MRSM BALING, KEDAH

Mathematics Form 1 (IGCSE)Homework

Semester 2 2012

Name: Siti Mazenah Dienta Rochani binti Mohamed Tajudin

Matric Number : 12094

Class: 1D ( Jauzi)

Teacher's Name : Mrs. Wan Siti Hajar Wan Hashim

Parent's Name : Mohamed Tajudin bin Alias ( 016-6006045)

Date:

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XTajudin Alias

Manager

TRIGONOMETRY

1.

Calculate the value of cos θ in the following triangle.

Solution:

2.

Calculate the length of the side x, given that sin θ = 0.6.

Solution:

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3.

Calculate the length of the side x, given that tan θ = 0.4

Solution:

4.

3cm

5cm

Calculate x.

Solution:

Using Pythagoras’ theorem:

x2= 32+52

x = √32+52

= √9+25

= √34

= 5.83 cm

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X

5.

12 m

Calculate y.

Solution:

Using Pythagoras’ theorem:

122 = y2+92

y2 = 122−92

y = √144−81

= √63 = 7.94 m

6.

Calculate a.

Solution:

The missing side is the opposite side and we have the adjacent and angle, hence,

7. Find x and H in the right triangle below. 

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9m

Y

a

34°

1.2 cm

tanΘ = oppadj

tan34Θ = a

1.2a = 1.2 x tan34Θ = 0.809 cm

Solution:

x = 10

tan 51

x = 13 (2 significant digits) 

8. Find the lengths of all sides of the right triangle below if its area is 400. 

Solution:

Area = (1/2)(2x)(x) = 400 

Solution for x:

x = 20x2 = 40 H = x sqrt(5)

H= 20 sqrt(5)

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H = 10

sin 51

H= 8.1 (2 significant digits)

Pythagoras theorem: (2x)2 + (x)2 = H2 

9. BH is perpendicular to AC. Find x the length of BC. 

Solution:

BH perpendicular to AC, means that triangles ABH and HBC are right triangles. Hence,

tan(39o) = 11 / AH or AH = 11 / tan(39o) 

HC = 19 - AH = 19 - 11 / tan(39o) 

Solution for x and substitution for HC:

x = sqrt [ 112 + (19 - 11 / tan(39o) )2 ] 

= 12.3 (rounded to 3 significant digits)

10.

The area of a right triangle is 50. One of its angles is 45o. Find the lengths of the sides and hypotenuse of the triangle.

Solution:

The triangle is right and the size one of its angles is 45o; the third angle has a size 45o and therefore the triangle is right and isosceles. Let x be the length of one of the sides and H be the length of the hypotenuse. 

Area = (1/2)x2 = 50 , solve for x: x = 10 

We now use Pythagoras theorem to find:

H: x2 + x2 = H2 

Solution for H:

H = 10 sqrt(2)

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Pythagoras theorem applied to right triangle HBC: 112 + HC2 = x2 

11. ABC is a right triangle with a right angle at A. Find x the length of DC.

Solution:

Since angle A is right, both triangles ABC and ABD are right and therefore we can apply Pythagoras theorem. 

142 = 102 + AD2 , 162 = 102 + AC2 

Also x = AC - AD 

= sqrt( 162 - 102 ) - sqrt( 142 - 102 )

= 2.69 (rounded to 3 significant digits)

12. Given the following right triangle, find tan A.

Solution:

tan A = 32

tan = 0

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13. Find the missing side of the following triangle.

Solution:

Using Pythagoras theorem,

=

14. Find the area of the following triangle.

Solution:

A = 12

xy

15. The angle of repose for sand is typically about 35°. What is the sine of this angle? 

Solution:

1. Type 35 into your calculator

2. Press the sin button.

3. Your calculator should read 0.574. 

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√ x2+ y2

ALGEBRAIC EXPRESSIONS

1. Simplify the expression −3 (4 x−2 )−2 (5 x−4 ).

Solution:

= −12 x+6 -10 x+8= −12 x−10 x+6+8= −22 x+14

2. Simplify the following algebraic expression.

-2x + 5 + 10x - 9 

Solution:

= (10x - 2x) + (5 - 9) = 8x - 4

3. Evaluate 8x + 7 given that x - 3 = 10.

Solution:

= 8 (13 )+7=104+7=111

4. Simplify the following algebraic expression.

3(x + 7) + 2(-x + 4) + 5x 

Solution:

= 3x + 21 - 2x + 8 + 5x = (3x - 2x + 5x) + (21 + 8) = 6x + 29

5. Evaluate: -18 + 4(6 ÷ 2)2

Solution:

= -18 + 4(6 ÷ 2)2 = -18 + 4(3)2 = -18 + 4*9 = -18 + 36 = 18

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6. Simplify: 12x3 - 3(2x3 + 4x -1) - 5x + 7 .

