Part 1: Linear Systems and Line Segments Create and attach a 1 page ‘cheat’ sheet for the Linear Systems Unit.

Solve each system by graphing:

y = x+2, y = 4-1

2x – y + 9 = 0, x + y – 3 = 0

Solve by substitution:

x – 2y = 4, 2x – 3y = 7

Solve by elimination:

3x + 8y = -3, x + 8y = -5

4x – 5y = -22, 5x +6y = -3

Find the length of the line segment joining each pair of points. (Express to nearest 10th of a unit)

(2,5) and (5,7)

(-1, -3) and (-2, -4)

Find the midpoint of each line segment.

M (-2, 2), P (4, 8)

A (3, 3), J (-9, 3)

The Midpoint of AB is M(-2, 1). IF A(-7, 3), what are the coordinates of B?

Page Break

Part 2: Trigonometry

Create and attach a 1 page ‘cheat’ sheet for the Trigonometry Unit.

Solve each right triangle:

Shape E

Shape F 62⁰ D

150 cm

ShapeText BoxA

Shape15⁰

ShapeB C

Page Break

Solve each triangle:

Shape L

58⁰

56⁰

M 19 cm N

ShapeJ

58⁰

18 mm 15 mm

K L

Khalid wants to be sure his boat is safely anchored. He knows that the angle of depression that the boat makes with the horizontal when anchored should be less than 12⁰. The boat is 100 m above the seabed and the anchor cable is 440 m long. Is Khalid safely anchored? Explain.

Part 3: Polynomials

Create and attach a 1 page ‘cheat’ sheet for the Polynomial Unit.

Simplify

(4x - 3) + (5x+4)

(2x2 – 3x + 4) – (x2 + 4x - 1)

(-6x3 yz2 )(-2xy2 z)

Factor, if possible

7x + 42

X2 + 7x + 12

3x2 = 12x = 9

X2 + 9

5x2 – 20x + 20

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Part 4: Quadratics

Create and attach a 1 page ‘cheat’ sheet for the Quadratics Unit.

Graph the data and write the relation in the form of
y=a(x−h)2
y=ax−h2

X

Y

-6

32

-5

18

-4

8

-3

2

-2

0

-1

2

Graph the data and write the relation in the form of
y=a(x−h)2
y=ax−h2

X

Y

-1

-8

0

-4.5

1

-2

2

-0.5

3

0

4

-0.5

Graph the relation by plotting the vertex and 2 other points.

y=(x+1)2−1
y=x+12−1

y=−2(x−6)2+2
y=−2x−62+2

For each, write the relation in standard form:
y=ax2+bx+c
y=ax2+bx+c

a=2, b=5, y intercept = 11

y=5x2+bx+c
y=5x2+bx+c

, vertex at (1, 4)

a=−3
a=−3

, passes through (2, 11) and (0, -5)

Factor:

x2−17x+66
x2−17x+66

x2−3x−18
x2−3x−18

x2+13x
x2+13x

x2−9
x2−9

Find the zeros.

y=(x+3)(x−3)
y=x+3x−3

y=−10x2
y=−10x2

y=8(x−8)(x+0.5)
y=8(x−8)(x+0.5)

y=8(x−8)(x+0.5)

Step-by-step explanation:

Part 1: Linear Systems and Line Segments

Create and attach a 1 page ‘cheat’ sheet for the Linear Systems Unit.

Solve each system by graphing:

y = x+2, y = 4-1

2x – y + 9 = 0, x + y – 3 = 0

Solve by substitution:

x – 2y = 4, 2x – 3y = 7

Solve by elimination:

3x + 8y = -3, x + 8y = -5

4x – 5y = -22, 5x +6y = -3

Find the length of the line segment joining each pair of points. (Express to nearest 10th of a unit)

(2,5) and (5,7)

(-1, -3) and (-2, -4)

Find the midpoint of each line segment.

M (-2, 2), P (4, 8)

A (3, 3), J (-9, 3)

The Midpoint of AB is M(-2, 1). IF A(-7, 3), what are the coordinates of B?

Page Break

Part 2: Trigonometry

Create and attach a 1 page ‘cheat’ sheet for the Trigonometry Unit.

Solve each right triangle:

Shape E

Shape F 62⁰ D

150 cm

ShapeText BoxA

Shape15⁰

ShapeB C

Page Break

Solve each triangle:

Shape L

58⁰

56⁰

M 19 cm N

ShapeJ

58⁰

18 mm 15 mm

K L

Khalid wants to be sure his boat is safely anchored. He knows that the angle of depression that the boat makes with the horizontal when anchored should be less than 12⁰. The boat is 100 m above the seabed and the anchor cable is 440 m long. Is Khalid safely anchored? Explain.

Part 3: Polynomials

Create and attach a 1 page ‘cheat’ sheet for the Polynomial Unit.

Simplify

(4x - 3) + (5x+4)

(2x2 – 3x + 4) – (x2 + 4x - 1)

(-6x3 yz2 )(-2xy2 z)

Factor, if possible

7x + 42

X2 + 7x + 12

3x2 = 12x = 9

X2 + 9

5x2 – 20x + 20

Page Break

Part 4: Quadratics

Create and attach a 1 page ‘cheat’ sheet for the Quadratics Unit.

Graph the data and write the relation in the form of

y=a(x−h)2

y=ax−h2

X

Y

-6

32

-5

18

-4

8

-3

2

-2

0

-1

2

Graph the data and write the relation in the form of

y=a(x−h)2

y=ax−h2

X

Y

-1

-8

0

-4.5

1

-2

2

-0.5

3

0

4

-0.5

Graph the relation by plotting the vertex and 2 other points.

y=(x+1)2−1

y=x+12−1

y=−2(x−6)2+2

y=−2x−62+2

For each, write the relation in standard form:

y=ax2+bx+c

y=ax2+bx+c

a=2, b=5, y intercept = 11

y=5x2+bx+c

y=5x2+bx+c

, vertex at (1, 4)

a=−3

a=−3

, passes through (2, 11) and (0, -5)

Factor:

x2−17x+66

x2−17x+66

x2−3x−18

x2−3x−18

x2+13x

x2+13x

x2−9

x2−9

Find the zeros.

y=(x+3)(x−3)

y=x+3x−3

y=−10x2

y=−10x2

y=8(x−8)(x+0.5)

y=8(x−8)(x+0.5)

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: