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MFM 2P1 UNIT 1- ALGEBRA - REVIEW NOTE Date:
I. Simplify. 5x2 + 7x —7— 9x2 — 4x —2 (Collect Like Terms)
=5x2 —9x2 + 7x—4x—7—2=—4x2 +3x—9
2. Simplify. (— 5x2y3 4xy2) (Numbers with Numbers, Powers with Powers)(Multiply powers with the same base, keep the base, add exponents.)
=—20x3y5
3. Simplify. (lox
— 30x3y4
5 xy2
= 6x4y2
.4. Simplify using the Distributive Property.
a) 2x(5x—3) b) (x—4Xx+9) c) (2x+5)2
2x(ox—3) (x—4Xx+9)=(2x+5X2x+5)2 =x2+9x—4x—36
2=12x —6x =4x +lOx+lOx+25=x2+5x—362=4x +20x+25
5. Simplify. —4c—3X: x—5)li tip the binomial rs and r t the answer in brackets.)
=_4(2x2 —S -‘ i
= 4(2x2 —I Lv
= _1 +44r —‘,()
MFM 2P1 UNIT i-ALGEBRA - WVTEWI
1. Simplify. c?ncncLa) —12x2 —4x+1l+7x2 —lOx—6 b) 14x2 —5x+1O—8x2 —6x—19. 1 2 2’ 2 22c) 4r —xy+3xy—3x +7xy d) óx y—2x y —lix y+5x y —3xy
2. Simplify.
a) (&g’—5y) b) (_3x2y(2xy2) c) (3.y_2x2y_4xy2)
3. Simplify.
a)35x4y3
b)(8x4y3x2y)
c)—48x8y6
—5y2 6x2y2 48x5y3
4. Simplify.
a) 3(2x—5) b) 4x(5x+7) c) _6x(x2 5x)
ci) 7—2(8x—5) e) (x—9Xx+7) f) (2x÷9)2g) (3x2 +l7xy)_(12x2 —3y) h) (3? _5y)_(3xy_7y2)
1) 2x(x + y) — 3x(2x — 3y) j) 2x(3x — 5y)— x(2y + 3x)
5. Simplify.
• a) (x+iXx+4) b) (x+3Xx—3) c)ci) (2x+lXx+3) e) (x÷2 U (3x—2X5x+4)
6. Simplify.
a) 3(x—lXx—4) b) —4(x—3X2x—5) c) —3(2x—yX2x+y) d) 3(x—5)2
Answers:
1. a) —5x2 —14x+5 b) 6? —llx—9 c) x2 +9xy d) —Sx2y+3x1y2 —Jxy
2. a) — 40xy1 b) — 6x3y3 c) 24x3y3
1 4 313. a) —7rv b) 4x c) —xy
4. a) 6x—15 20x2 +28x c) —6? +30x2 d) 17—i6x e) x2—2x—63P) 4? +16x tSl L) —9? +20.rv ‘i) 3? —Sxy+7y2 1) —4? +1 En j) 3? —l2xy
u v2 -5x+3 h) . x -IOx+25 d) 2x2+7x t3?+4x+4 fl l5x1+2x—
- b)14i2-5x+lo-23—Li-)9- l2yj±3zZ_ ti- LDL+)V/c 14)c St SX-(oX +jo-I9
tDx’-)!x-9
C
a
t )e
C
Sc&) (*l)&-{) b) (+3)(x-3) c (-s)(x.S)xt’txtitI
l9 ‘-IOz-f2S
• ci) (,c÷lX,c-t-3) e) (jtl)(x*2)(*l)
=
c) C3i-z)(sH)J5I2u-IOtz2
Z
a) 3(xJ)&Lk)= (3z-J)(j-H) z-’-t(dx2- 5-&c IS)
-
•c3(z-j)(vj)
1
.
MFM 2P1 UNIT 2- SOLVING EQUATIONSREVIEW NOTE
1. Solve. Sx+4=2x+16
8x+4—2X=2x+16—2X
6x + 4 = 16
6x ÷4—4=16—46x = 12
6x 12
6=6x=2
2. Solve. 3(x —4)— 5(x + 2) = —2(3x — i)
3ç—4)—5(x÷2)=—2(3x—1)
3x—12—Sx—1O=—6x+2
3x—5x-12...Io=—6x÷2
—2X-22=—6x+2
4x—22=2
4x = 24
x=6
x x—13. Solve. —÷--——=122 5
i o3) + 1 O(z-i) = 1 0(12) *MultipIy each term by the lowest common denominaEor.
