点P1分线段AB为5:7,P2分线段AB为5:11,已知P1P2=10cmAB长
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1.102.(-3,-2)3.-3
A——D—C————B∵AC:BC=5:7∴AC=AB×5/(5+7)=5/12AB∵AD:BD=5:11∴AD=AB×5/(5+11)=5/16AB∴CD=AC-AD=5/12AB-5/16AB=(
MA=5/12ABNA=5/16ABMN=MA-NA=5/48AB=10所以AB=96
因为cd=ac-ad=5/12ab-5/16ab=5/48ab所以ab=48/5cd=48/5*10=96cm
10/(11/(11+5)-7/(5+7))=9610/(7/(5+7)-5/(5+11))
AP1=5/(5+7)*AB=5/12*96=40(cm)AP2=5/(5+11)*AB=5/16*96=30(cm)P1P2=AP1-AP2=40-30=10(cm)
(-1+x*3)/(1+x)=-7/3得出x=-1/4;-6/(1-1/4)=y,所以y=-8
因为P1,P2是黄金分割,所以AP2=BP1=(√5-1)/2AB,所以P1P2=2AP2-AB即a=(√5-1)AB-AB解得AB=(√5+2)a
设AB的长为xcm,根据题意得:512x-516x=2,解得x=19.2.∴AB的长为19.2cm.
∵直线m过点M(-2,0)∴设直线m:y=k1(x+2),联立方程得:(1+2k²1)x²+8k²1x+8k²1-2=0由韦德定理:x1+x2=-8k²
p1是指向数组a的首地址的指针,p2为空指针.因为p2为空指针,所以p2取反为真.(即!p2的值为真).p1不为空,所以不管他指向哪,都为真.所以p1,!p2,为真,p2,p1&&p2为假.p1指向数
AM=5/(5+7)*AB=5AB/12AN=5/(5+11)*AB=5AB/16MN=AM-AN=5AB/12-5AB/16=(5*4AB-5*3AB)/48=5AB/48=10AB/48=2AB=
设AB长为x(5/12-5/16)x=10解得x=96
设:AB=x,则AO=(5/12)x,BO=(7/12)x,AP=(5/16)x,PB=(11/16)xAP+PO+BO=AB(5/16)x+10+(7/12)x=x解得x=96即AB=96答:AB长
设AD=aDC=b=10CB=cAB=dd=a+10+c(a+10)/c=5/7a/(10+c)=5/11a=30b=10c=56AB=d=a+b+c=30+10+56=96
点C分线段AB5:7∴AC=5/12AB点D分线段AB为5:11∴AD=5/16AB∵CD=AC-AD=5㎝∴5/12AB-5/16AB=5㎝AB=48㎝
AC=5/12ABAD=5/16ABCD=AC-AD =(5/12-5/16)AB =5/48AB =5则,5/48
A——D—C——B∵AC:BC=5:7∴AC=5/(5+7)×AB=5/12AB∵AD:BD=5:11∴AD=5/(5+11)×AB=5/16AB∴CD=AC-AD=5/12AB-5/16AB=(20
由已知可得:ac=5/12ab,cb=7/12ab,ad=5/16ab,db=11/16ab.ac-ad=cd=10=5/12ab-5/16ab得ab=96