Mathematics Assignment For Students

Mathematics 1
MATHS
by [NAME]
Course
Professor’s Name
Institution
Location of Institution
Date

Mathematics 2
Question 1
a) 4
√ √ y5
Written in power form;
[ [ y5 ] 1
2 ] 1
4
Applying power rules;
y5 × 1
2 × 1
4
¿ y
5
8
b) (−4 x−5 y2 ) ( 18 x10 y−3 )
−3 x−3 y
¿ (−4 x−2 y2 ) ( 6 x10 y−4 )
−1
¿ ( 4 x−2 y2 ) ( 6 x10 y−4 )
¿ ( 2 4 x8 y−2 )
c) ( 36 m−6 n−1
16 m2 n−5 ) 3
2
¿ ( 9 m−8 n4
4 )3
2
¿ ( 27 m−12 n6
8 )
d) ( 54 x−5 y6
6 x2 y−3 ) 1−2
¿ ( 9 x−7 y9
1 )−2

Mathematics 3
¿ 9 x14 y−18
Question 2
a) (−81 )
1
4
Assume that −81=a+bi where i= √−1
Plotting polar coordinates we find that;
a=r cos θ
b=r sin θ
Therefore, a+ bi=r (cos θ+isin θ)
Using Euler’s formula;
cos θ+ isin θ=eiθ
Therefore, a+ bi=r eiθ ⟹−81=r eiθ
So if r =81then eiθ=−1
For eiθ=−1 then θ= ( 2 k−1 ) π wherek =1,2,3,4 ... .
Assume that ¿
Therefore, α 4 =81 ei ( 2 k−1 ) π
α 2=± 9 e
i ( 2 k−1 ) π
2
¿
¿
α =± 3 e
i ( 2 k−1 ) π
4
¿
∨α=± i 3 e
i ( 2k−1 ) π
4
¿
¿
¿
Going back to Euler’s method
α =± 3 ¿
Starting with the positive value first, α = (3 (± √2
2 )+ ± √ 2
2 )=1.5 √2+ 1.5i √2

Mathematics 4
For negative value, α =− ( 1.5 √2+1.5 i √2 )
For α =± i3 e
i (2 k−1 ) π
4
¿
¿
Starting with the positive value first, α =−1.5 √ 2+1.5 i √2
For negative value, α = ( 1.5 √ 2−1.5 i √ 2 )
Therefore (−81 )
1
4 has four possible values .
1.5 √2+1.5 i √2
− ( 1.5 √ 2+ 1.5i √ 2 )
−1.5 √ 2+ 1.5i √2
( 1.5 √ 2−1.5 i √ 2 )
b) ( 16
49 )
−1.5
¿ 1
( 16
49 ) 3
2
= ( 49
16 ) 3
2
This is like finding the square root then cubing the result.
¿ ( 7
4 )
3
¿ ( 343
64 )
c) 5−7 × ( 52 ) 4
5−3
¿ 5−7 ×58
5−3 According to the law of indices,
¿ 5−7+ 8
5−3 = 51
5−3 =54