ast 1

course phy201

Professor Smith I think I may have resent you work from last night but i was afraid to delete it beause i was afraid it would do something to the file. Can you let me know if it does.

You have at least two alternatives:

Just highlight the part of the file that doesn't duplicate what has already been sent, right-click to copy, and paste into the Submit Work Form.

Copy the entire file into the submit work form, then deleted the part you don't want.

Whatever you do, do NOT save the SEND file after opening it. That will preserve the original SEND file in case you need it later.

This ends the first assignment.

ӌ\̀gHцҰ

assignment #001

Since it's already been posted I'm just going to delete the qa rates for you.

assignment #001

001. Areas

qa areas volumes misc

01-21-2008

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22:54:30

`q001. There are 11 questions and 7 summary questions in this assignment.

What is the area of a rectangle whose dimensions are 4 m by 3 meters.

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RESPONSE -->

area is length times width so

4*3=12meters

confidence assessment: 3

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22:55:14

A 4 m by 3 m rectangle can be divided into 3 rows of 4 squares, each 1 meter on a side. This makes 3 * 4 = 12 such squares. Each 1 meter square has an area of 1 square meter, or 1 m^2. The total area of the rectangle is therefore 12 square meters, or 12 m^2.

The formula for the area of a rectangle is A = L * W, where L is the length and W the width of the rectangle. Applying this formula to the present problem we obtain area A = L * W = 4 m * 3 m = (4 * 3) ( m * m ) = 12 m^2.

Note the use of the unit m, standing for meters, in the entire calculation. Note that m * m = m^2.

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RESPONSE -->

I didnt square my meters

self critique assessment: 1

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22:57:47

`q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?

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RESPONSE -->

well we are either looking at 12 meters ^3

or we are looking at 3*3*4=36

or 4+3+3=13 meters

confidence assessment: 1

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22:58:38

01-21-2008 22:58:38

A right triangle can be joined along its hypotenuse with another identical right triangle to form a rectangle. In this case the rectangle would have dimensions 4.0 meters by 3.0 meters, and would be divided by any diagonal into two identical right triangles with legs of 4.0 meters and 3.0 meters.

The rectangle will have area A = L * W = 4.0 m * 3.0 m = 12 m^2, as explained in the preceding problem. Each of the two right triangles, since they are identical, will therefore have half this area, or 1/2 * 12 m^2 = 6.0 m^2.

The formula for the area of a right triangle with base b and altitude h is A = 1/2 * b * h.

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NOTES ------->

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22:59:47

`q003. What is the area of a parallelogram whose base is 5.0 meters and whose altitude is 2.0 meters?

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RESPONSE -->

l*w

5*2=10meters^2

confidence assessment: 0

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23:00:20

A parallelogram is easily rearranged into a rectangle by 'cutting off' the protruding end, turning that portion upside down and joining it to the other end. Hopefully you are familiar with this construction. In any case the resulting rectangle has sides equal to the base and the altitude so its area is A = b * h.

The present rectangle has area A = 5.0 m * 2.0 m = 10 m^2.

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RESPONSE -->

YES

self critique assessment: 2

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23:02:03

`q004. What is the area of a triangle whose base is 5.0 cm and whose altitude is 2.0 cm?

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RESPONSE -->

ok

1/2*5*2=5

confidence assessment: 0

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23:02:25

It is possible to join any triangle with an identical copy of itself to construct a parallelogram whose base and altitude are equal to the base and altitude of the triangle. The area of the parallelogram is A = b * h, so the area of each of the two identical triangles formed by 'cutting' the parallelogram about the approriate diagonal is A = 1/2 * b * h. The area of the present triangle is therefore A = 1/2 * 5.0 cm * 2.0 cm = 1/2 * 10 cm^2 = 5.0 cm^2.

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RESPONSE -->

holy cow i got it right

self critique assessment: 2

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23:06:54

`q005. What is the area of a trapezoid with a width of 4.0 km and average altitude of 5.0 km?

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RESPONSE -->

4*5=20km^3

confidence assessment: 0

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23:07:14

Any trapezoid can be reconstructed to form a rectangle whose width is equal to that of the trapezoid and whose altitude is equal to the average of the two altitudes of the trapezoid. The area of the rectangle, and therefore the trapezoid, is therefore A = base * average altitude. In the present case this area is A = 4.0 km * 5.0 km = 20 km^2.

