Reflection in the line x - 2 = 0, x = 2

k = 2

(-7, 8) → (2 x 2 - (-7), 8)

(-7, 8) → (4 + 7, 8)

(-7, 8) → (11, 8)

Note:
1. -- = + or -(-) = +
2. -(-7) = + 7

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Pick any two points from the table to calculate the gradient.

Using the first two points, (0, 1) and (2, 2).

Y₂ = 2
Y₁ = 1
X₂ = 2
X₁ = 0

Gradient = $\frac{\mathrm{2\; -\; 1}}{\mathrm{2\; -\; 0}}$ = $\frac{1}{2}$

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Pick any two points from the table to calculate the gradient.

Using the first two points, (0, 1) and (2, 2).

Y₂ = 2
Y₁ = 1
X₂ = 2
X₁ = 0

Gradient = $\frac{\mathrm{2\; -\; 1}}{\mathrm{2\; -\; 0}}$ = $\frac{1}{2}$

Using the first point (0, 1)

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

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

2 x y - 1 x 2 = 1(x - 0)

2y - 2 = x

2y = x + 2

y = x² - 5x + k

From the point (3, 1), x = 3 and y = 1

Substitute the values of x and y into the equation and solve for k

1 = 3² - 5 x 3 + k

1 = 3 x 3 - 15 + k

1 = 9 - 15 + k

1 = -6 + k

1 + 6 = k

7 = k

∴ k = 7

Discount % = $\frac{\mathrm{Discount}}{\mathrm{Actual\; Price}}$ x 100%

Discount = Actual Price - Selling Price

Actual price for the 5 tins = GH₵ 15.00 x 5
Actual price for the 5 tins = GH₵ 75.00

Discount = GH₵ 75.00 - GH₵ 66.00
Discount = GH₵ 9.00

Discount % = $\frac{\mathrm{GH₵\; 9.00}}{\mathrm{GH₵\; 75.00}}$ x 100%

Discount % = 12%

Method I

The angle for the entire circle is 360°

If 140° = 40 cm²
360° = $\frac{\mathrm{40\; x\; 360}}{140}$ = 102.86 cm²

Method II

Area of a sector = $\frac{\mathrm{\theta }}{360}$ x πr²

But πr² = area of the circle

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

Area of a sector = 40 cm²
θ = 140

40 = $\frac{140}{360}$ x area of circle

Make area of circle the subject.

Multiply both sides by 360

40 x 360 = 140 x area of circle

Divide both sides by 140

$\frac{\mathrm{40\; x\; 360}}{140}$ = area of circle

Area of circle = 102.86 cm²

Gradient = $\frac{\mathrm{Change\; in\; Y}}{\mathrm{Change\; in\; X}}$ = $\frac{{\mathrm{Y}}_2-{\mathrm{Y}}_1}{{\mathrm{X}}_2-{\mathrm{X}}_1}$

Choose any two points from the table.

P₁(6.20, 3.90) and P₂(6.85, 5.20)

Gradient = $\frac{\mathrm{5.2-3.9}}{\mathrm{6.85-6.20}}$ = $\frac{1.3}{0.65}$ = 2

Each small grid on the x axis = $\frac{\mathrm{Interval\; on\; the\; x\; axis}}{10}$

Each small grid on the x axis = $\frac{1}{10}$ = 0.1

Each small grid on the y axis = $\frac{\mathrm{Interval\; on\; the\; y\; axis}}{10}$

Each small grid on the y axis = $\frac{5}{10}$ = 0.5

Intersection 1 = (2, 17 x 0.5) = (2, 8.5)

Note: Point = Number of small grids from 0 x value of small grid

Intersection 2 = (-15 x 0.1, 4 x 0.5) = (-1.5, 2)

Each small grid on the x axis = $\frac{\mathrm{Interval\; on\; the\; x\; axis}}{10}$

Each small grid on the x axis = $\frac{1}{10}$ = 0.1

Each small grid on the y axis = $\frac{\mathrm{Interval\; on\; the\; y\; axis}}{10}$

Each small grid on the y axis = $\frac{5}{10}$ = 0.5

Number of small grids of a point on an axis = $\frac{\mathrm{Point\; value}}{\mathrm{Value\; of\; each\; small\; grid\; on\; that\; axis}}$

Number of small grids on the x-axis for -2.5 = $\frac{-2.5}{0.1}$ = -25 from 0 at the left

Number of small grids on the y axis of the corresponding -2.5 on the curve = 18 from 0

Each small grid = 0.5

Point = 18 x 0.5 = 9

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

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

3y + bx - 6 = 0

Since the line passes through (3, 4), when you substitute the x value, 3 and y value 4 into the equation, the result is supposed to be 0. Then you can find the b value.

