y2=2px Y=2PY0 (YO2-P2)(X-P 2)

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y2=2px Y=2PY0 (YO2-P2)(X-P 2)
[ y2=2x, [ x2+y2=8. 如何解这道方程组

因为y2=2x所以x2+2x=8x=2或-4因为y2为正数所以x=2y2=4y=2或-2

已知两圆x2+y2=1,(x-2)2+y2=4,求两圆公切线方程

√3X-3Y+2√3=0或√3X+3Y+2√3=0过程很难写,只能把答案写上去了,其实用平几很容易算出来的

已知随机变量x,y 有D(X)=4 ,D(Y)=1,pxy=0.5 则D(2x-3y)=?

D(x-y)=D(X)+D(Y)-2cov(x,y)cov(x,y)=pxy*√(D(X)*D(Y))=0.5*√4*1=1代入:D(2x-3y)=4D(X)+9D(Y)-2*2*3cov(x,y)=

已知2x=3y,求xy/(x2+y2)-y2/(x2-y2)的值

已知2x=3y,求xy/(x^2+y^2)-y^2/(x^2-y^2)的值2x=3y-->x=(3/2)yx^2=(9/4)y^2xy/(x^2+y^2)-y^2/(x^2-y^2)==(3/2)y*

解方程组:y2=x3-3x2+2x;x2=y3-3y2+2y

y^2=x^3-3x^2+2xx^2=y^3-3y^2+2y两式相减得:y^2-x^2=(x^3-y^3)-3(x^2-y^2)+2(x-y)(x-y)(x^2+xy+y^2-2x-2y+2)=0所以

设随机变量X~B(10,0.5),N(2,10),又E(XY)=14,则X与Y的相关系数Pxy=

因为X~B(10,0.5),N(2,10),所以EX=10×0.5=5,DX=10×0.5×0.5=2.5,EY=2,DY=10,又E(XY)=14,所以X与Y的协方差为cov(X,Y)=14-5×2

已知(x2+y2+3)(x2+y2-2)-6=0,求x2+y2的值

(x²+y²)²+(x²+y²)-6-6=0(x²+y²)²+(x²+y²)-12=0(x²

若|p+2|与q-8q+16互为相反数,分解因式:(x²+y²)-(pxy+q)

|p+2|>=0q-8q+16=(q-4)2>=0|p+2|与q-8q+16互为相反数所以p=-2q=4x²+y²+2xy-4=(x+y)2-4=(x+y-2)(x+y+2)

设Var(X)=25,Var(y)=36,pXY=0.4,求Var(X+Y)和Var(X-Y).

var(x+y)=var(x)+var(y)+0.4*5*6*2=85var(x-y)=25+36-24=37再问:为什么这么做呢?能写详细一点么,要交作业呢,求教了。比如什么公式之类的?

已知x,y为实数,且(x2 +y2)(x2 +y2+2)=3.求x2 +y2的值

设t=x2+y2(t大于等于0)则t(t+2)-3=0(t+3)(t-1)=0t=-3(舍去)或t=1所以,x2+y2=1

解方程组x2+y2=20,2x2-3xy-2y2=o

由2x²-3xy-2y²=0得2-3(y/x)-2(y/x)²=0(2+y/x)*(1-2y/x)=0得y/x=1/2或-2即y=1/2x或y=-2x代入x²+

DX=4,DY=9,Pxy=0.4,求D(x+y)及D(x-y)?

D(x+y)=Dx+Dy+2*cov(x,y)=Dx+Dy+2*Pxy*Dx(开方)*Dy(开方)=4+9+2*0.4*2*3=17.8D(x-y)=Dx+Dy-2*cov(x,y)=Dx+Dy+2*

若:根号【(x+3)2+y2】+根号【(x-3)2+y2】=10,则x2/16+y2/25=____

可以看到没有根号时,那两个分别是以(—3,0)和(3,0)为圆心的圆,即条件要求两个圆的相交点正好半径和等于10.根据两圆关于y轴对称时正好可以得到一个特殊点(0,4)或者(0,—4)满足条件.所以最

已知实数x.y满足(x2+y2)(x2+y2-1)=2,求x2+y2的值

可设x²+y²=t.则t(t-1)=2.===>t²-t-2=0.===>(t-2)(t+1)=0.===>t=2.即x²+y²=2.

若|p+2|与q2-8q+16互为相反数,分解因式(x2+y2)-(pxy+q)=______.

依题意得|p+2|+(q2-8q+16)=0,即|p+2|+(q-4)2=0,∴p+2=0,q-4=0,解得p=-2,q=4,∴(x2+y2)-(pxy+q),=(x2+y2)-(-2xy+4),=x

1.已知y1=-x+2,y2=3x+4,当x分别取何值时,y1=y2,y1<y2,y1>y2?

1.y1=y2:-x+2=3x+4=>4x=-2=>x=-1/2y1-1/2y1>y2:-x+2>3x+4=>xx=1y=1带入y=ax+7=>a=-6

若|p+2|与q²+8q+16互为相反数,分解因式(x²+y²)-(pxy+q)=————

|P+2|>=0q²-8q+16=(q-4)^2>=0|P+2|与q²-8q+16互为相反数则P+2=0P=-2(q-4)^2=0q=4(X²+y²)-(pxy

已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x

因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²