標題:
f.3 math
發問:
圖片參考：http://img815.imageshack.us/img815/4334/img210.jpg 圖片參考：http://img256.imageshack.us/img256/4232/img207w.jpg 圖片參考：http://img3.imageshack.us/img3/1443/img211gl.jpg 圖片參考：http://img687.imageshack.us/img687/3200/img209a.jpg thx a lot ~~~
最佳解答:
1.Let RM=y,QR=2y,PS=x area of PMR= (1/2)(xy) area of PQR=(1/2)(2y)(x) area of PMR/area of PQR=1/2 2. 4/3 √3 =4*3^(-1)√3 =4*3^(-1/2) =√(16/3) 3.角ECD=角ACB=90o 角ABC=角CED ∴triangle DCE～triangle ACB (A.A.A.) X/12=12/20 x=6 102-62=y2(Pyth.thrm) y=8 4.2/√2 =2/√2 *√2/√2 =2√2 / 2 =√2 2/√2- √2/2 =√2 - √2/2 =√2/2 5. √75-√27-√12 =5√3-3√3-2√3 =0 3√8+2√18-5√32 =6√2+6√2-20√2 =-8√2
Q2: (10/20)=x/12 x.........=6 ______________ Q5: sqrt75-sqrt27-sqrt12 5sqrt3-3sqrt3-2sqrt3 =0 ________________ Q5b: 3sqrt8+2sqrt18-5sqrt32 6sqrt2+6sqrt2-20sqrt2 -8sqrt2
f.3 math
發問:
圖片參考：http://img815.imageshack.us/img815/4334/img210.jpg 圖片參考：http://img256.imageshack.us/img256/4232/img207w.jpg 圖片參考：http://img3.imageshack.us/img3/1443/img211gl.jpg 圖片參考：http://img687.imageshack.us/img687/3200/img209a.jpg thx a lot ~~~
最佳解答:
1.Let RM=y,QR=2y,PS=x area of PMR= (1/2)(xy) area of PQR=(1/2)(2y)(x) area of PMR/area of PQR=1/2 2. 4/3 √3 =4*3^(-1)√3 =4*3^(-1/2) =√(16/3) 3.角ECD=角ACB=90o 角ABC=角CED ∴triangle DCE～triangle ACB (A.A.A.) X/12=12/20 x=6 102-62=y2(Pyth.thrm) y=8 4.2/√2 =2/√2 *√2/√2 =2√2 / 2 =√2 2/√2- √2/2 =√2 - √2/2 =√2/2 5. √75-√27-√12 =5√3-3√3-2√3 =0 3√8+2√18-5√32 =6√2+6√2-20√2 =-8√2
此文章來自奇摩知識+如有不便請留言告知
其他解答:Q2: (10/20)=x/12 x.........=6 ______________ Q5: sqrt75-sqrt27-sqrt12 5sqrt3-3sqrt3-2sqrt3 =0 ________________ Q5b: 3sqrt8+2sqrt18-5sqrt32 6sqrt2+6sqrt2-20sqrt2 -8sqrt2
文章標籤
全站熱搜
留言列表