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RESPONSE -->

k

self critique assessment: 2

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23:09:17

`q006. What is the area of a trapezoid whose width is 4 cm in whose altitudes are 3.0 cm and 8.0 cm?

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RESPONSE -->

4*3=12

4*8=32

12+32=44cm

maybe

confidence assessment: 1

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23:09:55

The area is equal to the product of the width and the average altitude. Average altitude is (3 cm + 8 cm) / 2 = 5.5 cm so the area of the trapezoid is A = 4 cm * 5.5 cm = 22 cm^2.

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RESPONSE -->

well if I had divided my answer by 2 I would have been right

self critique assessment: 1

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23:13:52

`q007. What is the area of a circle whose radius is 3.00 cm?

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RESPONSE -->

A=(pi)r^2

A=(pi)3^2

(pi)9

(pi)=3.1415927

so

A=28.2743343

confidence assessment: 3

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23:19:21

The area of a circle is A = pi * r^2, where r is the radius. Thus

A = pi * (3 cm)^2 = 9 pi cm^2.

Note that the units are cm^2, since the cm unit is part r, which is squared.

The expression 9 pi cm^2 is exact. Any decimal equivalent is an approximation. Using the 3-significant-figure approximation pi = 3.14 we find that the approximate area is A = 9 pi cm^2 = 9 * 3.14 cm^2 = 28.26 cm^2, which we round to 28.3 cm^2 to match the number of significant figures in the given radius.

Be careful not to confuse the formula A = pi r^2, which gives area in square units, with the formula C = 2 pi r for the circumference. The latter gives a result which is in units of radius, rather than square units. Area is measured in square units; if you get an answer which is not in square units this tips you off to the fact that you've made an error somewhere.

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RESPONSE -->

oops forgot to put my measurement

self critique assessment: 2

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23:24:02

`q008. What is the circumference of a circle whose radius is exactly 3 cm?

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RESPONSE -->

ok I think circum. is (pi)d

but you didnt give me the diameter and I know there is an equation for 2(pi)r so I'm going to go for that one but I really thought it was the first one.

2(pi)3=

2*3.14*3=

18.84 cm

confidence assessment: 2

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23:25:10

The circumference of this circle is

C = 2 pi r = 2 pi * 3 cm = 6 pi cm.

This is the exact area. An approximation to 3 significant figures is 6 * 3.14 cm = 18.84 cm.

Note that circumference is measured in the same units as radius, in this case cm, and not in cm^2. If your calculation gives you cm^2 then you know you've done something wrong.

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RESPONSE -->

YIPPEE

self critique assessment: 2

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23:31:09

`q009. What is the area of a circle whose diameter is exactly 12 meters?

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RESPONSE -->

ok if the diameter is twice the radius then the radius here is 6 meters

so the area would be (pi)r^2

so (pi)*6^2

(3.14)*36=

113.04cm^2

confidence assessment: 2

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23:32:07

The area of a circle is A = pi r^2, where r is the radius. The radius of this circle is half the 12 m diameter, or 6 m. So the area is

A = pi ( 6 m )^2 = 36 pi m^2.

This result can be approximated to any desired accuracy by using a sufficient number of significant figures in our approximation of pi. For example using the 5-significant-figure approximation pi = 3.1416 we obtain A = 36 m^2 * 3.1416 = 113.09 m^2.

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RESPONSE -->

lordy I only used the 3 numbers for pi

self critique assessment: 2

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23:41:19

`q010. What is the area of a circle whose circumference is 14 `pi meters?

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RESPONSE -->

C=2(pi)r

14=2(3.14)r

14=6.28r

2.23=r

A=(pi)r^2

(pi)*2.23^2

3.14*4.97=

15.61cm^2

confidence assessment: 1

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23:42:18

We know that A = pi r^2. We can find the area if we know the radius r. We therefore attempt to use the given information to find r.

We know that circumference and radius are related by C = 2 pi r. Solving for r we obtain r = C / (2 pi). In this case we find that

r = 14 pi m / (2 pi) = (14/2) * (pi/pi) m = 7 * 1 m = 7 m.

We use this to find the area

A = pi * (7 m)^2 = 49 pi m^2.

