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secondary 4 | E Maths
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Help Hanisah! Anyone can contribute an answer, even non-tutors.
help please!
Date Posted: 5 years ago
Views: 244
(144p²)^(5/2) ÷ (24p‾²)^(3/2)
= 144^(5/2) ÷ 24^(3/2) x (p²)^(5/2) ÷ (p‾²)^(3/2)
= (2⁴ x 3²)^(5/2) ÷ (2³ x 3¹)^(3/2) x p^5 ÷ p‾³
(Trick is to apply prime factorisation on 144 and 24 )
= 2^(4 x 5/2 - 3 x 3/2) x 3^(2 x 5/2 - 1 x 3/2)
x p^(5 - (-3))
= 2^(11/2) x 3^(7/2) x p^8
So x = 11/2, y = 7/2 and z = 8
= 144^(5/2) ÷ 24^(3/2) x (p²)^(5/2) ÷ (p‾²)^(3/2)
= (2⁴ x 3²)^(5/2) ÷ (2³ x 3¹)^(3/2) x p^5 ÷ p‾³
(Trick is to apply prime factorisation on 144 and 24 )
= 2^(4 x 5/2 - 3 x 3/2) x 3^(2 x 5/2 - 1 x 3/2)
x p^(5 - (-3))
= 2^(11/2) x 3^(7/2) x p^8
So x = 11/2, y = 7/2 and z = 8
Thank you so much, J!
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