10Q2) tSimplify the fractions
ix+ ‘x—2= 120ix— 2= 120
a 7x=122w
MFM 2P1 UNIT 2- SOLVING UATI S - REVIEW
Solve.
a) 5x+4=2x+1O b) 8x—3=IOx—9 c) 4+3x=25—4x d) —6—3x=Sx+1O
2. Solve.
a) 3x+5x=32 b) —2x—4+6x=9 c) 3(2x—5)=3d) —3ç+2)=24 e) 5(x—3)=1O 1) 2(x÷4)=l—3x+3g) 3(x+I)÷2(x+2)=—32 h) (5x—i)—4=—2(x÷3)+6x i) 2(x—4)—5(x+fl=4(x—5)
3. Solve.
x x x x+2 x+I 2x+l x—2a) —=—5 b) —+—=12 c) +1=— d) =2 24 4 3 3 5
4. Solve and check.
a) 5x—2=—37 b) 3x+5=9x—7.Answers:
I. a) x=2 b) x=3 c) x=3 d) x=—2
2. a) x=4 b) c) x=3 d) x=—1O e) x=5
g) x=—— h) x=—1 i) x=1
3. a) x=—lO b) x=16 c) x=14 d)
4. a) x=—i b) =?
.
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t’1g1- MFM 2P1 UNIT 3- FACTORING - REVIEW
1. Common Factor. 12x3 — 20x3 + 24?
3. Factor the following Difference of Squares. A2—B2 =(A—BXA+B)
.
16x1—25
=(4x—5X4x+5)
Think: What is being suuared to give us 25? Put it here.
.4x3(3x_5+6x1L
__________________________________
* This is what’s left over.* This is what you have to multiply 4? by
to get what you started with.
The Greatest Common Factor
— the largest number that goes into 12, 20 and 24 evenly— x is common to all 3 terms, take out the one with the lowest exponent, x3
2. Factor the following Simple Trinomial (The Number Game).
x2—Sx—24 -
=(x—sXx+3)
Think: What is being squared to give us 16x2? Put it here. (4x)
Put one pfus and one minus behvce, he 4x and iIC 5.
MFM 2P1 UNIT 3- FACTORING - REVIEW(PRTZE
1. Factor..a) 2x+6 b) 6x—9 c) x2 +5x d) 2x +8x
e) 8x+4xy 0 lOx+15y—20z g) 12x5—20x+24x4
2. Factor.
a) x2+4x+3 b) x2—lOx+25 c) x2—9x+20d) x2+5x—36 e) x2—2x—48 fl x2+lóx+64g) x2 +3x—18 h) x2 —8x—20 i) x2 +5x+6
3. Factor.
a) x2—9 b) x2—36 c) x2—100y2 d) x2—4y2
e) 121x2 —144 f) 64x2 —1 g) 16x2 —25y2 i) 81x2 —49y2
4. Factor.
a) 4x—20 b) 9x2 +l8xy c) x2 —I d) x2 +2x+1
e) x2 —4x+4 0 8x3 —4x2 g) 4x2 —49y2 Ii) x2 —lOx+24
Answers:
1. a) 2(x+3) b) 3(2x—3) c) x(x+5) d) 2x(x+4)e) 4x(2+y) 0 5(2x+3y—4z) g) 4x(3x —5÷6x)
2. a) (x+3Xx+I) b) (x—SXx—5) c) (x—5Xx—4) d) (x+9Xx—4)e) (x—8Xx÷6) 0 (x+8Xx÷8) g) (x+oXx—3) h) (x—lOXx÷2)1) (x+JXx+2)
3. a) (x+3”(x—3) h) Iv+6Xx—6) c) (x÷1Oyx—1Oy) d) (x+2vXx—2v)(i lx±I 1: [v—I 2) I) 11xi- 1X8x—I) 0 (4x÷5y4x—5y) g) (QA+ 7yX9x—7y)
aj 4(x—)) iv) c) (x IXx—1) d) (x÷lXx+T)
-) (v—’(v—2) 2x—i) .) (: N’){2x_7y) (v—(,’tv—4)
-—4
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MFM 2P1
Solution: In AABC and i\DFE
AB BC AC
DFFEDE
118 x
12—
y — 16.5
‘Jr11_8
12 y
UNIT #4-SIMILAR TRIANGLESREVIEW NOTE
16.51’ll’l = 16.5L12)
çiij çs15.1 = x
(‘orresoonäura flNIe5 Qfc equaL1’
x 1116.5
S
F12
1. a) Why are the following triangles similar?c-i
b) Determine the values of the variablesx andy to 1 decimal place if necessary.
A
y
C B D
x
16.5
12 = y11 8
11 x
12 — 16.5
8.7 =
2. a) Why are the following triangles similar? cnrrspomde QflcthS flre33
ii 8 11 x
— x +45y
37— y
11 8(37)(x + 45)[!1] = (37)(x + 45)(
x +45]
=V
Slix 195 = 37x
.
x
b) Determine the values of the variables x andy to I decimal place if necessary.
