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C(5,3)等于多少从5个里选3个,不计顺序的概率是多少?大学学的忘了.是5乘4乘3 等于 好像不是哈,
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Cmn=m!/[n!(m-n)!]所以C53=5!/(2!×3!)=10
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C(5,3)=5乘4乘3再除以（1乘以2乘以3）=10A(5,3)=5*4*3=60
5*4*3/3*2*1=10如果是A(5,3)，从5个选3个，计顺序的话，才是等于60
Cmn=m!/[n!(m-n)!]5*4*3/1*2*3=10答：10.
C(5,3)=5*4*3/3*2*1=10
扫描下载二维码(1又2/3+2又3/4+3又3/5+4又5/6)/(3又1/3+5又2/4+7又3/5+9又4/6)=?_百度知道
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(1又2/3+2又3/4+3又3/5+4又5/6)/(3又1/3+5又2/4+7又3/5+9又4/6)=?
求解？过程？详细！~~~~
我有更好的答案
3-2/3，其他分数同理。分子=14-1/3-1/4-1/5-1/6分母=28-2/5-2/4-2&#47思路;3=1-1/6分母的每一项都是分子的2倍，所以原式=1/2：2&#47
亲，答案的过程~~
第2、3行就是过程
采纳率：42%
=改写为分数形式把(1-1/4)+(1-2/&6+(1-1/3)+(1-1/4)+(1-1/5)&/3）+（1-2/4)+(1-2/15+(1-2/3）+（1-2/5)&5)&4)+(1-1/3)+(1-1/5)和&(1-2/约为2和1 =8/16=1&#47=&lt
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In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?A. 3 B. 15/4C. 5 D. 16/3 E. 20/3
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Hi I need a quick clarification on the concept of perpendicular bisector.With a perpendicular bisector, the bisector always crosses the line segment at right anglesIf any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?So here BC should be the perpendicular bisector and the AC=CD=3 right ?Please let me know what am missing here.I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ?
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rvinodhini wrote:Hi I need a quick clarification on the concept of perpendicular bisector.With a perpendicular bisector, the bisector always crosses the line segment at right anglesIf any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?So here BC should be the perpendicular bisector and the AC=CD=3 right ?Please let me know what am missing here.I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ? A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.Complete solution:In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?A. 3 B. 15/4C. 5 D. 16/3 E. 20/3Attachment:
splittingtriangle.jpg [ 4.22 KiB | Viewed 146448 times ]
Important property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram). So, \(\frac{CD}{AC}=\frac{AC}{BC}\) --& \(\frac{CD}{4}=\frac{4}{3}\) --& \(CD=\frac{16}{3}\).Answer: D.For more on this subject please check Triangles chapter of Math Book: Hope it helps.
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Now I get it ..wonderful explanation !! thanks
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thankx for the explanation..
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MBAhereIcome wrote:all three triangles abc, acd & abd are similar.so, \(\frac{4}{3} = \frac{cd}{4}\) ... cd = \(\frac{16}{3}\)it can't be \(\frac{4}{3} = \frac{4}{cd}\) ... cd = 3, because in that case \(5^2+5^2 = 6^2\) is not trueIf the 3 triangles are proportional, why can't I solve using the ratio: \(\frac{AD}{AB} = \frac{AB}{(X+3)}\)\(\frac{4}{5} = \frac{5}{(X+3)}\)4(X+3) = 254X + 12 = 25X = \(\frac{13}{4}\)
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pubchum wrote:MBAhereIcome wrote:all three triangles abc, acd & abd are similar.so, \(\frac{4}{3} = \frac{cd}{4}\) ... cd = \(\frac{16}{3}\)it can't be \(\frac{4}{3} = \frac{4}{cd}\) ... cd = 3, because in that case \(5^2+5^2 = 6^2\) is not trueIf the 3 triangles are proportional, why can't I solve using the ratio: \(\frac{AD}{AB} = \frac{AB}{(X+3)}\)\(\frac{4}{5} = \frac{5}{(X+3)}\)4(X+3) = 254X + 12 = 25X = \(\frac{13}{4}\) Attachment:
splittingtriangle.jpg [ 4.22 KiB | Viewed 128869 times ]
In similar triangles the ratio of the corresponding sides are equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram).The ratios you are equating are not of corresponding sides. If you want to equate AD/AB then it should be AD/AB=AC/BC --& AD/5=4/3 --& AD=20/3. Also AD/AB=CD/AC --& (20/3)/5=CD/4 --& CD=16/3.I merged this thread with an earlier discussion of the same question, so check this post:
it might help to clear your doubts.Please ask if anything remains unclear.
