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maths (中3to中4) 15點!

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1.a rectangle which length is(2x+3) and width is(x-1)The area of the rectangular land is known to be no more than(2x^2+9)a) find the range of the values of x.b) find the maximum area of the piece of land.2. a theatre is designed to accommodate at least 500audiences and there are 15 seats each row.a)... 顯示更多 1.a rectangle which length is(2x+3) and width is(x-1) The area of the rectangular land is known to be no more than(2x^2+9) a) find the range of the values of x. b) find the maximum area of the piece of land. 2. a theatre is designed to accommodate at least 500audiences and there are 15 seats each row. a) what is the least no. of rows in the theatre? b)using the result in a), if three seats are added each row,can the theatre accommodate 600audiences? 3. a piece of wire of 5m long is cut into x pieces of 15cm long and 17 pieces of 10cm long. What is the maximum no. of pieces of wire of 15cm long? 4.Tom deposits a sum of money in a bank at a rate of 10%p.a. compounded yearly, for at least how many years are needed to triple the money? 5.the rateable value of a flat is $150000. if the rates percentage charge is 5%,find the rates payable for the first quarter of 2005. 4.suppose y=k/x^2, where k is a constant. What is the percentage change of y when x is decreased by 20%? 5. 4-4cos^θ 6.sin^2θ/2tanθ 7.sinθ/sin(90-θ)

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1a.(2x+3)(x-1)<2x2+9 2x2+x-3<2x2+9 x<12 b. maximum area=(2x11+3)(11-1) =250 2a. 500/15 = 33 1/3 Therefore, at least 34 rows. 2b. 34(15+3) = 612 Therefore, If three seats are added each row, the theatre can accommodate 600 audiences3. 5m = 500cm 10×17 +15x≦ 500 15x≦ 330 x≦ 22 Therefore,maximum no. is 224. Let the original of money is $y and z years be the years are needed to triple the money y(1+10%)^z≧3y 1.1yz≧3y z≧2.72... Therefore,at least how 3 years are needed to triple the money5.the rates =$150000×5%÷4 =$18754.new y=k/(0.8x)2 =k/0.64x2percentage change of y=(k/0.64x2- k/x2)/ (k/x2) x 100% =56.25% Therefore, y increase by 56.25% 5. 4-4cos2θ =4(1-cos2θ) =4sin2θ6.sin2θ/2tanθ = sin2θ× 1/2tanθ = sin2θ× 1/2(sinθ/cosθ) = sin2θ× (cosθ/2sinθ) = sinθcosθ/27.sinθ/sin(90-θ) =sinθ/cosθ =tanθ*1-cos2θ=sin2θ sin(90-θ)=cosθ sinθ/cosθ=tanθ

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