F.2 數學一問, hope someone can answer me plz

👨🏻‍🏫 學術台數學

[匿名]

2007-01-02 5:55 am

Question 1 : Three kinds of tea's brand A, B, C costing $50, $60 nad $80 per kg respectively, they are then mixed together so that their weights are in the ratio of 3:4:5. FInd the cost of 60kg of the mixture.

Question 2: Given that (a+1/a) = 4 Find
(a^6+1/(a^6))

THX!

更新1:

Question 3: A piece of wire, 100 cm long, is cut into two pieces. Each piece is then bent into the shape of a square. If the sum of the enclosed areas is 397 cm square, prove that x^2-100x+1824 =0 where x m is the length of one the pieces.

回答 (2)

Yau

2007-01-07 1:52 am

✔ 最佳答案

Q.1 Brand A佔:60x[3/(3+4+5)]
=15kg
Brand B佔:60x[4/(3+4+5)]
=20kg
Brand C佔:60x[5/(3+4+5)]
=25kg
The total cost:15x50+20x60+25x80
=750+1200+2000
=$3950
so it cost $3950 for 60kg of those mixture

Q.2 Don't trust the upper guy,2a/a=2 not a
so the upper expreesion can't written as 2a/a+1/a=4
I solved it like that:
given a+1/a=4 find a^6+1/(a^6)
a+1/a=4 i.e a+a^-1=4
a^6+1/(a^6)
i.e a^6+(a^-6)
=(a^3)^2+(a^-3)^2
=[a^3+(a^-1)^3]^2-2(a^3)(a^-3)
=[a^3+(a^-1)^3]^2-2
={(a+a^-1)[a^2-a(a^-1)+(a^-1)^2]}^2-2
={(a+a^-1)[a^2+(a^-1)^2-a(a^-1)]}^2-2
={(a+a^-1)[a^2+(a^-1)^2-1]}^2-2
={(a+a^-1)[(a+a^-1)^2-2-1]}^2-2
then we substitute a+a^-1=4 into the exprssion
={(4)[(4)^2-2-1]}^2-2
=(4*13)^2-2
=2704-2
=2702

參考: My brain

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[匿名]

2007-01-02 6:34 am

Solution
Q1:
( 60kg x 3/12 x $50 )+( 60kg x 4/12 x $60 )+( 60kg x 5/12 x $80 )
=$750+$1200+$2000
=$3950

Q2:
a+1/a=4
2a/a+1/a=4
(2a+1)/a=4
2a+1=4a
-2a=1
a=-1/2

then sub a=-1/2 into the equation(a^6+1/(a^6))
=64 1/64

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