Solution:

=12x3 - 3(2x3 + 4x -1) - 5x + 7 = 12x3 - 6 x3 - 12 x + 3 - 5x + 7  = 6 x3 - 17 x + 10

7. Solve the equation: -5x + 20 = 25

Solution:

= −5 x=25−20= −5 x=5= x=−1

8. Solve 4 y+8 x−16 y

Solution:

= 4 y−16 y+8 x= −12 y+8 x

9. Simplify 3x + 1 + 8x + 9.

Solution:

= 11x+1010. Simplify 2x + 5y - 7x + 8y.

Solution:

= 2 x−7 x+5 y+8 y= −5 x+13 y

11. Calculate 3(4y+9)-4

Solution:

= 12 y+27−4= 12 y+23

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¿3 x+8 x+1+9

12. Simplify 7(4+5)+8s

Solution:

= 28+35+8 s= 63+8 s

13. Evaluate 3x+5(7-4)

Solution:

= 3 x+35−20= 3 x+15

14. 5y+6+4y-6x

Solution:

= 5 y+4 y−6 x+6= 9 y−6x+6

15. 4(7y-6) x 8

Solution:

= = 28y - 192

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28 y−24 x8

INTEGERS

1. -3 + (-2)

Solution:

= −3−2= -5

2. -12 + (-7) + (-5)

Solution:

= -12 + (-7) = -19 = -19 + (-5) = -24

3. -8(-3)

Solution:

= -8 x -3= 24

4. (-4) 5

Solution:

= -1024

5. Add 17 + (-3) + (-2) + 5.

Solution:

= 17−3−2+5= 12 + 5= 17

6.  Multiply (–9) (6).

Solution:

= -9 x 6= -54

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7. Divide (–22) ÷ (–11).

Solution:

= -22 ÷ -11= 2

8. Simplify 8 – 4×2 + 6.

Solution:

= 8 - 8 + 6= 6

9. Simplify 5 × (13 – 6) 3 + 96 ÷ 6.

Solution:

= 5 x21+16= 1731

10. Simplify 8 – 3 + 2 – 7 – 5.

Solution:

= 5 + (-5-5)= 5 + (-10)= -5

11. Multiply (–7) (+9).

Solution:

= -7 x 9= -63

12. Find (–9) 4.

Solution:

= -9 x 4= 6561

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13. Add -9 + 6.

Solution:

= -3

14. Divide (–48) ÷ (+6).

Solution:

= -48 ÷ 6= -8

15. Simplify 5×2 2 + 4 – 3.

Solution:

= 5 x 4 + 1= 21

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LINEAR EQUATION

1. Solve for x.

x - 4 = 10

Solution:

X = 10+4X = 14

2. Solve x.

Solution:

=

= =

= = = = x = 3

3. Solve x.

2(3x - 7) + 4 (3 x + 2) = 6 (5 x + 9) + 3

Solution:

= = = =

=

=

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4. 2x - 4 = 10

Solution:

x = 2 x=10+4X = 2 x=14X = 7

5. 5x - 6 = 3x - 8

Solution:

X = 5 x−3 x=−8+6X = 2 x=−2X = -1

6. Solve x.

Solution:

X = 12( 34

x+ 56 )=12(5x−125

3)

X = 12( 34

x+ 56 )=12(5 x−125

3)

X= 12( 34

x)+12( 56 )=12 (5 x )−12

1253

X = 3(3x) + 2(5) = 60x - 500X = 9x + 10 = 60x - 500X = 10 = 51x - 500X= 510 = 51x

X = 51051

X= 10

7. Find x if: 2x + 4 = 10

Solution:

X = 2x + 4 - 4 = 10 - 4X = 2x = 6X = 2x / 2 X = 6 / 2

X = 3

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8. Find x if: 3x - 4 = -10.

