(2a + b) (a + 2b) = (2a + b) (a + 2b)

(2a + b) (a + 2b) = 2a(a + 2b) + b(a + 2b)

(2a + b) (a + 2b) = 2a² + 4ab + ab + 2b²

(2a + b) (a + 2b) = 2a² + 5ab + 2b²

Let x = the missing binary number.

1100110 - x = 111011

1100110 - 111011 = x

Note: in binary, the value of a number borrowed is 2.

x – 3 ≥ 10

x ≥ 10 + 3

x ≥ 13

This is read as x is greater than or equal to 13. Or equal to means the number is also included.

List from the set {1, 3, 5, 7, 9, 11, 13, 15} the numbers greater than or equal to 13.

The truth set is {13, 15}

Note:
1. equilateral triangle has 3 lines of symmetry
2. circle has an infinite number of lines of symmetry
3. isosceles triangle has 1 line of symmetry

of means multiplication.

2$\frac{1}{2}$% of ₵2,000.00 = 2$\frac{1}{2}$% x ₵2,000.00

Change the mixed fraction to improper fraction.

2$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 2\; +\; 1}}{2}$ = $\frac{\mathrm{4\; +\; 1}}{2}$ = $\frac{5}{2}$

2$\frac{1}{2}$% = $\frac{5}{2}$%

% means over 100

2$\frac{1}{2}$% = $\frac{5}{2}$ ÷ 100

Every whole number is being divided by 1

100 = $\frac{100}{1}$

2$\frac{1}{2}$% = $\frac{5}{2}$ ÷ $\frac{100}{1}$

Reciprocate the fraction (numerator becomes denominator and denominator becomes numerator) at the right and change the division (÷) to multiplication (x) sign.

2$\frac{1}{2}$% = $\frac{5}{2}$ x $\frac{1}{100}$

2$\frac{1}{2}$% = $\frac{\mathrm{5\; x\; 1}}{\mathrm{2\; x\; 100}}$ = $\frac{5}{200}$

2$\frac{1}{2}$% of ₵2,000.00 = $\frac{5}{200}$ x ₵2000 = 5 x ₵10 = ₵50

Note:
1. the zeros (0) at the end of 2000 and 200 cancel each other
2. 2 divides itself 1 time and 20, 10 times

Prime factors of 18

$$\begin{array}{cc}\mathrm{}& 18\\ 2& 9\\ 3& 3\\ 3& 1\end{array}$$

Prime factors of 18 = 2 x 3 x 3

Prime factors of 18 = 2 x 3²

Prime factors of 36

$$\begin{array}{cc}\mathrm{}& 36\\ 2& 18\\ 2& 9\\ 3& 3\\ 3& 1\end{array}$$

Prime factors of 36 = 2 x 2 x 3 x 3

Prime factors of 36 = 2² x 3²

Prime factors of 120

$$\begin{array}{cc}\mathrm{}& 120\\ 2& 60\\ 2& 30\\ 2& 15\\ 3& 5\\ 5& 1\end{array}$$

Prime factors of 120 = 2 x 2 x 2 x 3 x 3

Prime factors of 120 = 2³ x 3 x 5

The Highest Common factor is the product (multiplication) of the common prime factors in their lowest powers.

Highest common factor of 18, 36 and 120 is 2 x 3 = 6

$\frac{3}{4}$, $\frac{2}{3}$, $\frac{3}{5}$

Arranging/Ordering Fractions

Before you can compare fractions, their denominators must be the same. Once they are the same, you can arrnage the fractions in the magnitude of their numerators.

To make the denominator the same, express the denominator of each of the fractions to be the L.C.M of the denominators by multiplying the numerator and the denominator by the number of times the denominator can divide the L.C.M for each of the fractions.

The L.C.M of the denominators (4, 3, 5) is 60.

