Main PageNabteb: MATHEMATICS obj and Theory Answer available here
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1a). 36*(1/6)^n= 6*216^n-1 6^2 *6^-n=6*6^3(n-1) 6^2-n= 6^1*6^3n-3
6^2-n= 6^1 3n-3 6^2-n= 6^3n-2 2-n = 3n-2 4n= 4
Divide both side by 4 n= 1
1b). Logbase3(2x 1)-log3(x-3)=2 Logbase3 2x 1/x-3=log9 base 3 2x 1/x-3=9 2x 1=9(x-3) 2x 1=9x-27 9x-2x=27 1 7x=28 X=28/7 X=4
2a) 4.56*3.6/0.12 =456*10^-2*36*10^-1/12*10^-2 =456*36*10^-1 =1368*10^-1
2b)123base n= 81 1*n^2 2*n^1 3*n^o= 81 n^2 2n 3= 81 n^2 2n-78=0 ax2 bx c = 0 n=-b�√(b^2-4ac)/2a n= -2�√(2)^2-4(1)(-78)/2 n= -2�√4 312/2 n=-2�√316/2 n=-2�17.76/2 n=-2 17.76/2 or =-2-17.76/2 =15.76/2 or -19.76 =7.88 or -9.88 n≈ 8
3a). Housing = 900 Wears= 1200 Savings= 1050 Drinks = 850 Total= 900 1200 1050 850= 4000 Housing = 900*360/4000= 81� Wears = 1200*360/4000= 108� Savings = 1050*360/4000=94.5� Drinks = 850*360/4000= 76.5�
DRAW THE CIRCLE AND INSERT THE ANSWER WITH YOUR PROTRACTOR
3bi). Proportion (wears) = 1200/4000= 6/20= 3/10 Proportion (savings) = 1050/4000 = 21/80
4i). P(50�s,90�w), and Q (50�s,90�E) Distance along the circle latitude = Ѳ/360*2πRcos Ѳ =(90 90) 222*6400*cos50/360*7 = 180*2*22*6400*0.6428/2520 =32581618.36/2520 =12929.21km
4ii). distance along the great circle = Ѳ/360*2πR =(90 90)*2*22*6400/7 = 180/360 226400/7 = 20,114.28km
5a). Given P= 2i 3j, q= i-2j and r = 4i 3j 2p q 2(2i 3j) i-2j =4i 6j i-2j =5i 4j P q r =2i 3j i-2j 4i 3j =2i i 4i 3j-2j 3j =7i 4j 3q- 2r 3(i-2j)-2(4i 3j) =3i-6j-8i-6j =3i-8i-6j-6j =-5i -12j 4p�3q-2 = 4(2i 3j)-3(i-2j)-2 =8i 12j-3i 6j-2 =8i-3i 12j 6j-2 =5i 18j-2
6a). 16*8^x=32*4^2x-1 2^4*2^3x=2^5*2^2(2x-1) 2^4 3x= 2^5*2^4x-2 2^4 3x=2^5 4x-2 2^4 3x= 2^3 4x 4 3x=3 4x 4x-3x= 4-3 X=1
6b). DRAW THE DIAGRAM |AC|^2 = (2)^2 (3)^2 =4 9 = √13= 3.61km The bearing of C from A Sin Ѳ= 2/3.61 Ѳ= sin^-1(0.554)=33.6 sin Ѳ= 3/3.61 Ѳ= 56.2�
The bearing of C from A = 146�
8). 1 . 2 . 3. 4 . 5 1.(1,1). (1,2) .(1,3) .(1,4) .(1,5) 2. .(2,1). (2,2) .(2,3) .(2,4) .(2,5) 3 .( 3,1). ( 3,2) .(3,3) .(3,4) .( 3,5) 4.(4,1). (4,2) .(4,3) .(4,4) .(4,5) 5.(5,1). (5,2) .(5,3) .(5,4) .(5,5)
Total outcome = 36 The probability of obtaining 5 = 4/36 = 2/18= 1/9 Probability that a multiple of 3 is obtained = 12/36= 1/3 Probability that equal likely numbers is obtained =6/36 =1/6
8b). By considering the chord MN |OS|= distance from the centre |os|^2 = |om|^2-|ms|^2 = (12)^2- (15)^2 =144-25 |os|= √119= 10.9cm By considering the chord PQ | OR| = distance from the centre |OR|^2= |OP|^2- |PR|^2 =(12)^2 � (4)^2 =144-16= 128 |OR|= √128= 11.3cm
8c). q^2= p^2 r^2-2prcosѲ =(3)^2 (4)^2-2(3)(4)cos40 =9 16-24(-0.7660) =25 18.384 q^2 = 43.384 q^2= √43.384= 6.59km ≈6.6km
9a). 44n 55n =121n 4*n^1 4*n^0 5*n^1 5*n^0=1*n^2 2*n^1 1*n^0 4n 4 5n 5=n^2 2n 1 9n 9=N^2 2n 1 9n-2n 9-1=n^2 7n 8=n^2 n^2-7n-8=0 ax2 bx c = 0 n=-b�√(b^2-4ac)/2a when a=1, b=7, c=-8 =-7�√(7^2-4(1)(-8)/2a =-7�√49 32/2 =-7�√81/2 -7�9/2 = -7 9/2 = 2/2 =1 or -7-9/2 = -16/2 =-8 Since n can not be negative n= 1
9b). X,0,1,2,3,4,5,6 F,26,29,14,15,5,6,3=98 Fx,0,29,28,45,20,30,18=170
(x-x) ,-1.7,-0.7,-0.3.-1.3,-2.3,-3.3,-4.3
(x-x) ^2,2.89,0.49,0.09,1.69,5.29,10.89,18.49 F(x-x) ^2,75.13,14.21,1.26,25.35,26.45,65.34,55.47=263.22
Mean = �fx/�f= 170/98= 1.735 = 1.74
Mode . the modal score is 1 because it has highest number of frequency which is 29.
Standard deviation =√�f(x-x)^2/ �f =√263.22/98= √2.6859 =1.6389
10a). h=5t^2-20t 25=5t^2-20t 5t^2/5-20t/5-25/5=0 t^2-4t-5t=0 ax2 bx c = 0 -b�√(b^2-4ac)/2a , where a= 1,b=-4,c=-5 -(-4)�√(4^2-4(1)(-5)/2(1) =4�√16 20/2 =4�√36/2 =4�6/2=4 6/2=10/2=5 Or 4-6/2=-2/2=-1 :. t= 5 or -1 Since t cannot be negative t :. t = 5sec.
10b). Sum of interior angle = 1260 Sum of interior angle = (n-2) 180= 1260 n-2=1260/180 n-2=7 n=7 2=9 n=9 180(n-2)/n =180(9-2)/9 =180*7/9=140� Ext int = 180� Ext 140=180� Ext= 180-140= 40 Ext =40 The size of exterior angle = 9(40) =360� 10c. 3/4(x-3)≥9/2 =6(x-3)≥4*9 6x-18≥36 6x≥36 18 6x≥54==== x=54/6=9, x≥9
11a). R∞L R=KL 25=12L K=25/12 :.R=25/12L----this is the equation connecting R∞L together
11aii). R=25/12L Given R=40 ohms L=? L=12R/25=12*40/25 =19.2M
11b). Tbase4=a 3d=13--------1 Tbase2 = a d=3---------2 a 3d=13 -a-d=-3 2d=10 Divide bothside by 2 2d/2=10/2 d=5 from (ii) a d=3 a 5=3 a=3-5 a=-2 the first term =-2 and common difference =5 fifth term T base5= a 4d =-2 4(5) = -2 20= 18 T base5 = 18
11c). 3^x-2*9^x 1*27^-x/81^5x-3 = 3^x-2*3^2(x 1)*3^-3x/3^ 4(5x-3) =3^x-2*3^2x 2*3^-3x/3^20x-12=3^x-2 2x 2-3x/3^20x-10 =3^0/3^20x-12=1/3^20x-12=3^12-20x
NOTE * means multiply X,/ means divide ,-Minus,^ means raise to power
