credit maths
DESCRIPTION
Credit Maths. Revision Chapters 1 - 4. Standard Form/Scientific Notation. 50 200. 5.02 10 4 =. 4.7 10 7. 47 000 000 =. 0.00031. 3.1 10 -4 =. 7.04 10 -5. 0.0000704 =. Large Numbers. a. Change to standard form. 9.9 10 4. i. 99000 =. 6.104 10 8. - PowerPoint PPT PresentationTRANSCRIPT
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Credit Maths
Revision Chapters 1 - 4
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5.02 104 =
50 200
47 000 000 = 4.7 107
3.1 10-4 =
0.00031
0.0000704 = 7.04 10-5
Standard Form/Scientific Notation
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Large Numbers
a. Change to standard form
b. Change to normal form
i. 99000 = 9.9 104
i. 6.03 103 = 6.0360.3603.6030.
ii. 610400000 = 6.104 108
iii. 5223000 = 5.223 106
ii. 8.15 105 = 8.1581.5815.815000.8150.81500.iii. 2.7 106 = 2.7270.2700.2700000.27000.270000.27.
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Small Numbers
a. Change to standard form
b. Change to normal form
i. 0.0039 = 3.9 10-3
i. 3.07 10-3 = 3.070.3070.03070.00307
ii. 0.000502 = 5.02 10- 4
iii. 0.0000004 = 4 10-7
ii. 5.91 10-5 = 5.910.5910.05910.00005910.005910.000591iii. 8.6 10-6 = 8.60.0860.00860.00000860.000860.0000860.86
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5.07 104 =
50 700
457 000 000 =4.57 108
3.1 107 = 31 000 000
9 070 000 = 9.07 106
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E.g. Fractions Non calc
7
31
10
12. a
9
22
3
13. b
4
32
3
15. c
7
10
10
21
3
11
1
31
3
9
20
3
10
20
9
3
10
1
21
3
2
11
2
3
4
3
3
13
12
9
12
43
12
9
12
162
12
52
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Ex Fractions Non calc
5
22
6
14. a
15
22
3
22. b
4
14
5
43. c
5
12
6
25
5
11
2
101
10
15
32
3
8
32
15
3
8
1
41
5
4
11
4
5
4
1
5
47
20
5
20
167
20
217
20
18
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Similar Shapes
5cm
x cm
2cm 8cm
15cm
5 cm
y cm
6cm
6cm
Small
BigEnl s.f. = 4
2
8
x = 4 × 5 = 20cm
big
SmallRed s.f. = ..333.0
15
5
y = 0.333. × 6 = 2cm
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Ex 1 Find x
20m
40 m
x m 22 m
big
smallRed s.f. = 5.0
40
20
x = 0.5 × 22 = 11m
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Ex 2 Find x
big
smallRed s.f. = 75.0
8
6
x = 0.75 × 20 = 15cm
20cm x cm
6 cm8cm
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Find x to 1 d.p.
20 m
2 m
x m
12m
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Find x to 1 d.p.
22 m 20 m
x m
12m
small
bigEnl s.f. = 1.1
20
22
x = 1.1 × 12 = 13.2m
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Similar Areas and Volumes
8
5.
big
smallfsred
64
25
8
5.
2
fsArea
e.g.1 e.g.2
8m5m
21m2?m2
12m9m
?m3
60m3
22.82164
25? m
9
12.
small
bigfsEnl
27
102
9
12.
3
fsVol
32.1429
214260
27
102? m
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Enlargement
Linear Scale Factor = k =
Area Scale factor = k2 =
Missing Length = 2.4 x 10 = 24 cm
4.276.5
? cm
10 cm
25 cm2 144cm2
76.525
144
Find the missing dimension
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Enlargement
Linear Scale Factor = k =
Area Scale factor = k2 = 32 = 9
Missing area = 9 x 30 = 270 cm2
35
15
5cm30cm2
? cm2
15cm
1.
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Enlargement
Linear Scale Factor = k =
Area Scale factor = k2 =
Missing area = 4 x 80 = 320 mm2
215
30
15 mm
30mm
?mm2
2.
80 mm2
22 = 4
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Reduction
Linear Scale Factor = k =
Area Scale factor = k2 =
Missing area =
3
1
18
6
3.
? mm2
540 mm2
6 mm
18 mm
9
1
3
12
2mm605409
1
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Reduction
Linear Scale Factor = k =
Area Scale factor = k2 =
Missing Length = 0.3125 x 7.5 = 2.34 cm
3125.009765625.0
09765625.012.5
5.0
4.