$\frac{3}{4}$ = $\frac{\mathrm{3\; x\; 15}}{\mathrm{4\; x\; 15}}$ = $\frac{45}{60}$

$\frac{2}{3}$ = $\frac{\mathrm{2\; x\; 20}}{\mathrm{3\; x\; 20}}$ = $\frac{40}{60}$

$\frac{3}{5}$ = $\frac{\mathrm{3\; x\; 12}}{\mathrm{5\; x\; 12}}$ = $\frac{36}{60}$

Ascending means from the lowest to highest.

The ascending order of magnitude of the numerators are 36, 40, 45 hence the arranged fraction is $\frac{3}{5}$, $\frac{2}{3}$, $\frac{3}{4}$

$\frac{1}{\mathrm{a}}$ = $\frac{1}{\mathrm{b}}$ + c

Get rid of the fractions by multiplying both sides by the L.C.M of the denominators (a and b). The L.C.M is a x b.

a x b x $\frac{1}{\mathrm{a}}$ = a x b x $\frac{1}{\mathrm{b}}$ + c x a x b

b x 1 = a x 1 + c x a x b

b = a + abc

b - abc = a

Factorize b out

b(1 - ac) = a

Divide both sides by (1 - ac)

b = $\frac{\mathrm{a}}{\mathrm{1\; -\; ac}}$

Ratio = 9 years : 12 years : 24 years

Note:
1. 3 divides 9, 3 times, 12, 4 times and 24, 8 times
2. the years cancel out

Ratio = 3 : 4 : 8

The more the workers, the lesser the time.

Hence time (t) is inversely proportional to the number of workers (w).

t ∝ $\frac{1}{\mathrm{w}}$

Remove the proportionality by introducing a constant (k).

t = k x $\frac{1}{\mathrm{w}}$

Use the known information to solve for k

10 workers used 9 days.

9 = k x $\frac{1}{10}$

k = 9 x 10

The general equation is:

t = 9 x 10 x $\frac{1}{\mathrm{w}}$

t = $\frac{\mathrm{9\; x\; 10}}{\mathrm{w}}$

Calculate how many days it would take the 15 workers.

w = 15

t = $\frac{\mathrm{9\; x\; 10}}{15}$ = 3 x 2 = 6 days

Note:
1. 3 divides 9, 3 times and 15, 5 times
2. 5 divides itself 1 time and 10, 2 times

Before you can compare fractions, their denominators must be the same. Once they are the same, you can compare the fractions in the magnitude of their numerators.

To make the denominator the same, express the denominator of each of the fractions to be the L.C.M of the denominators by multiplying the numerator and the denominator by the number of times the denominator can divide the L.C.M for each of the fractions.

> means greater than.

I:  $\frac{5}{6}$ > $\frac{3}{4}$

The L.C.M of 6 and 4 is 12

$\frac{5}{6}$ = $\frac{\mathrm{5\; x\; 2}}{\mathrm{6\; x\; 2}}$ = $\frac{10}{12}$

$\frac{3}{4}$ = $\frac{\mathrm{3\; x\; 3}}{\mathrm{4\; x\; 3}}$ = $\frac{9}{12}$

Now since the denominators are the same, compare the numerators.

10 is greater than 9 hence $\frac{5}{6}$ > $\frac{3}{4}$ is true

II:  $\frac{3}{4}$ > $\frac{3}{5}$

The L.C.M of 4 and 5 is 20

$\frac{3}{4}$ = $\frac{\mathrm{3\; x\; 5}}{\mathrm{4\; x\; 5}}$ = $\frac{15}{20}$

$\frac{3}{5}$ = $\frac{\mathrm{3\; x\; 4}}{\mathrm{5\; x\; 4}}$ = $\frac{12}{20}$

15 is greater than 12 hence $\frac{3}{4}$ > $\frac{3}{5}$ is true

III:  $\frac{3}{8}$ > $\frac{1}{2}$

The L.C.M of 8 and 2 is 8.

$\frac{1}{2}$ = $\frac{\mathrm{1\; x\; 4}}{\mathrm{2\; x\; 4}}$ = $\frac{4}{8}$

3 is not greater than 4 hence $\frac{3}{8}$ > $\frac{1}{2}$ is false

First find the rule of the mapping.

