Significant figures, any of the digits of a number beginning with the digit farthest to the left that is not zero and ending with the last digit farthest to the right that is either not zero or that is a zero but is considered to be exact.

The significant figures are 9, 4, 5 and 3 and the third significant figure is 5. Since the next figure to the third is less than 5, 1 is not added to the third significant figure. The fourth significant figure becomes 0 hence the significant figure is 9450 x 10^-6

9450 x 10^-6 = 0.⌒0⌒0⌒9⌒4⌒5⌒0.0

Change the fraction and decimal to percentage so that they will all be in percentages.

Percentage is simply multiplying by 100%.

0.45 ⇒ 0.45 x 100% = 45%

$\frac{3}{4}$ ⇒ $\frac{3}{4}$ x 100% = 75%

25% is already in percentage

Now arrange in ascending order

Note: Ascending means from lowest to highest.

25%, 45%, 75%

45% = 0.45

75% = $\frac{3}{4}$

Hence the ascending order is 25%,0.45,$\frac{3}{4}$

Applying log_a^bc = c x log_a^b = clog_a^b

log₁₀^2y = y x log₁₀²

But log₁₀^2y = 1.8060

y x log₁₀² = 1.8060

But log₁₀² = 0.3010

Hence y x 0.3010 = 1.8060

Divide both sides by 0.3010

y = $\frac{1.8060}{0.3010}$ = 6

Express $\sqrt{\mathrm{72}}$ in the form $\sqrt{\mathrm{Perfect\; square\; x\; a\; number}}$

$\sqrt{\mathrm{72}}$ = $\sqrt{\mathrm{36\; x\; 2}}$ = $\sqrt{\mathrm{36}}$ x $\sqrt{2}$ = 6$\sqrt{2}$

Note:

$\sqrt{\mathrm{36}}$ = 6

Express $\sqrt{\mathrm{48}}$ in the form $\sqrt{\mathrm{Perfect\; square\; x\; a\; number}}$

$\sqrt{\mathrm{48}}$ = $\sqrt{\mathrm{16\; x\; 3}}$ = $\sqrt{\mathrm{16}}$ x $\sqrt{3}$ = 4$\sqrt{3}$

Note:

$\sqrt{\mathrm{16}}$ = 4

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{6\sqrt{2}}{4\sqrt{3}-\sqrt{3}}$

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{6\sqrt{2}}{3\sqrt{3}}$

Note: 3 divides itself 1 time and 6, 2 times.

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{2\sqrt{2}}{\sqrt{3}}$

Multiply both the numerator and the denominator by $\sqrt{3}$ to get rid of the denominator root

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{2\sqrt{2}x\sqrt{3}}{\sqrt{3}x\sqrt{3}}$

2$\sqrt{2}$ x $\sqrt{3}$ = 2$\sqrt{\mathrm{2\; x\; 3}}$ = 2$\sqrt{6}$

$\sqrt{3}$ x $\sqrt{3}$ = 3

$\frac{\sqrt{\mathrm{72}}}{\sqrt{\mathrm{48}}-\sqrt{3}}$ = $\frac{2\sqrt{6}}{3}$

Prime number: A number that is divisible only by itself and 1

Note: 1 is not a prime number

Integer: A whole number (not a fraction) that can be positive, negative or zero. e.g ...,-5,-4,-3,-2,-1,0,1,2,3...

P is a set of positive (No negative) integers and less than 7

Rational Number: A number such as 5/8 that can be expressed as quotient or fraction $\frac{\mathrm{p}}{\mathrm{q}}$ of two integers.

Real Number: Any positive or negative number. This includes all integers and all rational and irrational numbers.

x:x is read as x is such that x is ...

y varies inversely as x ⇒ y ∝ $\frac{1}{\mathrm{x}}$

Remove the proportionality sign (∝) by introducing a constant (k)

y = k x $\frac{1}{\mathrm{x}}$ = $\frac{\mathrm{k}}{\mathrm{x}}$

When x = 3, y = 6.

