1)^2 ) = 2. This distance is a diameter so that the radius is 1/2 (2) = 1.

The equation (x-h)^2 + (y-k)^2 = r^2 becomes

(x-1)^2 + (y-2)^2 = r^2.

Substituting the coordinates of the point (0, 1) we get

(0-1)^2 + (1-2)^2 = r^2 so that

r^2 = 2.

Our equation is therefore

(x-1)^2 + (y - 2)^2 = 2.

You should double-check this solution by substituting the coordinates of the point (2, 3). **

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RESPONSE -->

first off i got the radius of 1.2/5 then i found the genril form equation.

self critique assessment: 2

Be sure you understand the given solution here.

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17:20:23

2.3.16 (was 2.4.36). What is the standard form of a circle with (h, k) = (1, 0) with radius 3?

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RESPONSE -->

(x - h)^2 + (y - k)^2 = R^2

(x - 1)^2 + (y - 0)^2 = 3^2

(x - 1)^2 + y^2 = 9

confidence assessment: 2

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17:22:21

** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r.

In this example we have (h, k) = (1, 0). We therefore have

(x-1)^2 +(y - 0)^2 = 3^2.

This is the requested standard form.

This form can be expanded and simplified to a general quadratic form. Expanding (x-1)^2 and squaring the 3 we get

x^2 - 2x +1+y^2 = 9

x^2 - 2x + y^2 = 8.

However this is not the standard form.

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RESPONSE -->

finaly i got one

self critique assessment: 2

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17:30:07

2.3.24 (was 2.4.40). x^2 + (y-1)^2 = 1 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?

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RESPONSE -->

x^2 + (y-1)^2 = 1

center = (0,1)

radius =1

this is in theupper hemispheer.

intercepts x and y at (0,0) And (0,2)

confidence assessment: 2

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17:47:36

** The standard form of a circle is (x - h)^2 + (y - k)^2 = r^2, where the center is at (h, k) and the radius is r.

In this example the equation can be written as

(x - 0)^2 + (y-1)^2 = 1

So h = 0 and k = 1, and r^2 = 1. The center of the is therefore (0, -1) and r = sqrt(r^2) = sqrt(1) = 1.

The x intercept occurs when y = 0:

x^2 + (y-1)^2 = 1. I fy = 0 we get

x^2 + (0-1)^2 = 1, which simplifies to

x^2 +1=1, or

x^2=0 so that x = 0. The x intercept is therefore (0, 0).

The y intercept occurs when x = 0 so we obtain

0 + (y-1)^2 = 1, which is just (y - 1)^2 = 1. It follow that

(y-1) = +-1.

If y - 1 = 1 we get y = 2; if y - 1 = -1 we get y = 0. So the y-intercepts are

(0,0) and (0,2)

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RESPONSE -->

Wow i am still not sure how the 2 is negitive

The 2 shouldn't have been negative (I corrected it), and your answer was right. Good work.

self critique assessment: 2

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18:21:30

2.3.34 (was 2.4.48). 2 x^2 + 2 y^2 + 8 x + 7 = 0 **** Give the center and radius of the circle and explain how they were obtained. In which quadrant(s) was your graph and where did it intercept x and/or y axes?

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RESPONSE -->

This problem is being difficult please explain.

2 x^2 + 2 y^2 + 8 x + 7 = 0 minus 7

= 2 x^2 + 2 y^2 + 8 x = -7 divide all by 2

x^2 + y^2 + 4x = -7/2 rearange

x^2 + 4x + y^2 = -7/2 complete sqr in varibile

x^2 + 4x + 4 + y^2 = -7/2 +4 because (4/2)^2 = 4

but i can't figure out what to do with the y^2 (x+2)^2 + y^2 = 1/2

confidence assessment: 0

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18:26:28

** Starting with

2x^2+ 2y^2 +8x+7=0 we group x and y terms to get

2x^2 +8x +2y^2 =-7. We then divide by the common factor 2 to get

x^2 +4x + y^2 = -7/2. We complete the square on x^2 + 4x, adding to both sides (4/2)^2 = 4, to get

x^2 + 4x + 4 + y^2 = -7/2 + 4. We factor the expression x^2 + 4x + 4 to obtain

(x+2)^2 + y^2 = 1/2. From the standard form of the equation for a circle we see that

the center is (-2,0)

the radius is sqrt (1/2).

To get the intercepts:

We use (x+2)^2 + y^2 = 1/2

If y = 0 then we have

(x+2)^2 + 0^2 = 1/2

(x+2)^2 = 1/2

(x+2) = +- sqrt(1/2)

x + 2 = sqrt(1/2) yields x = sqrt(1/2) - 2 = -1.3 approx.

x + 2 = -sqrt(1/2) yields x = -sqrt(1/2) - 2 = -2.7 approx

If x = 0 we have

(0+2)^2 + y^2 = 1/2

4 + y^2 = 1/2

y^2 = 1/2 - 4 = -7/2.

y^2 cannot be negative so there is no y intercept. **

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RESPONSE -->

(x+2)^2 + y^2 = 1/2. is the answer I figured . how do you get the center is (-2,0) I under stood the rest

self critique assessment: 2

The x coordinate of the center is h, in the expression (x - h)^2. If the center was at (2, 0) then h would be 2 and the expression would contain (x - 2)^2.

However the expression contains (x + 2)^2.

The only way you can get that is if h = -2:

(x - h)^2 would be (x - (-2) )^2, which is (x + 2)^2.

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18:54:59

2.3.30 (was 2.4.54). General equation if diameter contains (4, 3) and (0, 1). **** Give the general equation for your circle and explain how it was obtained.

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RESPONSE -->

(4, 3) and (0, 1)

(0-4)^2+(1-3)^2 = R^2

= 8+4 = 12 12/2 = 6 R^2 = 6 so R = 2.4 or 12/5

confidence assessment:

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ܨЎ_؃̍kgbtW񁲴

assignment #001

001. Rates

qa rates

02-26-2007

"

r is half the distance from (4, 3) to (0, 1). That distance is sqrt(4^2 + 2^2) = sqrt(16 + 4) = sqrt(20) = 2 sqrt(5), not 12/5.

The radius is half the diameter, which would be 1/2 ( 2 sqrt(5)) = sqrt(5), so r^2 would be (sqrt(5))^2 = 5.

The center is halfway between these two points, at (2, 2).

So the equation would be

(x - 2)^2 + (y-2)^2 = 5.

You're stumbling over some of the algebra but you are getting most of your steps pretty much right, and you're doing a good job with self-critique. I recommend more practice on the problems in this section, but you're on the right track.

Be sure to study the fairly extensive notes I inserted into this document and let me know what, if anything, is not clear.