1a). 36*(1/6)^n= 6*216^n-1 6^2 *6^-n=6*6^3(n-1) 6^2-n= 6^1*6^3n-3
6^2-n= 6^1 3n-3 6^2-n= 6^3n-2 2-n = 3n-2 4n= 4
Divide both side by 4 n= 1
1b). Logbase3(2x 1)-log3(x-3)=2 Logbase3 2x 1/x-3=log9 base 3 2x 1/x-3=9 2x 1=9(x-3) 2x 1=9x-27 9x-2x=27 1 7x=28 X=28/7 X=4
2a) 4.56*3.6/0.12 =456*10^-2*36*10^-1/12*10^-2 =456*36*10^-1 =1368*10^-1
2b)123base n= 81 1*n^2 2*n^1 3*n^o= 81 n^2 2n 3= 81 n^2 2n-78=0 ax2 bx c = 0 n=-b�√(b^2-4ac)/2a n= -2�√(2)^2-4(1)(-78)/2 n= -2�√4 312/2 n=-2�√316/2 n=-2�17.76/2 n=-2 17.76/2 or =-2-17.76/2 =15.76/2 or -19.76 =7.88 or -9.88 n≈ 8
3a). Housing = 900 Wears= 1200 Savings= 1050 Drinks = 850 Total= 900 1200 1050 850= 4000 Housing = 900*360/4000= 81� Wears = 1200*360/4000= 108� Savings = 1050*360/4000=94.5� Drinks = 850*360/4000= 76.5�
DRAW THE CIRCLE AND INSERT THE ANSWER WITH YOUR PROTRACTOR
3bi). Proportion (wears) = 1200/4000= 6/20= 3/10 Proportion (savings) = 1050/4000 = 21/80
4i). P(50�s,90�w), and Q (50�s,90�E) Distance along the circle latitude = Ѳ/360*2πRcos Ѳ =(90 90) 222*6400*cos50/360*7 = 180*2*22*6400*0.6428/2520 =32581618.36/2520 =12929.21km
4ii). distance along the great circle = Ѳ/360*2πR =(90 90)*2*22*6400/7 = 180/360 226400/7 = 20,114.28km
5a). Given P= 2i 3j, q= i-2j and r = 4i 3j 2p q 2(2i 3j) i-2j =4i 6j i-2j =5i 4j P q r =2i 3j i-2j 4i 3j =2i i 4i 3j-2j 3j =7i 4j 3q- 2r 3(i-2j)-2(4i 3j) =3i-6j-8i-6j =3i-8i-6j-6j =-5i -12j 4p�3q-2 = 4(2i 3j)-3(i-2j)-2 =8i 12j-3i 6j-2 =8i-3i 12j 6j-2 =5i 18j-2
6a). 16*8^x=32*4^2x-1 2^4*2^3x=2^5*2^2(2x-1) 2^4 3x= 2^5*2^4x-2 2^4 3x=2^5 4x-2 2^4 3x= 2^3 4x 4 3x=3 4x 4x-3x= 4-3 X=1
6b). DRAW THE DIAGRAM |AC|^2 = (2)^2 (3)^2 =4 9 = √13= 3.61km The bearing of C from A Sin Ѳ= 2/3.61 Ѳ= sin^-1(0.554)=33.6 sin Ѳ= 3/3.61 Ѳ= 56.2�
The bearing of C from A = 146�
8). 1 . 2 . 3. 4 . 5 1.(1,1). (1,2) .(1,3) .(1,4) .(1,5) 2. .(2,1). (2,2) .(2,3) .(2,4) .(2,5) 3 .( 3,1). ( 3,2) .(3,3) .(3,4) .( 3,5) 4.(4,1). (4,2) .(4,3) .(4,4) .(4,5) 5.(5,1). (5,2) .(5,3) .(5,4) .(5,5)
Total outcome = 36 The probability of obtaining 5 = 4/36 = 2/18= 1/9 Probability that a multiple of 3 is obtained = 12/36= 1/3 Probability that equal likely numbers is obtained =6/36 =1/6