A
B
II
26
C
45
Dy
Solution:
E
In zXABC and AADE
AB BC ACADDEAE
11=8=
37 y x+45
37
= )5= 6x
MFM 2P1
1. 1) Mark all equal angles.
a)
b)
Is
M
30
Q
A
y
12P
R
>NQ14
. ii) Determine the value of x andy to I decimal place usin2 similar trian2les.
AN59 464
NB
y
10
T8
z
c) A
x
p
F 7
Answers:
Q)QABGnJL
59.4it h
Vlij(57.t+)(lt) Rs-I II
5%c
5.6 j s91
• ID) c)Arzr)LApp
72 z j?T ,p NW FFk• fi±
• 1510
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xcik) S t jo
II Lf5xrIS(stL)• xt 10 t5ct9-So4\5
___
.SO8
x=Js
MFM 2P1 UNIT 5- TRIGONOMETRY Date:REVIEW NOTE
1. a) Name the 3 sides of the following triangle relative to angle A
Adjacent
usin! the lull names.
b) Define the following ratios using the fult names of the sides above.
i) Cos(A) =Adjacent
HypotenuseH) Tan(A)
= OppositeAdjacent
Hi) Sin(A) =Opposite
Hypotenuse
2. Use your calculator to determine the following to 4 decimal places.
a) sin(3r)
3. Determine the value of angle A to the nearest degree.
4. Determine the value of x to 1 decimal place.
a) C
x
b)I
15
C
Tcin(A)= ¶21?..Idj
LSIfl(G) =
Hyp
(1’15
LcIfl(2l) - —
Opposite Hypotenuse
c) cos(27°) =
.5299
.89 10
b) sin (s°) = .1392
d) tan(81°) 6.3138
a) tan(A) = 1.327 A 530 b) tan(A) =1235
1000A= 51°
A35
B H
,iH )L = r
Cos(e)=2-14
Tan(C) =
Adj
Tan(26°)= q
5. Determine angle 9 to the nearest degree.
1
9
Solution: Cos(9)= -4Hyp
9=k14
9=5W
6. From the top of a fire tower the ANGLE OF DEPRESSION to a log cabin is 26°. Determine the distanceto the cabin from the base of the tower if the tower is 75m high. Include a labeled diagram in yoursolution.
Let x m be the distance from the cabin to the tower.
75m
C
26°Angle of Depression
xm75
X= Tan(26j
x=153.8
The distance i . 15. .8 m.
UNITS- TRIGONOMETRY - REVIE1S\(
a) sin(52°)
b) cos(68°
c) tan
d) cos(31°
e) tan (80°)
)
)
)
2. Determine the value of angle A to the nearest degree.
a) tan(A) = 1.327 b) cos(A) = 0.643
4. Determine angle 0 to the nearest degree.
1) sin(9°)_
g) tan(45°)
h) cos (57°).
i) n (76j_
j) cos(2°)_
* 829c) sin(A) =
_____
10001235
d) tan(A) =
____
1000
0
5. A 13 II ladder is placed against a wall. If the angle the ladder makes with the ground is 52°, how lhr upthy wail does U 2 ladder reach?
Ansnurs: 3. a) x’12.31. ;i) 0=390
b) x3l.1,) q=44°
x = 42.5
MFM 2P1
1. Use your calculator to determine the following to 4 decimal places.
.
. 3. Determine the value of x to 1
a) C
x
A B
15
decimal place.
b)
0
x
20
F Ic)
32
E H
c)
x
5b) 7
5
8
13
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MFM 2P1 UNIT 6- CONVERSIONS & MEASUREMENT - REVIE4 PRAcncConversions 1
1. Convert. 2. Convert. 3. Convert. 4. Convert.
a) 50 mto decimeters a) 1.5 fito inches a) 20 kg to centigrams a) 100 000 cm to in’
b) 2 450 ml to litres I,) 84 inches to feet I,) 5 dm to millimeters b) 32 in2 to cm2
Answers:
1. a) 500dm b) 2.45 L 2. a) 18 inches b) 7 feet3. a) 2 000 000 cg b) 500mm 4. a) 0.1 in’ b) 320 000 cm2
fl Area
1. Determine the area of the following shapes.
a) b) c)
/
6cm
/
14cm28cm
d) e) 32cm 4cm
•12m HlOcm
I I18cm
Answers: La) 154 cm2 b) 530.9 cm2 c) 84 cm2 d) 144 in2 e) 300 m2 Q 40 cm2
I Volume
I. Determine the volume of each of the following objects.
/1—i
a)
_______________8cm
28cm.
b)
Ii nI
‘N
//
/ 30cm
j_.__ /
6.’
c) d)
.32 cm I
12 m
IAnswers 1.a) 4032 cm3 b) 3360 cm3 c) 2M3.3 cm3 d) 1809.6 in3
I Surface Area 1
.
