*THEORY COMPLETED*✅✅✅
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Maths Obj
1-10: ACBCDDCBAA
11-20: CDCABCCCAC
21-30: DADBABCDAD
31-40: BADADBCACB
41-50: ADDDBDADCA
NO1) On February 28th 2012, value = (100-30/100) * #900,00.00
= 70/100 * #900,00
= #630,000.00
On february 28th 2013, value = (100-22/1000 * #630,00
= 78/100 8 #630,000
= #491,400
On february 28th 2014, value = 78/100 8 #491,400
=383,292
On february 28th 2015, value = 78/100 * #383,292
= #298,967.76
: (3a)Using Pythagoras theorem let the side of the rhombus = x
x^2= (10.2/2)^2 + (9.3/2)^2
x^2= 5.1^2 + 4.65^2
x^2= 26.01 + 21.62
x^2= 47.63
x= sqroot of 47.63
x= 6.9
Sides= 6.9cm
But perimeter of rhombus = 4x
Perimeter = 4x6.9
Perimeter = 27.6cm
(3b) Sin x= 3/5
Opp= 3, Hyp= 5
Let adjacent = x
From Pythagoras theoren
5^2= 3^2+x^2
25= 9+x^2
x^2= 25-9
x^2= 16
x= sqroot of 16
x= 4
5cosx-4tanx
=4 - 3= 1
5cosx-4tanx= 1
: (10a) Using Pythagoras theorem from SPQ
|SQ|^2 = 12^2 + 5^2
= 144+25
=169
SQ= sqroot of 169
= 13cm
Sin tita= 5/13 = 0.3846
Tita= Sin^-1(0.3846)
= 22.6degrees
From PRQ
Sin tita= |PR|/12
Sin 22.6 = PR/12
Sin 22.6= PR/12
PR= 12xsin 22.6
PR= 12x0.3843
PR= 4.61cm
(10bii)Let the height at which m touches the wall= y
Cos x^degrees= 8/10= 0.8
x^degrees= Cos^-1(0.8)
= 36.87degrees
Sin x^degrees = y/12
Sin 36.87= y/12
y= 12xsin36.87
y= 12x0.60000
y= 7.2m:
(8a)
Cost price for Lami= #300.00
Profit made by lami = x%
Ie selling price for lami=(100+x/100)×#300
=#3(100+x)
=#(300+3x)
Bola's cost price = #3(100+x)
Profit made by bola =x%
Selling price for bola =(100+x/100)×#3(100+x)
=#3/100(100+x)²
James cost price =#3/100(100+x)²=300+(6x+3/4)
expanding;
3/100(10000+200+x²) = 300+3/4+6x
3(10000+200x+x²)=30000+75+600x
30000+600x+3x²=30000+75+600x
3x²=75
X² = 75/3
X² = 25
X = square root 25
X = 5
(8b)
3x-2<10+x<2+5x
3x-2<10+x & 10+x<2+5x
3x-x<10+2 & 10-2<5x-x
2x<12 8<4x
X<12/2 4x>8
X<6 x>8/4
X>2
Also; 3x-2<2+5x
-4<2x
2x > -4
X > -2
Therefore; Range is -2<X<6
=======================
[NO2) Given that y = 2pxˆ² – p² x – 14
AT (3, 10)
10 = 2p(3)² – p² (3) – 14
10 = 18p – 3p² – 14
3p² – 18p + 24 = 0
p² – 6p + 8 = 0
using factor method,
p² – 2p -4p + 8 = 0
p(p-2) – 4(p-2) = 0
(p-4)(p-2) = 0
p-4 = 0 or p-4 = 0
p= 4 or p =2
2b) The lines must be solved simultenously
3y – 2x = 21 ——- (1)
4y + 5x = 5 ——-(2)
using elimination method,
(4) 3y – 2x = 21
(30 4y + 5x = 5
12y – 8y = 84 ——— (3)
12y + 15x = 15 ——-(4)
equ (4) minus equ(3)
23x = -69
x = -69/23
x = -3
Put this into equation (1)
3y -2(-3) = 21
3y = 6 = 21
3y = 21 -6
3y = 15
y =15/3
y = 5
coordinates of Q is (-3, 5)
: (3a)
The diagonal = 10.2m and 9.3cm
Using Pythagoras theory
Ac² = 10.2² + 9-3²
Ac² = 104.04 + 86.49
Ac² = 190.53
Ac² = √190.53
Ac² = 13.80
(3b)
DRAW THE DIAGRAM
Using Pythagoras theory
5² = 3² + x²
x² = 5² - 3²
X²= 25 - 9
X² = √16
X= 4cm
