Qing asked in 科學及數學數學 · 7 years ago

5條MATHS MOCK MC急

3 Answers

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  • yc
    Lv 5
    7 years ago
    Favorite Answer

    suggested solutions:

    圖片參考:http://imgcld.yimg.com/8/n/HA08446686/o/2013032321...

  • 7 years ago

    8.y=x^2+bx+c=0

    Ax=[-b+√(b^2-4ac)]/2

    Dx=[-b-√(b^2-4ac)]/2

    Mx=(Ax+Dx)/2=-2b/4a=-b/2

    y(0)=0+0+c=c => Cy=By=c

    Bx=2Mx-Cx=-b/a-0=-b

    area(MBC)=|Bx*Cy/2|

    =|(-b)*c/2|

    =|bc/2|.........ans

    18.螺旋=最短路線: w=角速度, Q=角位移=wt, v=rw

    (1) 基本公式:

    x=r*cosQ=r*cos(wt) => dx=-rw*sin(wt)dt

    y=r*sinQ=r*sin(wt) => dy=rw*cos(wt)dt

    z=vt=rwt => dz=rwdt

    (2) 求弧線長s=?

    (ds)^2=(dx)^2+(dy)^2+(dz)^2

    =[(rw)^2*(cosQ^2+sinQ^2)+(rw)^2]dt^2

    =[(rw)^2+(rw)^2]dt^2

    =2*(rw)^2*dt^2

    ds=rw√2dt

    s=∫rw√2dt

    =∫r√2d(wt)

    =∫r√2dQ........Q=0~3π

    =r√2*Q

    =3r√2π

    =6√(π^2+1)

    (3) 求半徑r=?

    兩邊平方: 18r^2*π^2=36(π^2+1)

    r^2=2(π^2+1)/π^2

    r=√[2(π^2+1)]/π

    =√2*a/π

    where a=√(π^2+1)

    (4) 求柱高h=?:

    繞1.5圈 => s=6a

    繞1圈 => s=4a

    商高定理:

    h^2=s^2-(2πr)^2

    =16a^2-4*2a^2

    =8a^2

    h=2√2*a

    area(ABCD)=2πrh

    =2π*(√2*a/π)*2√2*a

    =8a^2

    =8(π^2+1)

    =原題答案無法選擇

    26.ABCD=平行四邊形 => C=?

    A=(-1,4), B=(-3,-2), D=(2,-1)

    C=B對稱點 vs AD

    =2M-B

    =(1,3)-(-3,-2)

    =(1+3,3+2)

    =(4,5).......ans=(C)

    37.

    y1=log<a>x=log(x)/log(a)

    y2=log<b>x=log(x)/log(b)

    x=1; y1=y2=0

    x>1; y1>y2 => log(b)<log(a)

    0<b<a=分數<1........ans

    43.p=6/7 => q=1-p=1/7

    二項式分配: c(5,x)=5!/x!(5-x)!

    f(x)=c(5,x)*p^x*q^(5-x)

    直接改用商高定理求取:

    s^2=h^2+(3πr)^2........繞1.5圈=3π

    36(π^2+1)=6^2+(6π)^2

    =6^2+(3π2)^2

    => h=6, r=2

    A=2πrh

    =2π2*6

    =24π........ans=(B)

  • 麻辣
    Lv 7
    7 years ago

    8.y=x^2+bx+c=0Ax=[-b+√(b^2-4ac)]/2Dx=[-b-√(b^2-4ac)]/2Mx=(Ax+Dx)/2=-2b/4a=-b/2y(0)=0+0+c=c => Cy=By=cBx=2Mx-Cx=-b/a-0=-barea(MBC)=|Bx*Cy/2|=|(-b)*c/2|=|bc/2|.........ans

    18.螺旋=最短路線: w=角速度, Q=角位移=wt, v=rw(1) 基本公式:x=r*cosQ=r*cos(wt) => dx=-rw*sin(wt)dty=r*sinQ=r*sin(wt) => dy=rw*cos(wt)dtz=vt=rwt => dz=rwdt(2) 求弧線長s=? (ds)^2=(dx)^2+(dy)^2+(dz)^2=[(rw)^2*(cosQ^2+sinQ^2)+(rw)^2]dt^2=[(rw)^2+(rw)^2]dt^2=2*(rw)^2*dt^2ds=rw√2dts=∫rw√2dt=∫r√2d(wt)=∫r√2dQ........Q=0~3π=r√2*Q=3r√2π=6√(π^2+1)(3) 求半徑r=?兩邊平方: 18r^2*π^2=36(π^2+1)r^2=2(π^2+1)/π^2r=√[2(π^2+1)]/π=√2*a/πwhere a=√(π^2+1)(4) 求柱高h=?: 繞1.5圈 => s=6a繞1圈 => s=4a商高定理:h^2=s^2-(2πr)^2=16a^2-4*2a^2=8a^2h=2√2*aarea(ABCD)=2πrh=2π*(√2*a/π)*2√2*a=8a^2=8(π^2+1)=原題答案無法選擇

    26.ABCD=平行四邊形 => C=?A=(-1,4), B=(-3,-2), D=(2,-1)AD中點: M=(A+D)/2=(-1+2,4-1)/2=(1/2,3/2)C=B對稱點 vs AD=2M-B=(1,3)-(-3,-2)=(1+3,3+2)=(4,5).......ans=(C) 37.y1=log<a>x=log(x)/log(a)y2=log<b>x=log(x)/log(b)x=1; y1=y2=0x>1; y1>y2 => log(b)<log(a)0<b<a=分數<1........ans

    43.p=6/7 => q=1-p=1/7二項式分配: c(5,x)=5!/x!(5-x)!f(x)=c(5,x)*p^x*q^(5-x)f(3)=(5!/3!2!)*(6/7)^3*(1/7)^2=(5*4/2)*(6*36/7*49)/49=10*216/7*49^2=2160/16807=30*72/72*233.43=30/233.43........ans=(D)

    2013-03-20 12:42:54 補充:

    第18題修正:

    z=v*t 修改為: z=k*v*t

    多出未知數k.無法使用積分法求取: r.h.k

    直接改用商高定理求取:

    s^2=h^2+(3πr)^2........繞1.5圈=3π

    36(π^2+1)=6^2+(6π)^2

    =6^2+(3π2)^2

    => h=6, r=2

    A=2πrh

    =2π2*6

    =24π........ans=(B)

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