8b). By considering the chord MN |OS|= distance from the centre |os|^2 = |om|^2-|ms|^2 = (12)^2- (15)^2 =144-25 |os|= √119= 10.9cm By considering the chord PQ | OR| = distance from the centre |OR|^2= |OP|^2- |PR|^2 =(12)^2 � (4)^2 =144-16= 128 |OR|= √128= 11.3cm
8c). q^2= p^2 r^2-2prcosѲ =(3)^2 (4)^2-2(3)(4)cos40 =9 16-24(-0.7660) =25 18.384 q^2 = 43.384 q^2= √43.384= 6.59km ≈6.6km
9a). 44n 55n =121n 4*n^1 4*n^0 5*n^1 5*n^0=1*n^2 2*n^1 1*n^0 4n 4 5n 5=n^2 2n 1 9n 9=N^2 2n 1 9n-2n 9-1=n^2 7n 8=n^2 n^2-7n-8=0 ax2 bx c = 0 n=-b�√(b^2-4ac)/2a when a=1, b=7, c=-8 =-7�√(7^2-4(1)(-8)/2a =-7�√49 32/2 =-7�√81/2 -7�9/2 = -7 9/2 = 2/2 =1 or -7-9/2 = -16/2 =-8 Since n can not be negative n= 1
9b). X,0,1,2,3,4,5,6 F,26,29,14,15,5,6,3=98 Fx,0,29,28,45,20,30,18=170
(x-x) ,-1.7,-0.7,-0.3.-1.3,-2.3,-3.3,-4.3
(x-x) ^2,2.89,0.49,0.09,1.69,5.29,10.89,18.49 F(x-x) ^2,75.13,14.21,1.26,25.35,26.45,65.34,55.47=263.22
Mean = �fx/�f= 170/98= 1.735 = 1.74
Mode . the modal score is 1 because it has highest number of frequency which is 29.
Standard deviation =√�f(x-x)^2/ �f =√263.22/98= √2.6859 =1.6389
10a). h=5t^2-20t 25=5t^2-20t 5t^2/5-20t/5-25/5=0 t^2-4t-5t=0 ax2 bx c = 0 -b�√(b^2-4ac)/2a , where a= 1,b=-4,c=-5 -(-4)�√(4^2-4(1)(-5)/2(1) =4�√16 20/2 =4�√36/2 =4�6/2=4 6/2=10/2=5 Or 4-6/2=-2/2=-1 :. t= 5 or -1 Since t cannot be negative t :. t = 5sec.
10b). Sum of interior angle = 1260 Sum of interior angle = (n-2) 180= 1260 n-2=1260/180 n-2=7 n=7 2=9 n=9 180(n-2)/n =180(9-2)/9 =180*7/9=140� Ext int = 180� Ext 140=180� Ext= 180-140= 40 Ext =40 The size of exterior angle = 9(40) =360� 10c. 3/4(x-3)≥9/2 =6(x-3)≥4*9 6x-18≥36 6x≥36 18 6x≥54==== x=54/6=9, x≥9
11a). R∞L R=KL 25=12L K=25/12 :.R=25/12L----this is the equation connecting R∞L together
11aii). R=25/12L Given R=40 ohms L=? L=12R/25=12*40/25 =19.2M
11b). Tbase4=a 3d=13--------1 Tbase2 = a d=3---------2 a 3d=13 -a-d=-3 2d=10 Divide bothside by 2 2d/2=10/2 d=5 from (ii) a d=3 a 5=3 a=3-5 a=-2 the first term =-2 and common difference =5 fifth term T base5= a 4d =-2 4(5) = -2 20= 18 T base5 = 18
11c). 3^x-2*9^x 1*27^-x/81^5x-3 = 3^x-2*3^2(x 1)*3^-3x/3^ 4(5x-3) =3^x-2*3^2x 2*3^-3x/3^20x-12=3^x-2 2x 2-3x/3^20x-10 =3^0/3^20x-12=1/3^20x-12=3^12-20x
NOTE * means multiply X,/ means divide ,-Minus,^ means raise to power
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