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MAT 101
NAME: ADEYI HANNAH OLUWAKEMI
COLLEGE: ENGINEERING
DEPARTMENT: MECHATRONICS ENGINEERING
QUESTIONS
(1)
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘šπ‘–π‘‘π‘‘π‘™π‘’ π‘‘π‘’π‘Ÿπ‘š π‘œπ‘“ (2𝑝 βˆ’
1 10
2π‘ž
)
(b)
πΉπ‘œπ‘Ÿ π‘€β„Žπ‘Žπ‘‘ π‘£π‘Žπ‘™π‘’π‘’ π‘œπ‘“ 𝑛 π‘‘π‘œπ‘’π‘  3 . (𝑛+1
) = 7 . (𝑛2)?
3
(2)
1 5
𝐸π‘₯π‘π‘Žπ‘›π‘‘ (𝑐 βˆ’ ) 𝑒𝑠𝑖𝑛𝑔 π‘‘β„Žπ‘’ π‘π‘–π‘›π‘œπ‘šπ‘–π‘Žπ‘™ π‘ π‘’π‘Ÿπ‘–π‘’π‘ 
(3)
𝐸π‘₯π‘π‘Žπ‘›π‘‘
(4)
𝐸π‘₯π‘π‘Žπ‘›π‘‘
1
(4βˆ’π‘₯)2
1
𝑐
𝑖𝑛 π‘Žπ‘ π‘π‘’π‘›π‘‘π‘–π‘›π‘” π‘π‘œπ‘€π‘’π‘Ÿπ‘  π‘œπ‘“ π‘₯ 𝑒𝑝 π‘‘π‘œ π‘‘π‘’π‘Ÿπ‘š 𝑖𝑛 π‘₯ 3 , 𝑒𝑠𝑖𝑛𝑔 π‘‘β„Ž
√(1βˆ’2𝑑)
𝑖𝑛 π‘Žπ‘ π‘π‘’π‘›π‘‘π‘–π‘›π‘” π‘π‘œπ‘€π‘’π‘Ÿπ‘  π‘œπ‘“ 𝑑 𝑒𝑝 π‘‘π‘œ π‘‘π‘’π‘Ÿπ‘š 𝑖𝑛 𝑑 3 , 𝑒𝑠𝑖𝑛𝑔 π‘‘β„Ž
π΅π‘–π‘›π‘œπ‘šπ‘–π‘Žπ‘™ π‘‘β„Žπ‘’π‘œπ‘Ÿπ‘’π‘š. π‘†π‘‘π‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘™π‘–π‘šπ‘–π‘‘π‘  π‘œπ‘“ 𝑑 π‘“π‘œπ‘Ÿ π‘€β„Žπ‘–π‘β„Ž π‘‘β„Žπ‘’ 𝑒π‘₯π‘π‘Ÿπ‘’π‘ π‘ π‘–π‘œπ‘› 𝑖𝑠 π‘£π‘Žπ‘™π‘–π‘‘
(5)
(6)
πœ‹
π‘ƒπ‘Ÿπ‘œπ‘£π‘’ π‘‘β„Žπ‘Žπ‘‘ cos(𝑦 βˆ’ πœ‹) + sin (𝑦 + ) = 0
2
πœ‹
πœ‹
4
4
π‘œ)
π‘†β„Žπ‘œπ‘€ π‘‘β„Žπ‘Žπ‘‘ tan (π‘₯ + ) tan (π‘₯ βˆ’ ) = βˆ’1
(7)
π‘†π‘œπ‘™π‘£π‘’ π‘‘β„Žπ‘’ π‘’π‘žπ‘’π‘Žπ‘‘π‘–π‘œπ‘› 4 sin(π‘₯ βˆ’ 20 = 5π‘π‘œπ‘ π‘₯ .
(8)
𝐺𝑖𝑣𝑒𝑛 cos 𝐴 = 0.42 π‘Žπ‘›π‘‘ sin 𝐡 =
0.73, π‘’π‘£π‘Žπ‘™π‘’π‘Žπ‘‘π‘’ (π‘Ž) sin(𝐴 βˆ’ 𝐡), (𝑏) cos(𝐴 βˆ’ 𝐡), (c) tan(𝐴 +
𝐡)π‘π‘œπ‘Ÿπ‘Ÿπ‘’π‘π‘‘ π‘‘π‘œ 4 π‘‘π‘’π‘π‘–π‘šπ‘Žπ‘™ π‘π‘™π‘Žπ‘π‘’π‘ .