Note: If the mapping has equal interval (difference) for the y and x interval of 1, you can use the general linear sequence formula to find the rule of the mapping.

U_n = U₁ + (n-1)d

In the mapping, U_n = y and n = x and U₁ is the value of y when x = 1

U₁ = 3

y = 3 + (x-1)d

The d is the interval between each consecutive y

In this mapping, d = 5 - 3 = 7 - 5 = 2

We can substitute our d and simplify the expression for y

y = 3 + (x-1)2

y = 3 + 2x - 2

y = 2x + 3 - 2

y = 2x + 1

Now find the image when x = P

Image = 2 x P + 1 = 2P + 1

An hour means 1 hour.

Change the hour to minutes so the unit for time would be the same.

1 hour = 60 minutes.

If 60 minutes = 50 km

30 minutes = $\frac{\mathrm{50\; km\; x\; 30\; minutes}}{\mathrm{60\; minutes}}$ = 25 km

Note:
1. minutes cancel each other
2. 30 divides itself 1 time and 60, 2 times
3. 2 divides itself 1 time and 50, 25 times

Let f = the scale factor.

New Area = f² x Original Area

$\frac{\mathrm{New\; Area}}{\mathrm{Original\; Area}}$ = f²

New Area = Original Area x 144

$\frac{\mathrm{New\; Area}}{\mathrm{Original\; Area}}$ = 144

f² = 144

144 = 12 x 12 = 12²

f² = 12²

Since the powers are the same, the bases are also the same.

f = 12

Simple Interest = $\frac{\mathrm{Principal\; x\; Time\; x\; Rate}}{100}$

Principal = ₵2,500

Time = 4

Rate = 25

Interest = $\frac{\mathrm{₵2500\; x\; 4\; x\; 25}}{100}$ = ₵25 x 100 = ₵2500

Note:
1. the zeros (0) at the end of 2500 and 100 cancel each other
2. 4 x 25 = 100

Total number of goals = 0 x 1 + 1 x 7 + 2 x 6 + 3 x 4 + 4 x 1 + 5 x 1

Total number of goals = 0 + 7 + 12 + 12 + 4 + 5

Total number of goals = 40

Mean = $\frac{\mathrm{\Sigma fx}}{\mathrm{\Sigma f}}$ = $\frac{40}{20}$ = 2

Total number of matches played = Sum of frequency

Total number of matches played = 1 + 7 + 6 + 4 + 1 + 1

Total number of matches played = 20

Probability of is + Probability of not = 1

P(Head) + P(Tail) = 1

$\frac{1}{3}$ + P(Tail) = 1

P(Tail) = 1 - $\frac{1}{3}$

Every whole number is being divided by 1

1 = $\frac{1}{1}$

P(Tail) = $\frac{1}{1}$ - $\frac{1}{3}$

P(Tail) = $\frac{\mathrm{3\; x\; 1\; -\; 1\; x\; 1}}{3}$

P(Tail) = $\frac{\mathrm{3\; -\; 1}}{3}$

P(Tail) = $\frac{2}{3}$

a = $\left(\begin{array}{c}-2\\ 3\end{array}\right)$ and b = $\left(\begin{array}{c}2\\ -5\end{array}\right)$

a + 2b = $\left(\begin{array}{c}-2\\ 3\end{array}\right)$ + 2$\left(\begin{array}{c}2\\ -5\end{array}\right)$

a + 2b = $\left(\begin{array}{c}-2\\ 3\end{array}\right)$ + $\left(\begin{array}{c}\mathrm{2\; x\; 2}\\ \mathrm{-5\; x\; 2}\end{array}\right)$

a + 2b = $\left(\begin{array}{c}-2\\ 3\end{array}\right)$ + $\left(\begin{array}{c}4\\ -10\end{array}\right)$

a + 2b = $\left(\begin{array}{c}\mathrm{-2\; +\; 4}\\ \mathrm{3\; +\; -10}\end{array}\right)$

a + 2b = $\left(\begin{array}{c}2\\ -7\end{array}\right)$

143_five

3 = 3 x 5⁰ = 3 x 1 = 3

4 = 4 x 5¹ = 4 x 5 = 20

1 = 1 x 5² = 1 x 5 x 5 = 25

Note:
1. any number/variable raised to the power 0 is 1
2. any number/variable raised to the power 1 is the same number