0·5 m2
5.12 m2
? m
7·5 m
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Enlargement
Linear Scale Factor = k =
Volume Scale factor = k3 =
Missing Volume = 20 x 4.096 = 81.92cm3
6.15
8
5cm 8cm
V= 20cm3 V= x cm3
5.
096.46.15
8 33
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Distance Speed Time
D
S T
T
DS
S
DT TSD
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STARTER QUESTIONS
S
DT
sm
m
sm
kmT
/5
2000
/5
2
400T secs
T
DS
hrs
km
hrs
kmS
25.4
272
smin154
272
64S km/h
a. D = 2km and S = 5m/s T= ?
b. D = 272km and T = 4 hrs 15 mins S = ?
c. T = 3mins and S = 4m/s D = ?
STD
smin3/4 smD
ssmD 180/4
720mD
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1. A bus leaves the airport at 19.24 and arrives in the city centre at 20.11. If the journey is 17 miles calculate its average speed in mph to 1 decimal place
T
DS
47mins
17m hrs6047
17m
0.783
17m
21.7mphS
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2. Stewart drove 40 km from Laurencekirk to the golf course in Montrose. If he averaged 72km/h calculate the time of his journey to the nearest minute
S
DT
hkm
kmT
/72
40 hrsT 55.0 33minsmins 33.3
3. Mr Sim drove to Dover in the holidays. He left at 06.30 and arrived at 14.15. If he averaged 70mph, how far did he travel?
TSD mins 45 7hrs mph 70 D
hrs 7.75 mph 70 D542.5D miles
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Positive and Negative Numbers
With Algebra
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1. Basic Rules: Pos and Neg = NegNeg and Neg = Pos
Examples
a. 2 + (5) ( 4) b. 4 × (2) c. 24 ÷ (6)
= 2 5 + 4
= 1
= 8 = 4
Exercise
a. 3 (1) + ( 7) b. 42 ÷ (7) c. 6 +(15) ÷ 3
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2. Simplify: Pos and Neg = NegNeg and Neg = Pos
Examples
a. 2a + (5b) ( 4a) b. 2a2 +(3a)(4a2)(5a)
= 2a 5b + 4a
= 6a 5b
= 2a2 3a + 4a2 + 5a
Exercise
a. 3x + (y) (5x) b. 5m2 (3n2)(4m2)
= 6a2 + 2a
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3. Simplify: Pos and Neg = NegNeg and Neg = PosMultiplying makes powersExamples
a. 2a × 3a × a b. 3c × (c) × (2c) × 4c × (c)
= 6 a3 = 24
Exercise
a. 2y × 3y × y × y × y b. 2m × (5m) × (7m) × 2
c5
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4. Multiply out: Pos and Neg = NegNeg and Neg = Pos
Examples
a. –3(x + 4) b. – 4(2y – 3) c. –p(3 – p)
= –3x – 12 = –8y + 12
Exercise
a. –5(w + 2) b. 3(2m 5) c. 3p(p 4)
= –3p + p2
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5. Simplify: Pos and Neg = NegNeg and Neg = Pos
Examples
a. –3(x + 2) + 8 b. 2(y + 2) + 4(2y – 3) c. 2p(4p – 3) – 3p(p – 3)
= –3x – 6 + 8 = 2y + 4 + 8y – 12
Exercise
= 8p2 – 6p – 3p2 + 9p
a. 2(3c – 5) + 4(2 + c) b. 3(p – 4) – 2(p 5)
= –3x + 2 = 10y – 8 = 5p2 + 3p
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6. Evaluate: a = 3, b = -2, c = -5
ExamplesA) a2 – b B) abc + 2c C) b(2a – c)
= 32 – (–2) = 3 × (–2) × (–5) + (–10)
Exercise
A) 4a + c B) c2 – b C) ab – 3b
= –2(6 – (–5))= 9 + 2 = 30 – 10 = –2(6 + 5)= 11 = 20 = –2(11)
= –22
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7. Solve
Examples
a. 5x + 2 = x + 26 b. 2(3y – 1) = 2y + 10 c. 3m – 4 = –2(m – 4)
5x x = 26 – 2 6y – 2 = 2y + 10
Exercise
a. 7w + 2 = w + 14 b. 3(2m – 1) = m + 16
3m – 4 = –2m + 8
c. 2(3c – 5) = 4(2 + c) d. 3(p – 4) = –2(p 5)
4x = 24 6y – 2y = 10 + 2 3m + 2m = 8 + 4 x = 6 4y = 12
y = 3 5m = 12 m = 12/5