6 = $\frac{\mathrm{k}}{3}$

Solve for k.

k = 6 x 3 = 18

y = $\frac{18}{\mathrm{x}}$

When x = 9

y = $\frac{18}{9}$ = 2

Geometric sequence is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed number called the common ratio

Common ratio = $\frac{{\mathrm{a}}_{\mathrm{n}}}{{\mathrm{a}}_{\mathrm{n\; -\; 1}}}$

Common ratio = $\frac{\mathrm{Fourth\; term}}{\mathrm{Third\; term}}$ = $\frac{\mathrm{Third\; term}}{\mathrm{Second\; term}}$ = $\frac{\mathrm{Second\; term}}{\mathrm{First\; term}}$

First term = x

Second term = 4

Third term = 16

Fourth term = y

$\frac{\mathrm{y}}{16}$ = $\frac{16}{4}$ = 4

$\frac{\mathrm{y}}{16}$ = 4

Multiply both sides by 16

y = 4 x 16 = 64

$\frac{4}{\mathrm{x}}$ = $\frac{16}{4}$ = 4

$\frac{4}{\mathrm{x}}$ = 4

Multiply both sides by x

x x 4 = 4

Divide both sides by 4

x = $\frac{4}{4}$ = 1

x : y = 1 : 64

E = $\frac{\mathrm{k}{\mathrm{x}}^2}{\mathrm{2y}}$ + z

Multiply both sides the denominator 2y

E x 2y = $\frac{\mathrm{k}{\mathrm{x}}^2}{\mathrm{2y}}$ x 2y + z x 2y

2Ey = kx² + 2yz

kx² = 2Ey - 2yz

Divide both sides by k.

x² = $\frac{\mathrm{2Ey\; -\; 2yz}}{\mathrm{k}}$

Factorize 2y out.

x² = $\frac{\mathrm{2y(E\; -\; z)}}{\mathrm{k}}$

Take square root of both sides.

x = $\sqrt{\frac{\mathrm{2y(E\; -\; z)}}{\mathrm{k}}}$

Change the mixed fractions to improper fractions.

1$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 1\; +\; 1}}{2}$ = $\frac{\mathrm{2\; +\; 1}}{2}$ = $\frac{3}{2}$

2$\frac{2}{3}$ = $\frac{\mathrm{3\; x\; 2\; +\; 2}}{3}$ = $\frac{\mathrm{6\; +\; 2}}{3}$ = $\frac{8}{3}$

$1\frac{1}{2}2\frac{2}{3}$ = $\frac{\mathrm{x}}{96}$

$\frac{3}{2}$ ÷ $\frac{8}{3}$ = $\frac{\mathrm{x}}{96}$

Reciprocate the fraction at the right of the ÷ and change ÷ to multiplication (x)

Note: reciprocate means the numerator becomes the denominator and the denominator becomes the numerator.

$\frac{3}{2}$ x $\frac{3}{8}$ = $\frac{\mathrm{x}}{96}$

$\frac{\mathrm{3\; x\; 3}}{\mathrm{2\; x\; 8}}$ = $\frac{\mathrm{x}}{96}$

$\frac{9}{16}$ = $\frac{\mathrm{x}}{96}$

Cross multiply.

16 x x = 9 x 96

Divide both sides by 16

x = $\frac{\mathrm{9\; x\; 96}}{16}$ = 54

If a and b are the roots, (x - a)(x - b) = 0

Change the mixed fraction to improper fraction.

1$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 1\; +\; 1}}{2}$ = $\frac{\mathrm{2\; +\; 1}}{2}$ = $\frac{3}{2}$

For the roots -$\frac{1}{2}$ and $\frac{3}{2}$,

(x --$\frac{1}{2}$)(x - $\frac{3}{2}$) = 0

-- = +

(x + $\frac{1}{2}$)(x - $\frac{3}{2}$) = 0

x(x - $\frac{3}{2}$) + $\frac{1}{2}$(x - $\frac{3}{2}$) = 0

x² - $\frac{\mathrm{3x}}{2}$ + $\frac{\mathrm{x}}{2}$ - $\frac{\mathrm{1\; x\; 3}}{\mathrm{2\; x\; 2}}$ = 0

x² + $\frac{\mathrm{x\; -\; 3x}}{2}$ - $\frac{3}{4}$ = 0

x² - $\frac{\mathrm{2x}}{2}$ - $\frac{3}{4}$ = 0

Get rid of the fraction by multiplying both sides by the L.C.M of the denominators 2 and 4.

The L.C.M is 4.