Interior angle of a regular polygon = $\frac{\mathrm{(n\; -\; 2)\; x\; 180°}}{\mathrm{n}}$

120° = $\frac{\mathrm{(n\; -\; 2)\; x\; 180°}}{\mathrm{n}}$

Get rid of the fraction by multiplying both sides of the equation by the denominator (n).

120 x n = $\frac{\mathrm{(n\; -\; 2)\; x\; 180}}{\mathrm{n}}$ x n

120n = (n - 2) x 180

120n = 180n - 360

360 = 180n - 120n

360 = 60n

Divide both sides by 60.

n = $\frac{360}{60}$ = 6

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = $(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$

Change the mixed fractions to improper fractions.

3$\frac{1}{2}$ = $\frac{\mathrm{2\; x\; 3\; +\; 1}}{2}$ = $\frac{\mathrm{6\; +\; 1}}{2}$ = $\frac{7}{2}$

4$\frac{1}{3}$ = $\frac{\mathrm{3\; x\; 4\; +\; 1}}{3}$ = $\frac{\mathrm{12\; +\; 1}}{3}$ = $\frac{13}{3}$

Every whole number is being divided by 1

7 = $\frac{7}{1}$

3 = $\frac{3}{1}$

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = $(\frac{7}{2}+\frac{7}{1})÷(\frac{13}{3}-\frac{3}{1})$

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = $\left(\frac{\mathrm{1\; x\; 7\; +\; 2\; x\; 7}}{2}\right)÷\left(\frac{\mathrm{1\; x\; 13\; -\; 3\; x\; 3}}{3}\right)$

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = $\left(\frac{\mathrm{7\; +\; 14}}{2}\right)÷\left(\frac{\mathrm{13\; -\; 9}}{3}\right)$

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = $\left(\frac{21}{2}\right)÷\left(\frac{4}{3}\right)$

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = $\frac{21}{2}÷\frac{4}{3}$

Reciprocate the fraction (numerator becomes denominator and denominator becomes numerator) at the right and change the division (÷) to multiplication (x) sign.

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = $\frac{21}{2}x\frac{3}{4}$

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = $\frac{\mathrm{21\; x\; 3}}{\mathrm{2\; x\; 4}}$

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = $\frac{63}{8}$

$(3\frac{1}{2}+7)÷(4\frac{1}{3}-3)$ = 7$\frac{7}{8}$

n - $\frac{2}{3}$ > $\frac{1}{3}$ - n

Get rid of the fracttions by multiplying both sides by the L.C.M of the denominators (3).

3 x n - 3 x $\frac{2}{3}$ > 3 x $\frac{1}{3}$ - n x 3

3n - 2 > 1 - 3n

3n + 3n > 1 + 2

6n > 3

Divide both sides by 6

n > $\frac{3}{6}$

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

n > $\frac{1}{2}$

∴ the truth set is {n:n > $\frac{1}{2}$ }

∆ CDE is an equilateral triangle. All angles are equal (60°)

∠ CED = ∠ DCE = ∠ CDE = 60°

∠ CED + ∠ CEA = 180°(Sum of angles on a straight line)

60° + ∠ CEA = 180°

∠ CEA = 180° - 60°

∠ CEA = 120°

∆ ACE is an isosceles triangle. The base angles are the same.

