some init probs

some init probs

Obviously you are well prepared for the course.

For now, especially since you say your keyboard skills aren't that extensive, you can abbreviate your responses a little more, especially if you are 100% sure you understand the process. You will quickly enough encounter work where you aren't 100% sure and for now I'll leave it to you to judge how much detail to include.

If you want to learn some calculus while the text is in the review chapters you are welcome to move ahead on the q_a_ for the actual course. There are 16 q_a_ assignments, and they cover a large part of first-semester calculus, sometimes in a lot of depth, though not in the depth you will encounter in the text. Those q_a_'s stand pretty much independent of the other work in the course and make the text much easier to understand.

In any case, in order for me to see how well prepared you are need to complete the remaining calculus problems from the qa_init_probs, which increase quickly in difficulty beyond the point where you quit. I can't tell from the part you submitted just where your limits are, and not knowing that I can't tell you which topics you might be able to skip, or at least to do in less detail. This will be directly relevant in the first few modeling projects, in which I expect you will be able skip some of the redundancy.

There is no need for you to complete the remaining couple of questions in the q_a_areas ... , but when you are assigned the part about volumes and density you will need to do it. It will save you time in the long run.

Also be sure you read my note about the relevance of these ideas. They will be developed rather quickly into the central concepts you need in order to apply, as well as understand the foundations of, the subject.

Notes will be in large boldface, like this one.

x´É«´ßݼùåéäêÁcýÒÌìØìv”‡’€±þ Student Name: assignment #005 005. Calculus

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17:33:41 `questionNumber 50001 `q001. The graph of a certain function is a smooth curve passing through the points (3, 5), (7, 17) and (10, 29).

Between which two points do you think the graph is steeper, on the average?

Why do we say 'on the average'?

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RESPONSE --> between (7, 17) and (10, 29).

because the radius of the line varies, and there are only 3 points given to determine this. it is hard to tell exactly where the steepest part of the curve is without a graph. so, 'on the average' is said in reference to the only known reference points; the 3 points given, and therefore is not exact. hence it is said.

btw, i am really bad at typing. i wish i could go straight to the actual calculus part.

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17:35:51 `questionNumber 50001 Slope = rise / run.

Between points (7, 17) and (10, 29) we get rise / run = (29 - 17) / (10 - 7) =12 / 3 = 4.

The slope between points (3, 5) and (7, 17) is 3 / 1. (17 - 5) / (7 -3) = 12 / 4 = 3.

The segment with slope 4 is the steeper. The graph being a smooth curve, slopes may vary from point to point. The slope obtained over the interval is a specific type of average of the slopes of all points between the endpoints.

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RESPONSE --> neeeeeeeext.....

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18:49:25 `questionNumber 50001 2. Answer without using a calculator: As x takes the values 2.1, 2.01, 2.001 and 2.0001, what values are taken by the expression 1 / (x - 2)?

1. As the process continues, with x getting closer and closer to 2, what happens to the values of 1 / (x-2)?

2. Will the value ever exceed a billion? Will it ever exceed one trillion billions?

3. Will it ever exceed the number of particles in the known universe?

4. Is there any number it will never exceed?

5. What does the graph of y = 1 / (x-2) look like in the vicinity of x = 2?

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RESPONSE --> 2) 10, 100, 1000, 10000

1) every increasing digit to the right of the decimal and to the left of the 1 will be another 0 in the left (opposite) side of the decimal, to the right of the 1.

2) if the calculation is performed with the x value as 2.0000000001, then it will exceed a billion. with a numerically smaller # for x, the resulting quotent will get increasingly larger

3) if x is close enough to 2, then yes, it could.

4) if the calculation is performed with a decimal close enough to 2; then no.

5) if x = 2, then y =0. if the x axis is not streched, the slope in the line would be so steep (streching up and to the right from the y intercept) that there would be no visual slant; it would appear to be straight up and down.

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18:50:13 `questionNumber 50001 For x = 2.1, 2.01, 2.001, 2.0001 we see that x -2 = .1, .01, .001, .0001. Thus 1/(x -2) takes respective values 10, 100, 1000, 10,000.

It is important to note that x is changing by smaller and smaller increments as it approaches 2, while the value of the function is changing by greater and greater amounts.

As x gets closer in closer to 2, it will reach the values 2.00001, 2.0000001, etc.. Since we can put as many zeros as we want in .000...001 the reciprocal 100...000 can be as large as we desire. Given any number, we can exceed it.