4 x x² - $\frac{\mathrm{2x}}{2}$ x 4 - 4 x $\frac{3}{4}$ = 0 x 4

4x² - 4x - 3 = 0

Pick a point on the line of the relation: y = d - x and substitute into the relation to find d.

y = d - x

Using the point (1, 0)

x = 1 and y = 0

0 = d - 1

1 = d

d = 1

OR

Using the point (0, 1)

x = 0, y = 1

1 = d - 0

1 = d

d = 1

OR

Using the point (-2, 3)

x = -2 and y = 3

3 = d - (-2)

Note: -- or -(-) = +

3 = d + 2

3 - 2 = d

1 = d

d = 1

Since 3 can go into 6 and 9, split -6pqr into -3pqr and -3pqr

p²q² - 6pqr + 9r² = p²q² - 3pqr - 3pqr + 9r²

p²q² - 6pqr + 9r² = pq(pq - 3r) - 3r(pq - 3r)

p²q² - 6pqr + 9r² = (pq - 3r)(pq - 3r)

Since pq - 3r = pq - 3r, the multiplication makes it square

p²q² - 6pqr + 9r² = (pq - 3r)²

The equation above is similar to the general factorization rule of (a - b)²

(a - b)² = a² - 2ab + b²

Thus the first term square then the sign (either + or -) and 2 times the first and second terms + the second term square

Similarly (a + b)² = a² + 2ab + b²

The first term for the equation, p²q² - 6pqr + 9r² is pq and the second term is 3r and the sign is -

Let h = height of the water

Volume of a cuboid = Length x Breadth x Height

Since the length and breadth are in cm, convert the litres to cm³

1 litre = 1000 cm³

180 litres = 180 x 1000 cm³

Length = 250 cm, breadth = 120 cm

Volume = 180 litres = 180 x 1000 cm³

250 cm x 120 cm x h cm = 180 x 1000 cm³

Divide both sides by 250 cm x 120 cm

h = $\frac{\mathrm{180\; x\; 1000}}{\mathrm{250\; x\; 120}}$ cm = 6 cm

Let h = distance between the parallel sides

a = 6 cm and b = 7 cm

$\frac{1}{2}$ x (6 cm + 7 cm) x h = 49 cm²

$\frac{1}{2}$ x 13 cm x h = 49 cm²

Get rid of the fraction by multiplying both sides by the denominator 2.

2 x $\frac{1}{2}$ x 13 cm x h = 2 x 49 cm²

13 cm x h = 98 cm²

Divide both sides by 13 cm.

h = $\frac{98}{13}$ cm = 7.53846 cm ≈ 7.5 cm (one decimal place)

The angle (θ) substended are the same.

Area of a sector = $\frac{\mathrm{\theta }}{360}$ x π x radius x radius

Area = 231 cm²

$\frac{\mathrm{\theta }}{360}$ x π x radius x radius = 231

Make the angle (θ) the subject.

Get rid of the fraction by multiplying both sides by the denominator 360.

360 x $\frac{\mathrm{\theta }}{360}$ x π x radius x radius = 360 x 231

θ x π x radius x radius = 360 x 231

Divide both sides by π x radius x radius

θ = $\frac{\mathrm{360\; x\; 231}}{\mathrm{\pi \; x\; radius\; x\; radius}}$ --- eqn 1

Length of an arc = $\frac{\mathrm{\theta }}{360}$ x 2 x π x radius

Length = 66 cm

$\frac{\mathrm{\theta }}{360}$ x 2 x π x radius = 66

Make the angle (θ) the subject.

Get rid of the fraction by multiplying both sides by the denominator, 360.

360 x $\frac{\mathrm{\theta }}{360}$ x 2 x π x radius = 360 x 66

θ x 2 x π x radius = 360 x 66

Divide both sides by 2 x π x radius

θ = $\frac{\mathrm{360\; x\; 66}}{\mathrm{2\; x\; \pi \; x\; radius}}$ --- eqn 2

--- eqn 1 = --- eqn 2

$\frac{\mathrm{360\; x\; 231}}{\mathrm{\pi \; x\; radius\; x\; radius}}$ = $\frac{\mathrm{360\; x\; 66}}{\mathrm{2\; x\; \pi \; x\; radius}}$

Cross multiply.

360 x 231 x 2 x π x radius = 360 x 66 x π x radius x radius

Divide both sides by 360 x π x radius to cancel out.

$\frac{\mathrm{360\; x\; 231\; x\; 2\; x\; \pi \; x\; radius}}{\mathrm{360\; x\; \pi \; x\; radius}}$ = $\frac{\mathrm{360\; x\; 66\; x\; \pi \; x\; radius\; x\; radius}}{\mathrm{360\; x\; \pi \; x\; radius}}$

231 x 2 = 66 x radius

Divide both sides by 66

Radius = $\frac{\mathrm{231\; x\; 2}}{66}$ = 7 cm

Let h = height at which the ladder touches the building.