∠ CAE = ∠ ACE

∠ CEA + ∠ CAE + ∠ ACE = 180°(Sum of interior angles)

120° + ∠ CAE + ∠ ACE = 180°

∠ CAE + ∠ ACE = 180° - 120°

∠ CAE + ∠ ACE = 60°

∠ CAE = ∠ ACE = $\frac{\mathrm{60°}}{2}$ = 30°

x + ∠ ACE + ∠ DCE = 180° (Sum of angles on a straight line)

∠ DCE = 60°

x + 30° + 60° = 180°

x + 90° = 180°

x = 180° - 90°

x = 90°

Five times means 5 x (Multiplied by 5)

More than means if you add the number more than, you will get the other number.

is means equal to.

Let n = the number.

Five times a number ⇒ 5 x n = 5n

Five times a number is ⇒ 5n =

four more than the number ⇒ 4 + n

In equation,

5n = 4 + n

Solve for the number.

5n - n = 4

4n = 4

Divide both sides by 4

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

∴ the number is 1

Each means every 1.

If 1 = ₵15

450 = $\frac{\mathrm{₵15\; x\; 450}}{1}$ = ₵6,750

The percentage for the cost is 100%

If 100% = ₵200

2% = $\frac{\mathrm{₵200\; x\; 2\%}}{\mathrm{100\%}}$ = ₵2 x 2 = ₵4

Note: the zeros (0) at the end of 200 and 100 cancel each other.

The ₵4 commission is per a bottle.

If 1 = ₵4

24 = $\frac{\mathrm{₵4\; x\; 24}}{1}$ = ₵96

Equivalent means the value is the same so you can equate the two.

$\frac{9}{21}$ = $\frac{\mathrm{1\; +\; x}}{7}$

Note: 3 divides 9, 3 times and 21, 7 times.

$\frac{3}{7}$ = $\frac{\mathrm{1\; +\; x}}{7}$

Since the denominators (7) at both the left and right are the same, the numerators are also the same.

1 + x = 3

x = 3 - 1

x = 2

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 third significant figure ends at 39.9. The next significant figure to it is 7 which is more than 4 hence 1 is added to the third significant figure (9 + 1 = 10). The 0 is written and the 1 is carried forward to 39 (39 + 1 = 40.0)

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

Get rid of the fractions by multiplying both sides of the equation by the L.C.M of the denominators (x and 3). The L.C.M is 3 x x (3x).

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

3 + x = 3x

3 = 3x - x

3 = 2x

Divide both sides by 2

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

∠ NAB = ∠ SBA = 55°(Alternate angles)

The right-angles are 90°

Bearing starts from the north clockwise.

Bearing of Bebeka from Aboku is the ∠ from N to B.

Bearing of Bebeka from Aboku = 90° + 90° + 55° = 235°

Perimeter is the sum of all the sides of a shape.

Perimeter = 11_two cm + 11_two cm + 10_two cm + 10_two cm

Method I

Adding in binary (base two)

Note:

Converting the 5 to base 2

$$\begin{array}{cc}\mathrm{}& 5\\ 2& \mathrm{2\; r\; 1}\\ 2& \mathrm{1\; r\; 0}\\ 2& \mathrm{0\; r\; 1}\end{array}$$

You list the remainders from the bottom to the top.

5_ten = 101_two

Method II

Convert the binary to base 10 and add and convert back to base 2

11_{two = 1 x 2¹ + 1 x 2⁰}

Any number/variable raised to the power 0 is 1

Any number/variable raised to the power 1 is the same number/variable.

2⁰ = 1

2¹ = 2

11_two = 1 x 2 + 1 x 1

11_two = 2 + 1

11_two = 3

10_two = 1 x 2¹ + 0 x 2⁰

10_two = 1 x 2 + 0 x 1

10_two = 2 + 0

10_two = 2

11_two cm + 11_two cm + 10_two cm + 10_two cm = 3 cm + 3 cm + 2 cm + 2 cm

11_two cm + 11_two cm + 10_two cm + 10_two cm = 10 cm

Convert 10_{ten to binary (base two)}

$$\begin{array}{cc}\mathrm{}& 10\\ 2& \mathrm{5\; r\; 0}\\ 2& \mathrm{2\; r\; 1}\\ 2& \mathrm{1\; r\; 0}\\ 2& \mathrm{0\; r\; 1}\end{array}$$

You list the remainders from the bottom to the top.