Note that the function is simply not defined for x = 2. We cannot divide 1 by 0 (try counting to 1 by 0's..You never get anywhere. It can't be done. You can count to 1 by .1's--.1, .2, .3, ..., .9, 1. You get 10. You can do similar thing for .01, .001, etc., but you just can't do it for 0).

As x approaches 2 the graph approaches the vertical line x = 2; the graph itself is never vertical. That is, the graph will have a vertical asymptote at the line x = 2. As x approaches 2, therefore, 1 / (x-2) will exceed all bounds.

Note that if x approaches 2 through the values 1.9, 1.99, ..., the function gives us -10, -100, etc.. So we can see that on one side of x = 2 the graph will approach +infinity, on the other it will be negative and approach -infinity.

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RESPONSE --> ok...

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19:13:20 `questionNumber 50003 `q003. One straight line segment connects the points (3,5) and (7,9) while another connects the points (10,2) and (50,4). From each of the four points a line segment is drawn directly down to the x axis, forming two trapezoids. Which trapezoid has the greater area? Try to justify your answer with something more precise than, for example, 'from a sketch I can see that this one is much bigger so it must have the greater area'.

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RESPONSE --> A) the trapezoid formed from connecting the points (10,2) and (50,4) has the greater area than the trapezoid formed from connecting the points (3,5) and (7,9).

B) (10,2) and (50,4) are a lot farther apart than (3,5) and (7,9). this means that the trapezoid formed from connecting the long line to the short line is several times wider than the trapezoid formed from connecting the short line to the axis. without performing actual calculations one can see that the height of both trapezoids are roughly the same size, when compared to the width, so in other words although they are about the same in hight, one of them is definately a lot longer to the other.

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19:16:00 `questionNumber 50003 Your sketch should show that while the first trapezoid averages a little more than double the altitude of the second, the second is clearly much more than twice as wide and hence has the greater area.

To justify this a little more precisely, the first trapezoid, which runs from x = 3 to x = 7, is 4 units wide while the second runs from x = 10 and to x = 50 and hence has a width of 40 units. The altitudes of the first trapezoid are 5 and 9,so the average altitude of the first is 7. The average altitude of the second is the average of the altitudes 2 and 4, or 3. So the first trapezoid is over twice as high, on the average, as the first. However the second is 10 times as wide, so the second trapezoid must have the greater area.

This is all the reasoning we need to answer the question. We could of course multiply average altitude by width for each trapezoid, obtaining area 7 * 4 = 28 for the first and 3 * 40 = 120 for the second. However if all we need to know is which trapezoid has a greater area, we need not bother with this step.

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RESPONSE --> i hope i don't have to go into the amount of detail i have been doing for the actual CALCUALUS problems.

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19:17:08 `questionNumber 50004 `q004. If f(x) = x^2 (meaning 'x raised to the power 2') then which is steeper, the line segment connecting the x = 2 and x = 5 points on the graph of f(x), or the line segment connecting the x = -1 and x = 7 points on the same graph? Explain the basisof your reasoning.

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RESPONSE --> i answered the first 3 problems, so....

You need to complete this set. The difficulty progresses and the most important concepts of calculus are addressed in these quesitons. While you are obviously well prepared it is rare that anyone can answer them all correctly.

005. Calculus äZ³Üþpúùž±w…ˆQ¸Ãí“Á¿ÚrïºóCy Student Name: assignment #001 001. Areas

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16:10:03 `questionNumber 10001 `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 --> width X height = 2 dimesional area. so, 3m X 4m = 12m.

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16:51:39 `questionNumber 10001 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 --> oops. i forgot to put the 2nd dim. on the meters part.

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17:07:19 `questionNumber 10002 `q002. What is the area of a right triangle whose legs are 4.0 meters and 3.0 meters?

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RESPONSE --> the triangle is half the volume of the this polygon. so ( 12 m^2 )/2 = 6 m^2

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17:08:07 `questionNumber 10002 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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RESPONSE --> ok

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17:15:31 `questionNumber 10003 `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 --> if the paralellogram is square, (4 right angles) then: 5 X 2 =10 m^2

if it is not square, then ( parallel sides are obviously in a parallegram) so one angle needs to be given

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17:18:12 `questionNumber 10003 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 --> funny, " rearranged into a rectangle" wasn't mentioned in the problem

The given solution explains why the area of a parallelogram is what it is. You will do well to note it.

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17:26:46 `questionNumber 10004 `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 --> -again, not enough info.

-I WILL ASSUME this is a 90 degree (right) triangle.

This is an unnecessary assumption.

-5 X 2 = 10 m^2

-10/2 =5 m^2

-5 m^2 is the area of the assumed right triangle.