The angle the ladder makes with the ground is the same for both ladders.

The length of the ladder is the hypotenuse.

sin(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Hypothenuse}}$

10 cm long ladder

Opposite side to θ = 8 cm

Hypothenuse = 10 cm

sin(θ) = $\frac{\mathrm{8\; cm}}{\mathrm{10\; cm}}$ = $\frac{4}{5}$

12 cm long ladder

Opposite side to θ = h cm

Hypothenuse = 12 cm

sin(θ) = $\frac{\mathrm{h\; cm}}{\mathrm{12\; cm}}$ = $\frac{\mathrm{h}}{12}$

Since the angles are the same, the sin for both cases would be the same.

$\frac{\mathrm{h}}{12}$ = $\frac{4}{5}$

Cross multiply.

h x 5 = 12 x 4

Divide both sides by 5

h = $\frac{\mathrm{12\; x\; 4}}{5}$ = 9.6 cm

∆ MNT is an isosceles triangle (Two sides equal).

Isosceles triangle has the base angles the equal.

∠ MNT = ∠ MTN = 32°

∠ LMN = ∠ MNT + ∠ MTN (Sum of two opposite interior angles equal to exterior angle)

∠ LMN = 32° + 32° = 64°

∠ MLN = $\frac{1}{2}$ x ∠MON

|OM| = |ON| (Radii)

∆ MON is an isosceles triangle hence the base angles are the same.

∠ OMN = ∠ ONM = 20°

∠MON = 180° - (20° + 20°) (Sum of interior angles of a triangle is 180°)

∠MON = 180° - 40°

∠MON = 140°

∠ MLN = $\frac{1}{2}$ x ∠MON = $\frac{1}{2}$ x 140° = 70°

∠ MLN + ∠ LMN + ∠ LNM = 180°(Sum of interior angles of a triangle is 180°)

64° + 70° + ∠ LNM = 180°

134° + ∠ LNM = 180°

∠ LNM = 180° - 134°

∠ LNM = 46°

Draw an imaginary third parallel line.

∠ SRH = ∠ UHR = 42° (Alternate angles)

∠ HQP = ∠ QHU = y (Alternate angles)

∠ RHQ = ∠ UHR + ∠ QHU = 65°

42° + y = 65°

y = 65° - 42°

y = 23°

∠ QRS = ∠ PQT = 40° (Corresponding angles)

∆ PQT is an isosceles triangle (Two sides equal).

Isosceles triangle has base angles equal.

∠ RPT = ∠ PQT = 40° (Base angles equal)

|OP| = $\sqrt{{(4-0)}^2+{(-1-0)}^2}$

|OP| = $\sqrt{\mathrm{17}}$

|OQ| = $\sqrt{{(1-0)}^2+{(-4-0)}^2}$

|OQ| = $\sqrt{\mathrm{17}}$

|PQ| = $\sqrt{{(4-1)}^2+{(-1--4)}^2}$

Note: -- = +. --4 = + 4

|OP| = $\sqrt{\mathrm{18}}$

|OP| = |OQ| hence ∆OPQ has two sides equal and therefore is an isosceles triangle.

y = x² + 1

When x = 2$\sqrt{2}$

y = ${\left(2\sqrt{2}\right)}^2$ + 1

Note: when you square a square root, the square root disappears.

y = 2² x 2 + 1

y = 4 x 2 + 1

y = 8 + 1

y = 9

Let h = height of tree.

At the same time of the day, the angle of elevation from any point on the ground to the sun is considered equal.

tan(θ) = $\frac{\mathrm{Opposite}}{\mathrm{Adjacent}}$

8 m tall tree

tan(θ) = $\frac{\mathrm{8\; m}}{\mathrm{10\; m}}$ = $\frac{4}{5}$

h m tall tree

tan(θ) = $\frac{\mathrm{h\; m}}{\mathrm{40\; m}}$ = $\frac{\mathrm{h}}{40}$

Since the angle, θ is the same for both trees, the tan(θ) is the same for both.

$\frac{\mathrm{h}}{40}$ = $\frac{4}{5}$

Cross multiply.

h x 5 = 40 x 4

Divide both sides by 5.

h = $\frac{\mathrm{40\; x\; 4}}{5}$ = 32

cos (3x + 28°) = sin (2x + 48°)

Use the trigonometric values to change either cos to sin or sin to cos and equate the angles to solve for x.

