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Theory-Answers

Note
Where ever u c
^ it means raise to power
/ division
* multiplication
X normal X
================================
SECTION A ANS ALL
================================
1a)
(y-1)log4^10= ylog16^10
log4^10 (y-1)= log16^y10
4^(y-1)=16y
4^y-1=4^2y
y-1=2y
-1=2y=y
-1=y
y= -y
1b)


let the actual time for 5km/hr be t 
for 4km/hr=30mint + t
4km/hr=0.5 + t
distance = 4(0.5+t)
=2*4t
for 5km/hr, time= t
distance =5t
1+4t=5t
t=2hrs
actual distance = 5*2=10km
================================
2a)
2/3(3x-5)-3/5(2x-3)=3/1
L C M =15
10(3x-5)-9(2x-3)=45
30x-50-18x+27=45
30x-18x=45+50-27
12x-23=45
12x=45+23
12x=68
x=68/12
x=34/6
x=17/3
2b)
U'aS=180-(n+88)
=180-n-88=92-n
also, u'TQ=18m
80degree + 92-n+180-m=180degree
80+92+180-n-m=180degree
352-n-m=180degree
-n-m=180-352
-n-m=-172
+(n+m)= +172
m+n=172dgree
================================
3a) 
Tan 23.6° = h/50
Cross multiply
Tan 23.6° x h/50
h = 50 tan 23.6°
= 21.844m‎
3b) 
Area of <TRU = 45cm^2 (Note: This ^ means Raise to power)
A = 1/2bh
45 = 1/2 x 10 x h
45 = 5h
h = 9cm
Area of < QTUS = 1/2 ( QT + US)h
= 1/2 ( 6 + 16)9
= 99cm^2
================================
4a)
T6=37
T6=a+(6-1)d
T6=a+5d
a+5d=37 -----(eq1)
s6=147
sn=n/2(2a+(n-1)d)
147=3(2a+5d)
49=2a+5d
2a+5d=49 ----(eq2)
a+5d=37 ---(eq1)
2a+5d=49 ---(eq2)
a=12
4b)
S15=15/2(2(12)+14d)
S15=15/2(24+14d)
from(1)
a+5d=37
12+5d=37
5d=37-12
5d=25
d=5
S15 = 15/2(24+14(15)
S15= 15/2(24+70)
S15=15/2*94
S15=15*42
S15=630
================================
5a)
draw
U=20
B= y-45
S= y-34
B=bag
S=shoe
let n(B)=y
n(S)=y+11
for bag only y-45
for shoe only y-11-45=y-34
5b)
y-45+45+y-34=120
2y-34=120
2y=154
y=154/2
y=77
number of customers who bought shoe = y+11
77+11=88
5c)
n(bag)=77customers
probability =77/120
=0.642
================================
SECTION B ANS 5 QUESTIONS ONLY
================================
10a)
Sin x = 5/13
Using pythagoras rule
M^2 = 13^2 - 5^2 (^ means Raise to power)
M^2 = 169 - 25
M ^2 = 144
M = √144
M = 12
Hence: 
Cos x - 2sin x / 2tan x
12/13 - 2(5/13) / 2(5/12)
‎= 12/13 - 10/23 / 5/6
FIND LCM
= 12 - 10/13 / 5/6
= 12/65
10bi)
Considering < LMB 
/MB/^2. = 12^2 - 9.6^2
/MB/^2 = 51.84
/MB/ = √51.84
/MB/ = 7.2m
From < AML
/LA/^2 = 2.8^2 + 9.6^2
/LA/ ^2 = 100
/LA/ = √100
/LA/ = 10m
‎10bii)
Let the angle be. θ
From <AML
Tanθ = 9.6/2.8
Tan θ = 3.4288
θ = Tan^-1 ( 3.4288)
= 73.74°
================================
13ai)
given
x(*)y=x+y/2
i)3(*)2/5=3+2/5/2
=(15+2/5)*1/2
=17/5*1/2
=17/10= 1,7/10
13aii)
8(*)y=8^1/4
=8+y/2 =33/4
32+4y=66
4y=66-32
4y=34
y=34/4
y=17/2
y=8^1/2
13b)
given DABC
AB=(^-4/6) and AC =(3/^-8)
so AP =1/2(^-4/6)
AP=(^-2/3)
hence
CP = CA + AP
CP= -(3/^8)+(^-2/3)
CP = (^-5/11).