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RESPONSE -->

Wow Did I Ever Mess This one Up

self critique assessment: 1

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23:55:44

`q011. What is the radius of circle whose area is 78 square meters?

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RESPONSE -->

A=(pi)r^2

78=(pi)r^2

78=3.14r^2

divide both sides by 3.14

24.84=r^2

square root of both sides

4.98=r

confidence assessment: 1

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23:57:43

Knowing that A = pi r^2 we solve for r. We first divide both sides by pi to obtain A / pi = r^2. We then reverse the sides and take the square root of both sides, obtaining r = sqrt( A / pi ).

Note that strictly speaking the solution to r^2 = A / pi is r = +-sqrt( A / pi ), meaning + sqrt( A / pi) or - sqrt(A / pi). However knowing that r and A are both positive quantities, we can reject the negative solution.

Now we substitute A = 78 m^2 to obtain

r = sqrt( 78 m^2 / pi) = sqrt(78 / pi) m.{}

Approximating this quantity to 2 significant figures we obtain r = 5.0 m.

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RESPONSE -->

I think I got that one right

self critique assessment: 2

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00:01:28

`q012. Summary Question 1: How do we visualize the area of a rectangle?

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RESPONSE -->

to visualize the area of a rectangle I guess you first have to know what the area is which is l*w so i guess a nice bold red line around the short and long side of the rectangle

confidence assessment: 0

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00:01:50

We visualize the rectangle being covered by rows of 1-unit squares. We multiply the number of squares in a row by the number of rows. So the area is A = L * W.

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RESPONSE -->

gotcha

self critique assessment: 1

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00:08:31

`q013. Summary Question 2: How do we visualize the area of a right triangle?

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RESPONSE -->

I have no idea how to visualize this one i know the path. therom and the equation for area of a triangle but to visulaize 1 unit squares filling a triangle and then counting them I dont get it.

confidence assessment: 0

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00:08:53

We visualize two identical right triangles being joined along their common hypotenuse to form a rectangle whose length is equal to the base of the triangle and whose width is equal to the altitude of the triangle. The area of the rectangle is b * h, so the area of each triangle is 1/2 * b * h.

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RESPONSE -->

gotcha

self critique assessment: 1

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00:10:53

`q014. Summary Question 3: How do we calculate the area of a parallelogram?

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RESPONSE -->

Ok so this one we cut the protruding edges off and line them up to form the rectangle and count 1 unit squares and count the rows and multiple by the number of units in each row. for L*W

confidence assessment: 2

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00:11:19

The area of a parallelogram is equal to the product of its base and its altitude. The altitude is measured perpendicular to the base.

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RESPONSE -->

so did i get that one right?

self critique assessment: 1

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00:12:18

`q015. Summary Question 4: How do we calculate the area of a trapezoid?

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RESPONSE -->

I would speculate its the same response I gave on the last question cut off the protruding edges and form a rectangle and then solve like its a triangle.

confidence assessment: 0

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00:12:43

We think of the trapezoid being oriented so that its two parallel sides are vertical, and we multiply the average altitude by the width.

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RESPONSE -->

these questions are making my mind hurt

self critique assessment: 0

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00:13:03

`q016. Summary Question 5: How do we calculate the area of a circle?

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RESPONSE -->

A=(pi)r^2

confidence assessment: 3

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00:13:14

We use the formula A = pi r^2, where r is the radius of the circle.

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RESPONSE -->

GOT IT

self critique assessment: 2

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00:14:21

`q017. Summary Question 6: How do we calculate the circumference of a circle? How can we easily avoid confusing this formula with that for the area of the circle?

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RESPONSE -->

C=2(pi)r

we could always use the diameter instead

C=(pi)d

confidence assessment: 3

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00:14:38

We use the formula C = 2 pi r. The formula for the area involves r^2, which will give us squared units of the radius. Circumference is not measured in squared units.

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RESPONSE -->

or we could do that

self critique assessment: 1

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00:15:16

`q018. Explain how you have organized your knowledge of the principles illustrated by the exercises in this assignment.

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RESPONSE -->

I love equations

and each equation is different to me

confidence assessment: 0

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You're doing fine. An error here and there, but you know what you're doing.

Let me know if there's anything you don't understand.