.
I. Determine the surface area of each of the following objects.
c)
24 cm
18cm
8 cm
Urn
a)
28cm
b)
10 cm
20cm
16cm
d)
16cm
18cm
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MFM 2P1 UNIT 7- EQUATION OF A LINE & SLOPEREVIEW NOTE
1. Determine the slope of the Line through the following point.
(—2,7) and (8,—4)
Solution:Ax
7-(-4)
—2—8
11In = —
—10
—It10
2. State the slope and y—intercept for each of the following lines: 5x + 7y —it = 0
Solution: * Must put the equation in the form y = mx + b.
Sx + 7y —11 = 0
7y=—Sx+1l
—5 II7 7
—5 . . 11Slope is ——, The y—intercept is —
3. State the equations of the line in the form, y = mx + b,
a) in = b = 7 b) a horizontal line through (—9,—3) c) the x — axis
Solution: a) y=x+7 h) ‘=—3 c) y=0
4 t ‘I the ‘u n U:
Ii 4,6 ) the i’—axis c) i is indefined passing through (—72)
.Solution:a) t=
.
Solution:
—3
Ar
m= —(—4)
in
4-HI)
9
8
—x+4 rnt —A
y = mx + b9
y=jx+b
Since (4,5) is on the line
5 =(4)+b
368
8(5) = 8[2) + 8(b)
40=36+8b4 = Sb
-Lb2
9 1y =—x+—
8 2
.
15. Determine the equation of the line with the given slope and passing through the given point
m = —6, A(3,—4)
Solution: y=mx+by = —óx ÷ bSince (3,—4) is on the line—4=—6(3)÷b—4= —1814=by = —6x ÷14
6. Determine the equation of the line passing through the following two points.
A(4,5), B(—4,—4)
7. Graph the following lines using slope and y—intercept.
,a y=3x—l
A
S
r MFM 2P1 UNIT 7- EOUATION OF A LINE & SLOPE - REVIE1nC
1. Determine the slope of the line through the following points.
2. State the slope and y—intercept for each of the following lines:
7a) y=x+3
2b) y=—x
3. State the equations of the line in the form,y = mx+b
a) m=4,b=3 b)
4. State the equation of the following:
a) a vertical line through (—5,—3)
c) y=—2x—4
c) a horizontal line through (3,4)
b) they—axis
Cu-)
d) 4x+3y+8=0
d) thex—axis
c) m is undefined passing through (6,1)
5. Determine the equation of the line with the given slope and passing through the given point
a) m3,A(2,5) b) m=—4,A(5,—3) c) m=-1,A(lO,1)
6. Determine the equation of the line passing through the following two points.
7. Graph the following lines using slope and y—intercept.
d) ,n=&A(4,5)
41. a) m=—
3
2. a) rn=1,b=3
3. a) y=4x+3
a) x=—5
5. a) i=3r—’
(.1) =
—10b) m=—
9
—3= —
— x +
a) (6,8) and (3,4) b) (4,9) and (5,—I) c) (3,4) and (-9,4) d) (2,—l) and (2,4)
.
. a) (2,3), (4,9) b) (3,4), (6,2) c) (4,5), (—4,4) d) (4,0), (0,3)
)
a) y=2x+4 mtl,)t&
b) y=2x—2 rn\JL
) S—1
c) y=—x+3
d) y=—3x—4’
Answers:
-bc) m=0 d)mis undefined
b)
d)3 3
c) rn=—2,b=4
b) y=x—2
ii) x=O
y=—4x+17
3L
c) y=4 d) y=O
c) x=6
c) y=_tx.5
6. a) y=3 — —l0 i)
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4. Golf ub A charges $80 each time you play a round of golf. Golf ub B charges $22 but you have tobe a m ber, which costs $580 per year.
Let C doll represent the cost to play golf for a year and the number of times golf is played in oneyear. A line system that represents the costs to play olf is:
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c) Exp7mn what the solution rewesents.
/d) $‘uppose a person playc 7 dmes a year. Which club is the cheapest? Show your work.
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she choose? Explain.
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5. Golf Cours charges S80.00 per round of golf with no membe fee.Golf Course B c ges S30.O0 per round plus a $500 a ye embership fee.Let C dollars repres the cost to play golf in a ye et x represent the number of rounds of golfplayed in onc year. A un r system that repre ts the cost to play golf in a year is:
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C Ox+50
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c) What does the y-intercept mean? o1 Irkm
A tennis ball is shot into the air from a tennis cannon up on a platform. The following graph shows the ball’sheight above the ground, in metres, compared to the time from launch, in seconds.
a) What is the maximum height of the ball?
b) How many seconds after the cannon is shotdoes the ball reach its maximum height?
c) From what height above ground is thetennis ball shot?
d) After approximately how many secondsdoes the hail hit the ground?
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