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141
Chapter 5
RS
1
RT
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.
?
ST
16.5
1
16.5
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.
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33
33
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5
33
Because the sum of two measures equals the
third measure, the sides cannot form a triangle
and so the coordinates cannot be the vertices of
a triangle.
45.
Consider all possible triples using the lengths
3, 4, 5, 6, and 12.
3
1
4
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.
?
54
1
5
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.
?
33
1
5
¬
.
?
4
7
¬
.
5
✓
9
¬
.
3
✓
8
¬
.
4
✓
3
1
4
¬
.
?
64
1
6
¬
.
?
1
6
¬
.
?
4
7
¬
.
6
✓
10
¬
.
3
✓
9
¬
.
4
✓
3
1
4
¬
.
12
7
¬
.
12
4
1
5
¬
.
?
1
6
¬
.
?
55
1
6
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.
?
4
9
¬
.
6
✓
10
¬
.
5
✓
11
¬
.
4
✓
4
1
5
¬
.
?
12
9
¬
.
12
3
1
5
¬
.
?
65
1
6
¬
.
?
1
6
¬
.
?
5
8
¬
.
6
✓
11
¬
.
3
✓
9
¬
.
5
✓
3
1
5
¬
.
?
12
8
¬
.
12
3
1
6
¬
.
?
12
9
¬
.
12
4
1
6
¬
.
?
12
10
¬
.
12
5
1
6
¬
.
?
12
11
¬
.
12
Of all possible triples, 4 of them satisfy the
triangle inequality, so there are 4 possible
triangles.
46.
3
1
4
1
5
5
12, which is divisible by 3
3
1
4
1
6
5
13, which is not divisible by 3
4
1
5
1
6
5
15, which is divisible by 3
3
1
5
1
6
5
14, which is not divisible by 3
Carlota could make 2 different triangles with a
perimeter that is divisible by 3.
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This note was uploaded on 12/19/2011 for the course MAC 1140 taught by Professor Dr.zhan during the Fall '10 term at UNF.
 Fall '10
 Dr.Zhan
 Calculus, PreCalculus

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