Method I

Changing the cos to sin.

cos (3x + 28°) = sin (90° - [3x + 28°])

cos (3x + 28°) = sin (2x + 48°)

sin (90° - [3x + 28°]) = sin (2x + 48°)

Since both sides are sin, the angles are the same.

90° - (3x + 28°) = 2x + 48°

90° - 3x - 28° = 2x + 48°

90° - 28° - 48° = 2x + 3x

90° - 76° = 5x

14° = 5x

Divide both sides by 5.

x = $\frac{\mathrm{14°}}{5}$ = 2.8°

Method II

Changing the sin to cos.

sin (2x + 48°) = cos(90° - [2x + 48°])

cos (3x + 28°) = sin (2x + 48°)

cos (3x + 28°) = cos(90° - [2x + 48°])

Since both sides are cos, the angles are the same.

3x + 28° = 90° - (2x + 48°)

3x + 28° = 90° - 2x - 48°

3x + 2x = 90° - 48° - 28°

5x = 14°

Divide both sides by 5.

x = $\frac{\mathrm{14°}}{5}$ = 2.8°

Notes:

1. Bearing starts from the north

2. 300° = 90° + 90° + 90° + 30°

Two of the sides are equal hence it forms an isosceles triangle.

The base angles are equal.

∠ ACB = ∠ CAB = θ

∠ ABC = 60° + 90° = 150°

∠ ACB + ∠ CAB + ∠ ABC = 180° (Sum of interior angles of a triangle)

θ + θ + 150° = 180°

2θ = 180° - 150°

2θ = 30°

Divide both sides by 2

θ = $\frac{\mathrm{30°}}{2}$ = 15°

Bearing of his initial position from his final position = 180° - (θ + 60°)

θ = 15°

Bearing of his initial position from his final position = 180° - (15° + 60°)

Bearing of his initial position from his final position = 180° - 75°

Bearing of his initial position from his final position = 105°

Mean (x̄) = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{38}{16}$ = 2.375 ≈ 2.4 (one decimal place)

The number of students who took the test is the sum of all the frequencies.

Calculate the value of each grid to determine the frequencies.

Grid value = $\frac{\mathrm{Interval}}{\mathrm{Number\; of\; grids\; per\; interval}}$

Grid value = $\frac{5}{20}$ = 0.25

Value of frequency = Number of grids x Grid value

Frequency for mark 1 = 12 x 0.25 = 3

Frequency for mark 2 = 8 x 0.25 = 2

Frequency for mark 3 = 20 x 0.25 = 5

Frequency for mark 4 = 32 x 0.25 = 8

Frequency for mark 5 = 56 x 0.25 = 14

Frequency for mark 6 = 64 x 0.25 = 16

Frequency for mark 7 = 40 x 0.25 = 10

Frequency for mark 8 = 28 x 0.25 = 7

Frequency for mark 9 = 24 x 0.25 = 6

Frequency for mark 10 = 4 x 0.25 = 1

Number of students = 3 + 2 + 5 + 8 + 14 + 16 + 10 + 7 + 6 + 1

Number of students = 72

The number of students who took the test is the sum of all the frequencies.

Calculate the value of each grid to determine the frequencies.

Grid value = $\frac{\mathrm{Interval}}{\mathrm{Number\; of\; grids\; per\; interval}}$

Grid value = $\frac{5}{20}$ = 0.25

Value of frequency = Number of grids x Grid value

Frequency for mark 1 = 12 x 0.25 = 3

Frequency for mark 2 = 8 x 0.25 = 2

Frequency for mark 3 = 20 x 0.25 = 5

Frequency for mark 4 = 32 x 0.25 = 8

Frequency for mark 5 = 56 x 0.25 = 14

Frequency for mark 6 = 64 x 0.25 = 16

Frequency for mark 7 = 40 x 0.25 = 10

Frequency for mark 8 = 28 x 0.25 = 7

Frequency for mark 9 = 24 x 0.25 = 6

Frequency for mark 10 = 4 x 0.25 = 1

Number of students = 3 + 2 + 5 + 8 + 14 + 16 + 10 + 7 + 6 + 1

Number of students = 72

Percentage = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 100%

Number of students who scored below 5 marks = 3 + 2 + 5 + 8 = 18

Percentage who failed = $\frac{18}{72}$ x 100% = 25%

The number of students who took the test is the sum of all the frequencies.

Calculate the value of each grid to determine the frequencies.