*WELCOME TO VECTOR MATHS ANSWERS*

================================
 SECTION A ANS ALL QUESTIONS
 ================================

 1 a)
 (y -1 ) log 4 ^ 10 = ylog 16 ^ 10
 log 4 ^ 10 ( y -1 )= log 16 ^ y 10
 4 ^ ( y -1 )= 16 y
 4 ^ y -1 = 4 ^ 2 y
 y- 1 = 2 y
 -1 = 2 y= y
 -1 = y
 y= -y

 1 b )
 let the actual time for 5 km / hr be t
 for 4 km /hr = 30 mint + t
 4 km /hr =0 . 5 + t
 distance = 4 (0 . 5 + t )
 = 2 * 4 t
 for 5 km /hr , time = t
 distance =5 t
 1 + 4 t = 5 t
 t= 2 hrs
 actual distance = 5 * 2 = 10 km

 ================================

 2a) 
 2/3 (3x - 5) - 3/5 (2x - 3) = 3
 (15) x 2/3 (3x - 5) - 3/5 (15) (2x - 3) = 3(15)
 10(3x -5) - 9(2x - 3) = 45
 30x - 50 - 18x + 27 = 45. 
 12x - 23 = 45
 12x = 45+23
 12x = 68
 X = 68/12
 X = 5.67

 2b)
 80 = n+r ( ext. < equal sum of opp int. <s )
  80 = n+r ------------ (i)
  <UQT = 180 - 88 - n = 92 - n
  M = 80 + 92 - n
  M = 172 - n ----------- (ii)
 80 + 92 - n + 180 - m = 180°‎
80 + 92 + 180 - n - m = 180°‎
352 - n - m = 180
  -n - m = 180 - 352‎
 - n - m = - 172°‎
   m + n = 172°‎

===============================

3a) ‎
Tan 23.6° = h/50
  Cross multiply
 Tan 23.6° x h/50
 h = 50 tan 23.6°
  = 21.844m‎
 ‎aprox. 22m
‎
3b) 
 Area of <TRU = 45cm^2 (Note: This ^ means Raise to power)
 A = 1/2bh
  45 = 1/2 x 10 x h
  45 = 5h
  h = 9cm
 Area of < QTUS = 1/2 ( QT + US)h
  = 1/2 ( 6 + 16)9
  = 99cm^2

 ================================

 4 a)
 T 6 =37
 T 6 =a + ( 6 - 1 )d
 T 6 =a + 5 d
 a + 5 d = 37 - --- -( eq1 )
 s 6 = 147
 sn = n / 2 (2 a + (n -1 ) d )
 147 = 3 (2 a + 5 d )
 49 = 2 a + 5 d
 2 a+ 5 d = 49 -- -- (eq 2 )
 a + 5 d = 37 - --( eq1 )
 2 a+ 5 d = 49 -- -( eq2 )
 a =12

 4 b )
 S 15 = 15 /2 ( 2 (12 )+ 14 d )
 S 15 = 15 /2 ( 24 + 14 d )
 from(1 )
 a + 5 d = 37
 12 + 5 d = 37
 5 d =37 -12
 5 d =25
 d =5
 S 15 = 15 /2 ( 24 + 14 (15 )
 S 15 = 15 / 2 (24 + 70 )
 S 15 = 15 /2 * 94
 S 15 = 15 * 42
 S 15 = 630

 ================================

 5a)
 Let bag=B
 Shoe= S
 U=120
 n(BnS)=45, n(s)=x+11, n(b)=x
 n(SnB')=x+11-45
 =x-34
 n(BnS') = 45