(9)
𝐴𝑛 π‘œπ‘–π‘™ π‘π‘œπ‘šπ‘π‘Žπ‘›π‘¦ π‘π‘œπ‘Ÿπ‘’π‘  π‘Ž β„Žπ‘œπ‘™π‘’ 80π‘š 𝑑𝑒𝑒𝑝. πΈπ‘ π‘‘π‘–π‘šπ‘Žπ‘‘π‘’ π‘‘β„Žπ‘’ π‘π‘œπ‘ π‘‘ π‘œπ‘“ π‘π‘œπ‘Ÿπ‘–π‘›
π‘“π‘œπ‘Ÿ π‘‘π‘Ÿπ‘–π‘™π‘™π‘–π‘›π‘” π‘‘β„Žπ‘’ π‘“π‘–π‘Ÿπ‘ π‘‘ π‘šπ‘’π‘‘π‘Ÿπ‘’ π‘€π‘–π‘‘β„Ž π‘Žπ‘› π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’ 𝑖𝑛 π‘π‘œπ‘ π‘‘ π‘œπ‘“ £2 π‘π‘’π‘Ÿ π‘šπ‘’π‘‘π‘Ÿπ‘’ π‘“π‘œπ‘Ÿ π‘’π‘Žπ‘
(10)
𝐹𝑖𝑛𝑑 π‘‘β„Žπ‘’ π‘ π‘’π‘š π‘œπ‘“ π‘Žπ‘™π‘™ π‘‘β„Žπ‘’ π‘›π‘’π‘šπ‘π‘’π‘Ÿπ‘  𝑏𝑒𝑑𝑀𝑒𝑒𝑛 5 π‘Žπ‘›π‘‘ 250 π‘€β„Žπ‘–π‘β„Ž π‘Žπ‘Ÿπ‘’ 𝑒π‘₯π‘Ž
ANSWERS
No1a) (2p-1/2q)10
=[10(9)(8)(7)(6)/5! x (2p)10-5 x (-1/2q)5]
=10(9)(8)(7)(6)/5! x (2p)5 x (-1/2q)5
=252 x (32p5) x (-1/32q5)
=252p5/q5 ans
No1b) 3X(n+13) = 7X(n2)
3 x (n + 6)!/ (n-2)!3! = 7x n!/(n-2)!2!
3 x (n + 6)/6 = 7/2
6(n+6)=42, n+6 =7, n= 7-6, n= 1 ANS
No2) (c – 1/c)5
=(1*c5*c0) + (5*c4*c-1) +(10*c3*c-2) + (10*c2*c-3)+(c5*c1*c4)+(1*c0*c-5)
=c5+5c3+10c+(10/c)+(5/c3)+(1/c5) ANS
No 3)
1/(4-X)2
=(4-X)-2
Using (a+x)n = an + nan-1x + (n-1)/2! * an-2 x2 +{ n(n-1)(n-2)an-3x3}/3!
= 4-2 + (-2)(4)-2-1(-x) + (-2-1)/2! * 4-2-2 (-x2)+{ -2(-2-1)(-2-2)/3! * 4-2-3
(x)3}
=1/16 + x/32 + (3x2/256) + (x3/256) ANS
No 4) Using (a+x)n = an + nan-1x + (n-1)/2! * an-2 x2 +{ n(n-1)(n-2)an3x3}/3!