Grid value = $\frac{\mathrm{Interval}}{\mathrm{Number\; of\; grids\; per\; interval}}$

Grid value = $\frac{5}{20}$ = 0.25

Value of frequency = Number of grids x Grid value

Frequency for mark 1 = 12 x 0.25 = 3

Frequency for mark 2 = 8 x 0.25 = 2

Frequency for mark 3 = 20 x 0.25 = 5

Frequency for mark 4 = 32 x 0.25 = 8

Frequency for mark 5 = 56 x 0.25 = 14

Frequency for mark 6 = 64 x 0.25 = 16

Frequency for mark 7 = 40 x 0.25 = 10

Frequency for mark 8 = 28 x 0.25 = 7

Frequency for mark 9 = 24 x 0.25 = 6

Frequency for mark 10 = 4 x 0.25 = 1

Number of students = 3 + 2 + 5 + 8 + 14 + 16 + 10 + 7 + 6 + 1

Number of students = 72

The median is the middle number of an ordered distribution. If there are two numbers at the middle, add the two numbers and divide the sum by 2.

Median position = $\frac{\mathrm{\Sigma f\; +\; 1}}{2}$ = $\frac{\mathrm{72\; +\; 1}}{2}$ = $\frac{73}{2}$ = 36.5 position

Add the frequencies until you reach 36. The median is the average of the 36^th and 37^th marks.

3 + 2 + 5 + 8 + 14 = 32

The 36^th and 37^th falls on the 6 marks.

Median mark = $\frac{\mathrm{6\; +\; 6}}{2}$ = $\frac{12}{2}$ = 6 marks

Number of students who obtained 6 marks = 16

Total number of students = 72

P(Median mark) = $\frac{16}{72}$ = $\frac{2}{9}$

Note: 8 divides 16, 2 times and 72, 9 times.

Let x = number of students who liked both Music and Movies.

Coverig Music,

12 + 15 - x = 20

27 - 20 = x

7 = x

x = 7

Number of students who liked both Music and Movies = 7

Number of students = 20

P(Both Music and Movies) = $\frac{7}{20}$

Mean = $\frac{\mathrm{Sum\; of\; items}}{\mathrm{Number\; of\; items}}$

Mean = $\frac{\mathrm{2\; +\; 5\; +\; 2x\; +\; 7}}{4}$

The mean is not greater than 5 means the mean can be less than 5 or equal to 5.

$\frac{\mathrm{2\; +\; 5\; +\; 2x\; +\; 7}}{4}$ ≤ 5

$\frac{\mathrm{14\; +\; 2x}}{4}$ ≤ 5

Get rid of the fraction by multiplying both sides by the denominator, 4

4 x $\frac{\mathrm{14\; +\; 2x}}{4}$ ≤ 5 x 4

14 + 2x ≤ 20

2x ≤ 20 - 14

2x ≤ 6

Divide both sides by 2.

x ≤ $\frac{6}{2}$

x ≤ 3

3^-x = k

a^-b = $\frac{1}{{\mathrm{a}}^{\mathrm{b}}}$

3^-x = $\frac{1}{3^{\mathrm{x}}}$

$\frac{1}{3^{\mathrm{x}}}$ = k

Make 3^x the subject.

Get rid of the fraction by multiplying both sides by the denominator, 3^x

3^x x $\frac{1}{3^{\mathrm{x}}}$ = k x 3^x

1 = k x 3^x

Divide both sides by k.

$\frac{1}{\mathrm{k}}$ = 3^x

3^x = $\frac{1}{\mathrm{k}}$

The regular time to reach destination = $\frac{\mathrm{40\; km}}{\mathrm{50\; km/h}}$ = 0.8 hours

18 minutes = $\frac{18}{60}$ hours = 0.3 hours

Time left to get to destination = 0.8 hours - 0.3 hours = 0.5 hours.

The ferry must cover the 40 km in 0.5 hours.

Speed = $\frac{\mathrm{Distance}}{\mathrm{Time}}$ = $\frac{\mathrm{40\; km}}{\mathrm{0.5\; h}}$ = 80 km/h

∧ is logical conjunction (and)

→ is then or implies

Solve the inequality.

