数学题解答网站求解答

GMAT两大数学题求解答1.Each participant in a certain study was assigned a sequence of 3 different letters from the set {A,B,C,D,E,F,G,H}.If no sequence was assigned to more than one participant and if 36 of the possible sequences were not assigned,what was the number of participants in the study?(Note,for example,that the sequence A,B,C is different from the sequence C,B,A.)A.20B.92C.300D.372E.476Stations X and Y are connected by two separate,straight,parallel rail lines that are 250 miles long.Train P and train Q simultaneously left Station X and Station Y,respectively,and each train traveled to the other’s point of departure.The two trains passed each other after traveling for 2 hours.When the two trains passed,which train was nearer to its destination?(1) At the time when the two trains passed,train P had averaged a speed of 70 miles per hour.(2) Train Q averaged a speed of 55 miles per hour for the entire trip.A.Statement (1) ALONE is sufficient,but statement (2) alone is not sufficient.B.Statement (2) ALONE is sufficient,but statement (1) alone is not sufficient.C.BOTH statements TOGETHER are sufficient,but NEITHER statement ALONE is sufficient.D.EACH statement ALONE is sufficient.E.Statements (1) and (2) TOGETHER are NOT sufficient.第一题答案为C,第二题答案为A.
疯狂是神の1121
第一题大意是:从八个字母中选取三个,做排列组合,8*7*6=336,即总共能分给336个参与者,每个人的情况都不同.最后还有36种情况分不出去,显然一共有300个参与者.第二题就是两列车2小时相遇,(1)(2)两种情况那种可以确定哪列车更接近终点:情况(1):相遇时P车的平均速度是70mile/hour.说明相遇时已走过140miles,因为两车相对行驶,则Q车已走过110miles,显然P车距终点更近.情况(2):Q车在整个过程中平均速度是55miles,但并不清楚相遇前的平均速度,因而无法判断相遇时哪列车更接近终点.
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(1)延长AC到点P1,使得CP1=CA,连接BP1过P1作P1D1⊥X轴,交X轴于D1由AC=P1C,BC=BC,∠BCA=∠BCP1=90°,得△P1BC≌△ABC故P1为所求的P点。由AC=CP1,∠ACO=∠P1CD,∠AOC=∠P1D1C=90°得:△COA≌△CD1P1所以:P1D1=AO=1,CD1=CO=1,OD1=2CO=2由P1在第四象限,知P1的坐标为(2,-1)(2)过C作AB的平行线,过B作AC的平行线,两者交于点P2∵BP2//AC,CP2//AB∴四边形ABP2C是平行四边形,BC为对角线∴故有△ABC≌△P2CB∴P2也为所求的P点过B作BD2⊥X轴,交X轴于D2,过P2作P2F⊥BD2,交BD2于F∵OA=OC=1∴∠ACO=45°,AC=√2又∵∠ACB=90°,∠BAC=60°∴∠BCD2=45°,BC=√3AC=√6,△AOC∽△BD2C故BD2=AO·BC/AC=√3=CD2,OD2=√3+1即B点的坐标为(√3+1,√3)∵BD2//AO∴∠CBP2=90°,∠P2BD=45°=∠BP2F,△AOC≌△BFP2故FP2=BF=1,P2的横坐标比B大1,纵坐标比B小1即P2的坐标为(√3+2,√3-1)(3)作P2关于BC的对称点P3,由∠CBP2=90°,故P3在P2B的延长线上,且BP3=BP2∵BP3=BP2=AC,BC=CB,∠ACB=∠CBP3=90°∴△ABC≌△P3CB故P3也为所求的P点P3的横坐标比B小1,纵坐标比B大1故P3的坐标为(√3,√3+1)所以,符合条件的P点有3个,即(2,-1),(√3+2,√3-1),(√3,√3+1)
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左视图面积=1*3=38.俯视图面积=2*3=69.AB=1.2根号310.总个数是10个
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