3(4) + b(3) - 6 = 0

12 + 3b - 6 = 0

3b + 12 - 6 = 0

3b + 6 = 0

3b = -6

Divide both sides by 3

b = $\frac{-6}{3}$

b = -2

Rotation through 90^o anticlockwise

When rotating a point 90^o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-y,x).

In other words, switch x and y and make y negative

Rotation through 270^o anticlockwise

When rotating a point 270^o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(y,-x).

This means, we switch x and y and make x negative

Rotation through 180^o anticlockwise

When rotating a point 180^o anticlockwise(counterclockwise) about the origin our point A(x,y) becomes A'(-x,-y).

So all we do is make both x and y negative

Substitute y = x² + 4x - 6 into x² + 4x - 7 = 0

-7 = -6 - 1

x² + 4x - 7 = 0
x² + 4x - 6 - 1 = 0

But x² + 4x - 6 = y

y - 1 = 0
y = 0 + 1
y = 1

Each small grid on the x axis = $\frac{\mathrm{Interval\; on\; the\; x\; axis}}{10}$

Each small grid on the x axis = $\frac{1}{10}$ = 0.1

Each small grid on the y axis = $\frac{\mathrm{Interval\; on\; the\; y\; axis}}{10}$

Each small grid on the y axis = $\frac{5}{10}$ = 0.5

Intersection 1 = (2, 17 x 0.5) = (2, 8.5)

Note: Point = Number of small grids from 0 x value of small grid

Intersection 2 = (-15 x 0.1, 4 x 0.5) = (-1.5, 2)

Each small grid on the x axis = $\frac{\mathrm{Interval\; on\; the\; x\; axis}}{10}$

Each small grid on the x axis = $\frac{1}{10}$ = 0.1

Each small grid on the y axis = $\frac{\mathrm{Interval\; on\; the\; y\; axis}}{10}$

Each small grid on the y axis = $\frac{5}{10}$ = 0.5

Number of small grids of a point on an axis = $\frac{\mathrm{Point\; value}}{\mathrm{Value\; of\; each\; small\; grid\; on\; that\; axis}}$

Number of small grids on the x-axis for -2.5 = $\frac{-2.5}{0.1}$ = -25 from 0 at the left

Number of small grids on the y axis of the corresponding -2.5 on the curve = 18 from 0

Each small grid = 0.5

Point = 18 x 0.5 = 9

Use points on the graph to substitute the general equation y = ax² + bx + c.

Note: it is easier to use points that either x = 0 or y = 0

When x = -1, y = 0

0 = a(-1)² + b(-1) + c

a - b + c = 0 ------- eqn 1

When x = 2, y = 0

0 = a(2)² + b(2) + c

4a + 2b + c = 0 ------- eqn 2

When y = 2, x = 0

2 = a(0)² + b(0) + c

c = 2

Substitute c = 2 into ----- eqn 1

a - b + 2 = 0

a - b = -2 -------- eqn 3

Substitute c = 2 into ----- eqn 2

4a + 2b + 2 = 0

4a + 2b = - 2

Divide both sides by 2

2a + b = - 1 ------ eqn 4

Get rid of b. ----- eqn 3 + ----- eqn 4

3a = -3

Divide both sides by 3.

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

Substitute a = -1 into ---- eqn 4

2(-1) + b = -1

-2 + b = -1

b = -1 + 2

b = 1

Substitute a = -1, b = 1 and c = 2 into the general equation y = ax² + bx + c

y = ax² + bx + c

y = (-1)x² + 1(x) + 2

y = -x² + x + 2

y = 2 + x - x²

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

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