Add and Subtracting Vectors: A Mini-Investigation.
In the input line in the applet below, type in the following one at a time, and carefully observe what happens as you drag the "arrows" or vectors around. Click on the RESET icon to start over.

(1) u+w
(2) u-w
(3) w-u
(4) 2u+w
(5) u+2w

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Problem 1, Page 7
page 7 problem 1
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Problems 4 and 5, Page 7

Problem 7, Page 7
Page 7 Problem 7
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Problem 6, Page 7
454590.png
45-45-90 Special Right Triangle

The figure at left shows a picture of the SPECIAL RIGHT TRIANGLE known as the 45
o-45
o-90
oRIGHT TRIANGLE. It may also be referred to as an ISOSCELES RIGHT TRIANGLE. This triangle is one of the more important triangles you will run across, so you MUST commit the side relationships and angles to MEMORY!

How are the a and c related? By the Pythagorean Theorem, we have



but a represents a length, so a will always be positive. Therefore, we have



or, written is a more common format,





Problems 9 and 10, Page 7



Problem 8, Page 8

Because the length of the vector is [5,3], add 5 to all of the x values:
external image latex2png.2.php?z=100&eq=1%2B5%3D6%5C%5C%0A3%2B5%3D8%5C%5C6%2B5%3D11

Add 3 to all of the y values:
external image latex2png.2.php?z=100&eq=2%2B3%3D5%5C%5C-5%2B3%3D-2%5C%5C1%2B3%3D4

So the vertices of the triangle are:
A'=(6,5)
B'=(8,-2)
C'=(11,4)

These values become the vertices for the translated triangle shown below.
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Problem 9, Page 8

Find the difference between the point A (1,2) and point A' (-3,7)
external image latex2png.2.php?z=100&eq=-3-1%3D-4%5C%5C7-2%3D5

The vector is [-4,5]. Add this to the x values of the remaining points to find the vertices for the new triangle.
external image latex2png.2.php?z=100&eq=3-4%3D-1%5C%5C6-4%3D2

Add this to the y-values, too.
external image latex2png.2.php?z=100&eq=-5%2B5%3D0%5C%5C1%2B5%3D6

So the points are:
A'=(-3,7)
B'=(-1,2)
C'=(0,6)


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Problem 10, Page 8

The number of squares is infinite, because it can always be divided by itself into smaller squares.




Problem #4 Page 8 by Maddie and Emma
maggie g
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eq=x=-4+3t
eq=x=-4+3t

eq=y=1+2t
eq=y=1+2t

The slope of this line is 2/3. You can tell by using your equations or the line you graph. The two coefficents in front of t is your slope!