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17:28:13 `questionNumber 10004 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 --> i thought this was calculus...............

You'll be up to your neck in calculus before you know it. When you're 1/4 of the way through the course email me and ask me to enumerate the ways these geometric concepts have been used, and how necessary it is that you be able to use this reasoning in order to understand the fundamental concepts of differentiation and integration.

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17:49:40 `questionNumber 10005 `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 --> this is my trapozoid. the line intersecting the middle is what u mean by average altitude-??? _______ / \ _ _ / _ _ _ \ _ _ / \ /___________\

but i don't know the angle of the sides. -not completly sure, but i think the area is dependant on that-???

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17:54:11 `questionNumber 10005 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 --> ok. so the angle doesn't matter. i think i understand what you meant now.

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17:58:44 `questionNumber 10006 `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 --> using your method>>>>

the ave. of 3 & 8 is 5.5 so 5.5cm X 4cm = 22 cm^2

22 cm^2 is the area of the trapezoid.

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17:59:20 `questionNumber 10006 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 --> ok..................

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18:04:21 `questionNumber 10007 `q007. What is the area of a circle whose radius is 3.00 cm?

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RESPONSE --> pi X r^2 = area of circle

3 X 3 =9. 9 X pi = approx. 28.25

area of circle is approx. 28.25 cm ^2

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18:05:46 `questionNumber 10007 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 --> ok....

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18:09:37 `questionNumber 10008 `q008. What is the circumference of a circle whose radius is exactly 3 cm?

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RESPONSE --> dia = radius X 2 circum. = pi X dia

3 X 2 = 6 6 X pi = 8.8 cm

8.8 cm is s the circumference

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18:09:55 `questionNumber 10008 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 --> ok.......

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18:14:02 `questionNumber 10009 `q009. What is the area of a circle whose diameter is exactly 12 meters?

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RESPONSE --> this is getting old FAST.

dia=2r r^2Xpi=area circle

12\2=6 6^2=36 36Xpi=113.04 113.04 is approx. area of circle.

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18:14:53 `questionNumber 10009 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 --> oK........

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18:31:04 `questionNumber 10010 `q010. What is the area of a circle whose circumference is 14 `pi meters?

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RESPONSE --> don't know what the hell `pi meters is i guessing

crcm.\ pi= dia. (approx) dia\2=radius (radius^2) ()Xpi=area

14/3.14= aprx4.46 4.46/2=aprx2.23 2.23^2=aprx4.97 4.97Xpi=aprx15.61 area is =aprx15.61m or ll `pi meters-???

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18:32:43 `questionNumber 10010 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 --> thats wierd never seen that before........

pi stands for pi, so 14 pi stands for 14 * pi.

Is it easier to identify that the radius of a circle is 7 knowing that the area is 49 pi m^2 or from knowledge that the area, to ten significant figures, is 153.9380400?

The normalizing factor for the normal distribution is 1 / sqrt(2 pi). If you integrate 1 / sqrt(2 pi) * exp(-z^2 / 2) from z = -infinity to infinity you get 1. If you use a decimal approximation and integrate, say, 0.398942 * exp(-z^2 / 2) from z = -infinity to infinity you get 0.9999992971. The difference is important in some very relevant statistical situations.

There are cases where it's best to leave results as a multiply of pi, and of course situations in which you want the decimal. You need to have the flexibility to do it either way.

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18:38:31 `questionNumber 10011 `q011. What is the radius of circle whose area is 78 square meters?

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RESPONSE --> `q011- thought this was only 10 q's?

o god i just want to learn calculus.........

78/pi=24.8 sqrt24.8=5 5 m

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18:38:47 `questionNumber 10011 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 --> duh

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18:44:51 `questionNumber 10012 `q012. Summary Question 1: How do we visualize the area of a rectangle?

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RESPONSE --> um...as 2 ident. triangles????

as as the width x the hight???

as a quadraterial thats a paralelogram with 90 deg corners???

OR how do u see it ???????

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18:46:33 `questionNumber 10012 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 --> so how do we visualize the little squares???

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18:50:27 `questionNumber 10013 `q013. Summary Question 2: How do we visualize the area of a right triangle?

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RESPONSE --> either half of a line that is drawn diagonally from corner to corner of a rectangle.

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18:50:53 `questionNumber 10013 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 --> xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

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18:52:08 `questionNumber 10014 `q014. Summary Question 3: How do we calculate the area of a parallelogram?

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RESPONSE --> this is it. i'm sick of redundancy

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