Solution:

X = 3x - 4 + 4 = -10 + 4 X =3x X = -6 X = 3x / 3 X = -6 / 3 X = -2

9. Find x if: 4x - 4y = 8.

Solution:

X= 4x - 4y + 4y = 8 + 4yX= 4x = 8 + 4yX= 4x / 4 = (8 + 4y) / 4X= 2 + y

10. Find x if: x + 32 = 12

Solution:

X= x+32=12X= x + 9 = 12X= x + 9 - 9X= 12-9X= 3

11. Solve  4 x = 36..Solution:

X = 44

x=364

X = 9

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12. Work out 6(8 – 2 x )+ 25 = 5(2 – 3 x ).

Solution:

13. Explain 7(2 x – 3) – 4( x + 5)= 8( x – 1)+ 3.

Solution:

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14. Find x for .

Solution:

15. Solve x – 4(3 x – 2) – (– x + 6)= – 5 x + 8.

Solution:

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STRAIGHT LINE GRAPH

1. What are the coordinate of Point A marked on the following grid?

Answer :

= 1, 1

2. Find the equation of the straight line shown.

Solution:

The line passes through the points (0, 4) and (1, 1).So when the x-coordinate goes up by 1, the y-coordinate goes down by 3.

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-10-8-6-4-202468

10

-2 -1 0 1 2 3 4

8

6

4

2

0

-2

-4

-6

-8

So the gradient is

gradient =−3

1=−3

As the y-intercept is 4, the equation of the line must be y = -3x + 4.

3. Draw the lines y = 3x – 2 and y = 6 – 2x.

Solution:

Table of values for y = 3x – 2

x 0 1 2 3y 3×0 - 2

= -23×1 - 2

= 13×2- 2

= 43×3 - 2

= 7

Then, plot the points (0, -2), (1, 1), (2, 4) and (3, 7) and join them up.

Table of values for y = 6 – 2x

x 0 1 2 3y 6 - 2×0

= 66 - 2×1

= 46 - 2×2

= 26 - 2×3

= 0

We plot the points (0, 6), (1, 4), (2, 2) and (3, 0).

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Work out the y value corresponding to each x value using the formula

y = 3x - 2.

2 4 6 8

4. Write down the straight line equation from the information given.

M=3 C=4

Answer:

y = 3x+4

5. Write down the gradient and y-intercept of this equation.

y = 2(x-4)

Answer:

= 2, -8

6. Find the equation of line that passes trough point (-6, -2), (-4, -7)

Answer:

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y = −52

x−17

7. Write down the y-intercept of the following equation: 3 y=2(3−2 x)

Answer:

Y = 1 12

8. What is the y-intercept of this graph?

Answer:

= 3

9. Complete the table of values for y = x + 3, and then draw the graph.

Solution:

x -3 -2 -1 0 1 2 3Y= x + 3 0 1 2 3 4 5 6

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10. Write down coordinate P.

Answer:

= 2, 1

11. Find the y-intercept of the straight line that passes through (1, 5) and (5, 21).

Solution:

y = mx + c

5 = 4 × 1 + c

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m = (y2 – y1) ÷ (x2 - x1)

m = (21 - 5) ÷ (5 – 1)

m = 16 ÷ 4

5 = 4 + c

1 = c

12. Find the equation of the straight line that passes through (-3, -12) and (1,-4).

Solution:

y = mx + c

-12 = 2 × -3 + c

-12 = -6 + c

-6 = c

=  y = 2x - 6

13. For the straight line y = -2x + 3, what are: 

a) the slopeb) the y-intercept?

Answer:

= Slope = 2, y-intercept = 3

14. Find the equation of this linear graph.

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m = (y2 – y1) ÷ (x2 - x1)

m = (-4 - -12) ÷ (1 - -3) m = 8 ÷ 4

m = 2

Solution:

y = 3x – 5

15. Find the equation of this linear graph.

Solution:

= y = - 13

x + 3

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m = 6 ÷ 2 = 3

m = 3 ÷ 9 = −13

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