Download WBHS_Gr 10_Inv_170117 memo

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WBHS Mathematics
Grade 10 Algebra Patterns Investigation Memo– 17 Jan 2017
Marks: 45
Time: 1 hour
1. A calculator may be used in this investigation.
2. Part 1 and Part 2 must be detached and handed in with your script.
NAME: ______________________________
MATHS TEACHER: ___________________
PART 1 (to be completed on this page, detached and handed in with your script)
Below is a “number cell”. In a number cell, the first two numbers are added to get the third number. The
second and third numbers are added to get the fourth etc
3
1.1.
4
7
11
18
12√
19√
31√
Complete the following number cell:
5
7
(3)
1.2.
Complete the following number cell in which you are given only the first and last numbers.
12√
8
20√
32√
52
(3)
1.3.
Complete the following number cell:
x
x+ y
2x+y√
3x+2y√
5x+3y√
8x+5y√
13x+8y√
21x+13y√
(6)
1.4.
Describe fully the pattern that you observe in the coefficients of x and y.
The x-co-efiicients: the first two numbers are added to get the third number. The second and third
numbers are added to get the fourth etc. or it is the Fibonacci sequence.√
The y-co-efficients: follow the same pattern as for the x-co-efficients (Fibonacci).√
(2)
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PART 2 (to be completed on this page, detached and handed in with your script)
The numbers 1 to 64 are arranged in an 8 × 8 grid below.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
A 3 × 3 square is drawn on the grid. The opposite corners are multiplied:
34 × 20 = 680
36 × 18 = 648
The difference is found:
680 – 648 = 32.
2.1
Using the same grid above, do this with two more examples.
22 x 8 = 176√
54 x 40 = 2160√
6 x 24 = 144√
38 x 56 = 2128√
176 – 144 = 32√
2128 – 2160 = 32√
(Any 2 correct examples).
2.2
(any 2 correct egs.)
(6)
Each 8 x 8 has a general form of:
x
x+1
x+2
x+8
x+9
x+10
x+16
x+17
x+18
√
Now, (x + 16)(x + 2) = x² + 18x + 32√
And x(x + 18) = x² + 18x√
Therefore x² + 18x + 32 – (x² + 18x) = 32√
(3)
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PART 3
3.1
50cm 50cm 50cm50cm
100cm
50cm 50cm
100cm
100cm
50cm 50cm
50cm
100cm
√
50cm
100cm
100cm
100cm
√
√
100cm
√
(4)
3.2
You add the 2 preceding number of pavers √to get the next amount of pavers√ (Fibonacci)
(2)
3.3
89√√
(2)
4.1
Any 2 rows which are correct.
(2)
4.2
A number squared plus one more than this number squared√ plus the product of these two numbers
squared√ is equal to one more than the product of these two numbers squared.√
(3)
4.3
n2 + (n + 1)2 Ö + (n(n + 1))2 Ö = (n(n + 1) + 1)2 Ö
4.4
RTP:
(3)
4
n 2 + (n + 1) 2 + (n(n + 1)) 2 = (n(n + 1) + 1) 2 Ö
LHS = n 2 + n 2 + 2n + 1 + n 4 + 2n3 + n 2
= n 4 + 2n3 + 3n 2 + 2n + 1 Ö
RHS = (n 2 + n + 1) 2 Ö
(4)
= n 4 + n3 + n 2 + n3 + n 2 + n + n 2 + n + 1
= n 4 + 2n3 + 3n 2 + 2n + 1 Ö
\ LHS = RHS
4.5
192 + 202 + 3802 = 3812 Ö Ö
(2)