CosX = adjacent/hyp
= 4/5
Tan X = opp/adj. = 3/4
5cos x - 4tan x
5(4/5)- 4(3/4)
20/5 - 12/4
4-3= 1
= 70/100 * #900,00
= #630,000.00
On february 28th 2013, value = (100-22/1000 * #630,00
= 78/100 8 #630,000
= #491,400
On february 28th 2014, value = 78/100 8 #491,400
=383,292
On february 28th 2015, value = 78/100 * #383,292
= #298,967.76
: (3a)Using Pythagoras theorem let the side of the rhombus = x
x^2= (10.2/2)^2 + (9.3/2)^2
x^2= 5.1^2 + 4.65^2
x^2= 26.01 + 21.62
x^2= 47.63
x= sqroot of 47.63
x= 6.9
Sides= 6.9cm
But perimeter of rhombus = 4x
Perimeter = 4x6.9
Perimeter = 27.6cm
(3b) Sin x= 3/5
Opp= 3, Hyp= 5
Let adjacent = x
From Pythagoras theoren
5^2= 3^2+x^2
25= 9+x^2
x^2= 25-9
x^2= 16
x= sqroot of 16
x= 4
5cosx-4tanx
=4 - 3= 1
5cosx-4tanx= 1
: (10a) Using Pythagoras theorem from SPQ
|SQ|^2 = 12^2 + 5^2
= 144+25
=169
SQ= sqroot of 169
= 13cm
Sin tita= 5/13 = 0.3846
Tita= Sin^-1(0.3846)
= 22.6degrees
From PRQ
Sin tita= |PR|/12
Sin 22.6 = PR/12
Sin 22.6= PR/12
PR= 12xsin 22.6
PR= 12x0.3843
PR= 4.61cm
(10bii)Let the height at which m touches the wall= y
Cos x^degrees= 8/10= 0.8
x^degrees= Cos^-1(0.8)
= 36.87degrees
Sin x^degrees = y/12
Sin 36.87= y/12
y= 12xsin36.87
y= 12x0.60000
y= 7.2m:
(8a)
Cost price for Lami= #300.00
Profit made by lami = x%
Ie selling price for lami=(100+x/100)×#300
=#3(100+x)
=#(300+3x)
Bola's cost price = #3(100+x)
Profit made by bola =x%
Selling price for bola =(100+x/100)×#3(100+x)
=#3/100(100+x)²
James cost price =#3/100(100+x)²=300+(6x+3/4)
expanding;
3/100(10000+200+x²) = 300+3/4+6x
3(10000+200x+x²)=30000+75+600x
30000+600x+3x²=30000+75+600x
3x²=75
X² = 75/3
X² = 25
X = square root 25
X = 5
(8b)
3x-2<10+x<2+5x
3x-2<10+x & 10+x<2+5x
3x-x<10+2 & 10-2<5x-x
2x<12 8<4x
X<12/2 4x>8
X<6 x>8/4
X>2
Also; 3x-2<2+5x
-4<2x
2x > -4
X > -2
Therefore; Range is -2<X<6
=======================
[NO2) Given that y = 2pxˆ² – p² x – 14
AT (3, 10)
10 = 2p(3)² – p² (3) – 14
10 = 18p – 3p² – 14
3p² – 18p + 24 = 0
p² – 6p + 8 = 0
using factor method,
p² – 2p -4p + 8 = 0
p(p-2) – 4(p-2) = 0
(p-4)(p-2) = 0
p-4 = 0 or p-4 = 0
p= 4 or p =2
2b) The lines must be solved simultenously
3y – 2x = 21 ——- (1)
4y + 5x = 5 ——-(2)
using elimination method,
(4) 3y – 2x = 21
(30 4y + 5x = 5
12y – 8y = 84 ——— (3)
12y + 15x = 15 ——-(4)
equ (4) minus equ(3)
23x = -69
x = -69/23
x = -3
Put this into equation (1)
3y -2(-3) = 21
3y = 6 = 21
3y = 21 -6
3y = 15
y =15/3
y = 5
coordinates of Q is (-3, 5)
: (3a)
The diagonal = 10.2m and 9.3cm
Using Pythagoras theory
Ac² = 10.2² + 9-3²
Ac² = 104.04 + 86.49
Ac² = 190.53
Ac² = √190.53
Ac² = 13.80
(3b)
DRAW THE DIAGRAM
Using Pythagoras theory
5² = 3² + x²
x² = 5² - 3²
X²= 25 - 9
X² = √16
X= 4cm
CosX = adjacent/hyp
= 4/5
Tan X = opp/adj. = 3/4
5cos x - 4tan x
5(4/5)- 4(3/4)
20/5 - 12/4
4-3= 1
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