= 1-1/2 + -1/2(1)-1/2-1 (-2t) + -1/2(-1/2-1)/2! * 1-1/2-2 (-2t)2 +{ -1/2(1/2-1)(-1/2-
2)/3! * 1-1/2-3 (-2t)3}
=1 + t + (3/8 * 1 * 4t2) + (-5/16 *1* -8t3)
=1 + t + 12t2/8 + 40t3/16
=1 + t + 3t2/2 + 5t3 ANSWER
No 5) {cos(y- Ο€)} + {sin(y + Ο€/2)}
cos(y- Ο€) = { cos y cos Ο€ + sin y sin Ο€ }
Where Ο€ = 180
= (cos y)(cos180) + (sin y)(sin180)
= (cos y)(-1) + (sin y)(0)
sin(y + Ο€/2) = { sin y cos Ο€/2 + sin Ο€/2 cos y }
where Ο€/2= 180/2 = 90
= (sin y)(cos 90) + (sin 90)(cos y)
= (sin y)(0) + (1)(cos y)
Therefore, {cos(y- Ο€)} + {sin(y + Ο€/2)}
= (cos y)(-1) + (sin y)(0) + (sin y)(0) + (1)(cos y)
= -cos y + 0 + cos y + 0
= 0 Ans
No 6) tan(x + Ο€/4) tan(x - Ο€/4)
tan(x + Ο€/4) =tan x + tan 45/ 1-(tan x tan 45)
= tan x + 1/ 1- tan x
tan(x - Ο€/4) = tan x – tan 45/ 1+(tan x tan 45)
= tan x – 1/ 1 + tan x
= (tan x + 1/ 1- tan x)( tan x – 1/ 1 + tan x)
= (tan x + tan2x2 – tan x – 1) / (1+ tan x – tan x - tan2x2)
= (tan2x2 – 1) / (1 – tan2 x2)
=1(tan2x2 – 1) / -1(-1 + tan2 x2)
= 1/-1 = -1 ANS
No 7) 4 sin(x-200)= 5cos x
=4sin x * 4cos 20 – 4cos x * 4sin 20 =5cos x
=16sin x * cos20 – 16cos x sin 20 = 5cos x
Dividing through by cos x
=(16sin x * cos20 – 16cos x sin 20)/ cos x = (5cos x)/ cos x
=(16sin x * cos20)/ (cos x) – (16cos x sin 20)/(cos x) = 5
(16tan x * cos 20) = (5+ 16sin 20)
=(16 * 0.9397)tan x = 5+(16 * 0.3420)
Tan x = 10.472/15.0352 , Tan x = 0.6965
X = tan-10.6965, X= 34.86 ANS
No 8) cos A = 0.42 A = cos-1 0.42, A = 65.17
Sin B = 0.73 B = sin-1 0.73, B = 46.89
a) Sin(A+B) = (sin A cos B – sin B cos A)
=(sin 65.17 * cos 46.89) – (sin 46.89 * cos 65.17) =(0.9076 * 0.6834) –
(0.73 * 0.42) =(0.6203) – (0.3066) =0.3137 ANS
b) Cos (A-B) = (cos A cos B + sin a sin B)
=( cos 65.17 * cos 46.89) + (sin 65.17 * sin 46.89) =(0.4199 * 0.6834) +
(0.9076 * 0.7300) =(0.2870) + (0.6625) =0.9495 ANS
c) Tan (A+B) =( tan A + tan B)/ 1-(tan A tan B)
9. The cost of drilling 80m deep=?
First meter= £30
Second meter=£32
Third meter=£34
Using,
𝑠𝑛 =
𝑛
[2π‘Ž + (𝑛 βˆ’ 1)𝑑]
2
Where a= £30, n =80m, d= £2
𝑠80 =
80
[2(30) + (80 βˆ’ 1)2]
2
𝑠80 = 40[60 + 158]
𝑠80 = 40𝑆 × 218
𝑠80 = 8720
10. The numbers which are divisible by 4 between 5 and 250
8 , 12 , 16 , 20 , … , 248
a= 8 d=4 n=? L=248
𝐿 = π‘Ž + (𝑛 βˆ’ 1)𝑑
248 = 8 + (𝑛 βˆ’ 1)4
248 = 8 + 4𝑛 βˆ’ 4
248 βˆ’ 4 = 4𝑛
244 = 4𝑛
𝑛 = 61
𝑠𝑛 =
𝑠61 =
𝑛
[2π‘Ž + (𝑛 βˆ’ 1)𝑑]
2
61
[2(8) + (61 βˆ’ 1)4]
2
𝑠61 = 30.5[16 + 240]
𝑠61 = 30.5 × 256
𝑠61 =7808
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