$\frac{1}{3}$(2x - 9) ≥ 2(2x + 1)

Get rid of th fraction by multiplying both sides of the equation by the denominator, 3

3 x $\frac{1}{3}$(2x - 9) ≥ 3 x 2(2x + 1)

1 x (2x - 9) ≥ 6(2x + 1)

2x - 9 ≥ 12x + 6

2x - 12x ≥ 6 + 9

-10x ≥ 15

Divide both sides by -10. When dividing by a negative, the inequality sign changes.

x ≤ $\frac{15}{-10}$

x ≤ -1.5

Notes:

1. -1.5 is between -1 and - 2

2. For ≥ and ≤, the solid is shaded

3. For ≤, the arrow is to the left just as the direction of the sign and for ≥, the arrow is to the right

10101 + 1001 = 11110

***** = 111001 - 11110

***** = 11011

Note: In base 2, when you borrow, the value is two (2).

When the daughter was x years old, the woman is y + x years old. Thus if you add the daughters age to the mother's age when she gave birth to the daughtr, you will get the mother's current age.

When the daughter was x years old, the mother was n times as old as her daughter.

Times means multiplication (x)

y + x = n x x

y + x = nx

y = nx - x

Factorize x out.

y = x(n - 1)

∠ QRT = ∠ QST = 42° (Angles subtended by arc QT)

∠ QST + ∠ RSQ = 90°

42° + ∠ RSQ = 90°

∠ RSQ = 90° - 42°

∠ RSQ = 48°

∠ STR = ∠ SQR = 180° - 124°(Angles subtended by arc SR)

∠ STR = 180° - 124° (Sum of angles on straight line is 180°)

∠ STR = 56°

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{9\; x\; (2x\; +\; 1)\; -\; 2\; x\; (3x\; -\; 7)\; -\; 1\; x\; 5}}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{18x\; +\; 9\; -\; 6x\; +\; 14\; -\; 5}}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{18x\; -\; 6x\; +\; 9\; +\; 14\; -\; 5}}{18}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{12x\; +\; 18}}{18}$

Factorize 6 out at numerator.

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{6(2x\; +\; 3)}}{18}$

Note: 6 divides itself 1 time and 18, 3 times.

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{1\; x\; (2x\; +\; 3)}}{3}$

$\frac{\mathrm{2x\; +\; 1}}{2}$ - $\frac{\mathrm{3x\; -\; 7}}{9}$ - $\frac{5}{18}$ = $\frac{\mathrm{2x\; +\; 3}}{3}$

When you substitute x = root of the quadratic equation into the equation, the result should be 0.

Since the root is 2, when you substitute x = 2 into 2x² + (k + 2)x + k, the result should be 0.

2x² + (k + 2)x + k

When x = 2,

2(2)² + (k + 2)(2) + k = 0

2 x 4 + 2k + 4 + k = 0

8 + 2k + 4 + k = 0

8 + 4 + 2k + k = 0

12 + 3k = 0

12 = 0 -3k

12 = -3k

Divide both sides by -3.

k = $\frac{12}{-3}$ = -4

Isosceles triangles (two sides equal) has base angles equal.

For ∆XYZ, ∠YXZ = ∠YZX = a

a + a + 52° = 180°(Sum of interior angles of a triangle)

2a + 52° = 180°

2a = 180° - 52°

2a = 128°

Divide both sides by 2

a = $\frac{\mathrm{128°}}{2}$ = 64°

For ∆XZT, ∠ZXT = ∠ZTX = b

∠YZX + ∠XZT = 180°(Sum of angles on a straight line)

∠YZX = a = 64°

64° + ∠XZT = 180°

∠XZT = 180° - 64°

∠XZT = 116°

∠XZT + ∠ZXT + ∠ZTX = 180°(Sum of interior angles of a triangle)

116° + b + b = 180°

2b = 180° - 116°

2b = 64°

Divide both sides by 2.

b = $\frac{\mathrm{64°}}{2}$ = 32°

∠ZTX = 32°

Let y = area of the field

First day area cleared = 40% of y

First day area cleared = $\frac{40}{100}$ x y = 0.4y

Remaining uncleared area = y - 0.4y

Remaining uncleared area = 0.6y

Second day area cleared = 60% of remaining

Second day area cleared = $\frac{60}{100}$ x 0.6y

Second day area cleared = 0.36y

Remaining area after the second day = 0.6y - 0.36y

Remaining area after the second day = 0.24y

Percentage = $\frac{\mathrm{Number}}{\mathrm{Total}}$ x 100%

Percentage of the land remaining at the end of the second day = $\frac{\mathrm{0.24y}}{\mathrm{y}}$ x 100%

Note: the ys cancel each other.

