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)

XTajudin Alias Manager

Date:

Maths is easy and fun

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TRIGONOMETRY

1.

Calculate the value of cos in the following triangle. Solution:

2.

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

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3.

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

4.X

3cm5cm

Calculate x.Solution:

Using Pythagoras theorem: = x= = = = 5.83 cm

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5.9m 12 m

Y Calculate y.Solution:

Using Pythagoras theorem: = = = = = 7.94 m

6.

a

34 1.2 cm

Calculate a.Solution:

The missing side is the opposite side and we have the adjacent and angle, hence, tan = tan34 =

a = 1.2 x tan34 = 0.809 cm Maths is easy and fun Page 4

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

Solution:

x= x = 13 (2 significant digits)

H= H= 8.1 (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) Pythagoras theorem: (2x)2 + (x)2 = H2

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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(39 ) 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:o

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

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) Maths is easy and fun Page 6

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 = 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=

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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Page 8

ALGEBRAIC EXPRESSIONS 1. Simplify the expressionSolution:

(

)

(

).

= = =

-

2. Simplify the following algebraic expression. -2x + 5 + 10x - 9Solution:

= (10x - 2x) + (5 - 9) = 8x - 4 3. Evaluate 8x + 7 given that x - 3 = 10.Solution:

= ( ) =104+7 =111 4. Simplify the following algebraic expression. 3(x + 7) + 2(-x + 4) + 5xSolution:

= 3x + 21 - 2x + 8 + 5x = (3x - 2x + 5x) + (21 + 8) = 6x + 29 5. Evaluate: -18 + 4(6 2)2Solution:

= = = = =

-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 = 25Solution:

= = =

8. SolveSolution:

= =

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

= 11x+10

10. Simplify 2x + 5y - 7x + 8y.Solution:

= =

11. Calculate 3(4y+9)-4Solution:=

=

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12. Simplify 7(4+5)+8sSolution:

= =

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

= = 14. 5y+6+4y-6xSolution:

==

15. 4(7y-6) x 8Solution:

= = 28y - 192

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INTEGERS

1.

-3 + (-2)

Solution:

= = -5 2. -12 + (-7) + (-5)Solution:

= -12 + (-7) = -19 = -19 + (-5) = -24 3. -8(-3)Solution:

= -8 x -3 = 24 4. (-4) 5Solution:

= -1024 5. Add 17 + (-3) + (-2) + 5.Solution:

= = 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 42 + 6.Solution:=8

-8+6

=6 9. Simplify 5 (13 6) 3 + 96 6.Solution:

= = 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 52 2 + 4 3.Solution:

=5x4+1 = 21

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LINEAR EQUATION1. Solve for x.

x - 4 = 10Solution:

X = 10+4 X = 14 2. Solve x.

Solution:

= = = = = ==3. Solve x. 2(3x - 7) + 4 (3 x + 2) = 6 (5 x + 9) + 3Solution: = =

x = 3

= = = =

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4.

2x - 4 = 10

Solution:

x=X = X =7 5. 5x - 6 = 3x - 8Solution:

X= X= X = -1 6. Solve x.

Solution:

X= X=

( (

) )

( ( )

) )

( X= ( ) ( ) X = 3(3x) + 2(5) = 60x - 500 X = 9x + 10 = 60x - 500 X = 10 = 51x - 500 X= 510 = 51x X= X= 10 7. Find x if: 2x + 4 = 10Solution:

X = 2x + 4 - 4 = 10 - 4 X = 2x = 6 X = 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 + 4y X= 4x = 8 + 4y X= 4x / 4 = (8 + 4y) / 4 X= 2 + y 10. Find x if: x + 32 = 12Solution:

X= X= x + 9 = 12 X= x + 9 - 9 X= 12-9 X= 3 11. Solve 4 x = 36. .Solution:

X= 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 forSolution:

.

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, 12. Find the equation of the straight line shown.

-2

-1

10 8 6 4 2 0 -2 0 -4 -6 -8 -10

1

2

3

4

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. So the gradient is

gradient

3 3 1

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

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3. Draw the lines y = 3x 2 and y = 6 2x.Solution:

Table of values for y = 3x 2x y 0 30 - 2 = -2 1 31 - 2 =1 2 32- 2 =4 3 33 - 2 =7

Work out the y value corresponding to each x value using the formula y = 3x - 2.

Then, plot the points (0, -2), (1, 1), (2, 4) and (3, 7) and join them up. Table of values for y = 6 2xx y 0 6 - 20 =6 1 6 - 21 =4 2 6 - 22 =2 3 6 - 23 =0

We plot the points (0, 6), (1, 4), (2, 2) and (3, 0).8 6 4 2 0 -2 -4 -6 -8

8 6

4

2

-

2

4

6

8

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4. Write down the straight line equation from the information given. M=3 C=4Answer:

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:

y= 7. Write down the y-intercept of the following equation:Answer:

(

)

Y=1 8. What is the y-intercept of this graph?

Answer:

=3

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

-3 0

-2 1

-1 2

0 3

1 4

2 5

3 6

10. Write down coordinate P.

Answer:

= 2, 1

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

y = mx + c 5=41+c 5=4+c 1=c m = (y2 y1) (x2 - x1) m = (21 - 5) (5 1) m = 16 4 m=4 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 - 6m = (y2 y1) (x2 - x1) m = (-4 - -12) (1 - -3) m = 8 4 m=2

13. For the straight line y = -2x + 3, what are: a) the slope b) the y-intercept?Answer:= Slope

= 2, y-intercept =