 5b)
 Y - 45 + 45 + Y - 34 = 120
 2Y - 34 = 120
 2Y = 120 - 34
 2Y = 154
 Y = 154/2
 Y = 77
 11+x=77+11
 = 88
 Therefor 88 bought shoes costumer

 5c)
 n(bag)= 77 customers
 Pr. =77/120


 ================================
 SECTION B ANS 5 QUESTIONS ONLY
 ================================

 10) 
 Sin x = 5/13
 Using pythagoras rule
  M^2 = 13^2 - 5^2 (^ means Raise to power)
  M^2 = 169 - 25
  M ^2 = 144
  M = √144
  M = 12
 Hence: 
 Cos x - 2sin x / 2tan x
 12/13 - 2(5/13) / 2(5/12)
‎= 12/13 - 10/23 / 5/6
  FIND LCM
  = 12 - 10/13 / 5/6
  =  12/65

 10b) 
 Draw a triangle LACB
 in triangle LCB
 Hyp^2 = Opp^2 + Adj^2
 12^2 = 9.6^2 + |CB|^2
 144 = 92.16 = |CB|^2
 144 – 92.16 = |CB|^2
 51.84 = |CB|^2
 therefore, |CB| = √51.84
 |CB| = 7.2m
 |AC| + |CB| =|AB|
 |AC| + 7.2m = 10m
 |AC| = 10m – 7.2m

 |AC| = 2.8m
 In triangle LCA
 Hyp^2 = Opp^2 + Adj^2
 |LA|^2 = |AC|^2 + |LC|^2

 |LA|^2 = 2.8^2 + 9.6^2

 |LA|^2 = 7.84 + 92.16

 |LA|^2 =100

 |LA| = √100

 |LA| = 10m

 10bii) 
 in triangle LCA
 sinθ = Opp/Hyp
 sinθ = |LC|/|LA|
 sinθ = 9.6/10
 sinθ = 0.96
θ = sin^-1 (0.96)
θ = 73.74

 ===============================

 13ai)
 given
 x(*)y=x+y/2
 i)3(*)2/5=3+2/5/2
 =(15+2/5)*1/2
 =17/5*1/2
 =17/10= 1,7/10

 13aii)
 8(*)y=8^1/4
 =8+y/2 =33/4
 32+4y=66
 4y=66-32
 4y=34
 y=34/4
 y=17/2
 y=8^1/2

 13b)
 given 
 AB=(^-4/6) and AC =(3/^-8)
 so AP =1/2(^-4/6)
 AP=(^-2/3)
 hence
 CP = CA + AP
 CP= -(3/^8)+(^-2/3)

 CP = (^-5/11)

12a) Using completing the square method
3y^2-5y+2=0
y^2 - 5/3y + 2/3=0
y^2-5/3y=-2/3
y^2-5/3y+(^-5/6)^2=(-^5/6)^2-2/3
(y-5/6)^2=25/36-2/3
(y-5/6)^2=25/-24/36
(y-5/6)^2=1/36
(y-5/6)=+sqr1/36
y=5/6+1/6
y=5+1/6 or 5-1/6
y=6/6 or 2/3
y=1 or 2/3

12b)
given
M N = [2,3 1,4]
hence
[1,4 2,3] * [m,n x,y] =[2,3 1,4]
[m+2n, x*2y]
[4m+3n, 4x+5y] = [2,3, 1,4]
therefore
m+2n=2------(i)
4m+3n=3------(ii)
from ------(i)
m=2-2n
4(2-2n)+3n=3
8-8n+3n=3
8-5n=3
8-3=5n
5=5n
n=1
hence
m=2-2(1)
M=0
also
x+2y=1------(i)
4x+3y=4------(ii)
from ------(iii)
x=1-2y
4(1-2y)+3y=4
4-8y+3y=4
y=0
therefore x=1-2(0)
x=1

this N=[i i]