Percentage of the land remaining at the end of the second day = $\frac{\mathrm{0.24\; x\; 1}}{1}$ x 100% = 24%

Length of an arc = $\frac{\mathrm{\theta }}{360}$ x 2 x π x radius

Length of the arc = 11 cm

θ = 60

π = $\frac{22}{7}$

Substitute the values into the equation and solve for the radius.

$\frac{60}{360}$ x 2 x $\frac{22}{7}$ x radius = 11

Get rid of the fraction by multiplying both sides by the L.C.M of the denominators, 360 and 7 or (360 x 7).

360 x 7 x $\frac{60}{360}$ x 2 x $\frac{22}{7}$ x radius = 11 x 360 x 7

60 x 2 x 22 x radius = 11 x 360 x 7

Divide both sides by 60 x 2 x 22

radius = $\frac{\mathrm{11\; x\; 360\; x\; 7}}{\mathrm{60\; x\; 2\; x\; 22}}$

radius = 10.5 cm = 10$\frac{1}{2}$ cm

Pentagon has five sides hence n = 5.

Sum of the interior angles = (5 - 2) x 180°

Sum of the interior angles = 3 x 180°

Sum of the interior angles = 540°

y° + 2x° + 3x° + 2x° + y° = 540°

2y + 7x = 540 ---- eqn 1

But y = $\frac{\mathrm{3x}}{2}$

Make x the subject and substitute it into --- eqn 1

y = $\frac{\mathrm{3x}}{2}$

Get rid of the fraction by multiplying both sides by the denominator, 2

2 x y = 2 x $\frac{\mathrm{3x}}{2}$

2y = 3x

Divide both sides by 3.

x = $\frac{\mathrm{2y}}{3}$ --- eqn 2

Substitute --- eqn 2 into --- eqn 1

2y + 7 x $\frac{\mathrm{2y}}{3}$ = 540

Get rid of the fraction by multiplying both sides by the denominator, 3

3 x 2y + 3 x 7 x $\frac{\mathrm{2y}}{3}$ = 540 x 3

6y + 14y = 1,620

20y = 1,620

Divide both sides by 20

y = $\frac{1620}{20}$ = 81

Base is square. Area of a square is length x length

Let h = the height of the pyramid.

Length of the base = height of the pyramid.

Area of the base = Length x Length = h x h

Volume = $\frac{1}{3}$ x h x h x h

But volume = 114$\frac{1}{3}$ = $\frac{\mathrm{3\; x\; 114\; +\; 1}}{3}$ = $\frac{343}{3}$

$\frac{343}{3}$ = $\frac{1}{3}$ x h x h x h

Get rid of the fractions by multiplying both sides by the denominator, 3

3 x $\frac{343}{3}$ = 3 x $\frac{1}{3}$ x h x h x h

343 = 1 x h x h x h

h³ = 343

Take the cube root of both sides

h = $\sqrt[3]{\mathrm{343}}$ or ${343}^{\frac{1}{3}}$ = 7 cm

2x + 1 = 4(mod 7)

The least value which produces 4 in mod 7 is 1 x 7 + 4 = 11

2x + 1 = 11

2x = 11 - 1

2x = 10

Divide both sides by 2.

x = $\frac{10}{2}$ = 5

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$

Notes:

1. a² - b² = (a + b)(a - b)

2. 1 - x² = 1² - x² = (1 + x)(1 - x)

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{2x}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$ + $\frac{1}{\mathrm{1\; +\; x}}$

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{1\; x\; 2x\; +\; 1\; x\; (1\; -\; x)}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{2x\; +\; 1\; -\; x}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{2x\; -\; x\; +\; 1}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{x\; +\; 1}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

Note: x + 1 = 1 + x (the order of addition doesn't matter)

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{\mathrm{1\; +\; x}}{\mathrm{(1\; +\; x)(1\; -\; x)}}$

(1 + x) cancels each other 1 time.

$\frac{\mathrm{2x}}{1-{\mathrm{x}}^2}$ + $\frac{1}{\mathrm{1\; +\; x}}$ = $\frac{1}{\mathrm{1\; -\; x}}$

1. Find the number of times the cyclist complete one cycle by dividing the distance overed by the circumference.

Number of completed cycles = $\frac{\mathrm{302\; km}}{\mathrm{9\; km}}$ = 33.555

Number of completed cycles = 33

Note: completed cycle is when the cyclist reaches the starting point.

2. Distance from starting point = Distance covered - Number of completed cycles x the circumference

Distance from starting point = 302 km - 33 x 9 km

Distance from starting point = 302 km - 297 km

Distance from starting point = 5 km