10_ten = 1010_two

Perimeter = 1010_two cm

Area of a sector = $\frac{\mathrm{Angle}}{360}$ x area of the circle

Area of the circle = 120 cm²

Angle of the sector = 60°

Area of the sector = $\frac{60}{360}$ x 120 cm² = 20 cm²

Note:
1. 60 divides itself 1 time and 360, 6 times
2. 6 divides itself 1 time and 12, 2 times

∠ CBA = ∠ ADC = 110°(Opposite angles of a parallelogram are equal)

∠ ADC = ∠ EDF = 110° (Vertically opposite angles)

40° + ∠ EDF + θ = 180° (Sum of interior angles of a triangle)

40° + 110° + θ = 180°

150° + θ = 180°

θ = 180° - 150°

θ = 30°

Draw an imaginary parallel line to pass through the point T.

∠ BTQ = ∠ CBT = 40° (Alternate angles)

∠ DET + ∠ FET = 180° (Sum of angles on a straight line)

140° + ∠ FET = 180°

∠ FET = 180° - 140°

∠ FET = 40°

∠ FET = ∠ QTE = 40° (Alternate angles)

∠ BTE = ∠ BTQ + ∠ QTE

∠ BTE = 40° + 40°

∠ BTE = 80°

∠ BTE + y = 360° (Sum of angles at a point)

80° + y = 360°

y = 360° - 80°

y = 280°

∠ KGM = right-angle = 90°

73° + 90° + ∠ KMG = 180°(Sum of interior angles of a triangle)

163° + ∠ KMG = 180°

∠ KMG = 180° - 163°

∠ KMG = 17°

The base angles of an isosceles triangle are the same.

∠ ADC = ∠ ACD

∠ ADC + ∠ ACD + 55° = 180°(Sum of interior angles of a triangle)

∠ ADC + ∠ ACD = 180° - 55°

∠ ADC + ∠ ACD = 125°

∠ ADC = ∠ ACD = $\frac{\mathrm{125°}}{2}$ = 62.5°

∠ ADC = ∠ ABE = 62.5°(Corresponding angles)

x + ∠ ABE = 180°(Sum of angles on a straight line)

x + 62.5° = 180°

x = 180° - 62.5°

x = 117.5°

The sides of a right-angled triangle should obey Pythagorean theorem.

From Pythagorean theorem,

The hypothenuse (c) is the longest side. The side facing the right-angle (∟)

Analyze each measurements to see if it obeys pythagorean theorem.

Ama's measurements were 80 mm, 40 mm, 50 mm.

You can divide each of the measurements by 10 to make verification easier.

Ama's measurement = 8, 4, 5

The longest side is 8 hence the hypothenuse is 8

8² = 4² + 5²

8² = 8 x 8 = 64

4² = 4 x 4 = 16

5² = 5 x 5 = 25

64 = 16 + 25

64 = 41

64 is not equal to 41 hence Ama's measurement is incorrect.

Adjoa's measurements were 50 mm, 120 mm, 130 mm.

Adjoa's measurements = 5, 12, 13

The longest side is 13 hence the hypothenuse is 13

13² = 12² + 5²

13² = 13 x 13 = 169

12² = 12 x 12 = 144

5²

169 = 144 + 25

169 is equal to 144 + 25 (169) hence Adjoa's measurement is correct.

Abena's measurements were 20 mm, 30 mm and 40 mm.

Abena's measurements = 2, 3, 4

The longest side is 4 hence the hypothenuse is 4.

4² = 2² + 3²

2² = 2 x 2 = 4

3² = 3 x 3 = 9

16 = 4 + 9

16 = 13

16 is not equal to 13 hence Abena's measurement is incorrect.