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Hi Bunuel,How did you infer that angle BAC = angle ADC. Could you please explain that part.. I know this can be proved thro similarity, however I wanted to understand from your concept, which you have mentioned below. Pls explain.ThanksHBunuel wrote:rvinodhini wrote:Hi I need a quick clarification on the concept of perpendicular bisector.With a perpendicular bisector, the bisector always crosses the line segment at right anglesIf any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?So here BC should be the perpendicular bisector and the AC=CD=3 right ?Please let me know what am missing here.I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ? A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.Complete solution:In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?A. 3 B. 15/4C. 5 D. 16/3 E. 20/3Attachment:splittingtriangle.jpgImportant property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram). So, \(\frac{CD}{AC}=\frac{AC}{BC}\) --& \(\frac{CD}{4}=\frac{4}{3}\) --& \(CD=\frac{16}{3}\).Answer: D.For more on this subject please check Triangles chapter of Math Book: Hope it helps.
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imhimanshu wrote:Hi Bunuel,How did you infer that angle BAC = angle ADC. Could you please explain that part.. I know this can be proved thro similarity, however I wanted to understand from your concept, which you have mentioned below. Pls explain.ThanksHAttachment:
splittingtriangle.jpg [ 4.22 KiB | Viewed 128588 times ]
&B+&D+&A=180, since &A=90 then &B+&D=90;Similarly in triangle ABC: &B+&BAC=90 since &B=90-&D then (90-&D)+&BAC=90 --& &BAC=&D.Hope it's clear.
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Thanks for the solution. I just wanted to know whether it requires any calculation or is it just a corollary of the property described above. Its clear now.Bunuel wrote:imhimanshu wrote:Hi Bunuel,How did you infer that angle BAC = angle ADC. Could you please explain that part.. I know this can be proved thro similarity, however I wanted to understand from your concept, which you have mentioned below. Pls explain.ThanksHAttachment:splittingtriangle.jpg&B+&D+&A=180, since &A=90 then &B+&D=90;Similarly in triangle ABC: &B+&BAC=90 since &B=90-&D then (90-&D)+&BAC=90 --& &BAC=&D.Hope it's clear.
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Well i have a problem with similar triangles coz i sometimes make mistakes on the common sides. hence alternative approach for this problem..Considering Triangle ACD - AC^2 + CD^2 = AD^2Considering Triangle ABD - AB = sqrt(4^2 + 3^2) = 5 (Pythagorean triplet so you dont really have to do the math on the test)Considering Triangle BAD - AB^2 + AD^2 = BD^225 + ac^2 + cd^2 = (3 + cd)^2=& 25+16+cd^2= 9 + 6cd +cd^2=&32 = 6cdcd = 16/3
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enigma123 wrote:Attachment:Triangle.jpgIn triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?A. 3 B. 15/4C. 5 D. 16/3 E. 20/3Learn this property:AC^2=BC*DC=&DC=16/3
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In the triangles ABD, &CAB= &CDA----& tan &CAB = tan &CD----& 3/4=4/CD----& CD=16/3---&Answer D
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the given solution is neat and simple but it didnt strike me when solving anyways here is another alternate way though i admit its lengthy calculationlet us assume CD=xthus, AD=sqrt(x^2+16)also in traingle ABD, BD^2=AB^2+AD^2(x+3)^2=25+(x^2+16)x=32/6=16/3(D)
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Bunuel wrote:rvinodhini wrote:Hi I need a quick clarification on the concept of perpendicular bisector.With a perpendicular bisector, the bisector always crosses the line segment at right anglesIf any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?So here BC should be the perpendicular bisector and the AC=CD=3 right ?Please let me know what am missing here.I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ? A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.Complete solution:In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?A. 3 B. 15/4C. 5 D. 16/3 E. 20/3Attachment:splittingtriangle.jpgImportant property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram). So, \(\frac{CD}{AC}=\frac{AC}{BC}\) --& \(\frac{CD}{4}=\frac{4}{3}\) --& \(CD=\frac{16}{3}\).Answer: D.For more on this subject please check Triangles chapter of Math Book: Hope it helps.This part is clear:
triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles)However, how does one determine which angles are equal? Except 90 degree angles of both triangles, i could not seem to follow how exactly other angles became equal?
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earnit wrote:Bunuel wrote:rvinodhini wrote:Hi I need a quick clarification on the concept of perpendicular bisector.With a perpendicular bisector, the bisector always crosses the line segment at right anglesIf any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?So here BC should be the perpendicular bisector and the AC=CD=3 right ?Please let me know what am missing here.I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ? A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.Complete solution:In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?A. 3 B. 15/4C. 5 D. 16/3 E. 20/3Attachment:splittingtriangle.jpgImportant property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram). So, \(\frac{CD}{AC}=\frac{AC}{BC}\) --& \(\frac{CD}{4}=\frac{4}{3}\) --& \(CD=\frac{16}{3}\).Answer: D.For more on this subject please check Triangles chapter of Math Book: Hope it helps.This part is clear:
triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles)However, how does one determine which angles are equal? Except 90 degree angles of both triangles, i could not seem to follow how exactly other angles became equal?Both smaller triangles are similar to the large triangle. So they are similar to each other too.In triangles BAD and BCA, Angle BAD = BCA (90 degrees)and angle B is common in bothSo by AA, triangles BAD and BCA are similarSimilarly, in triangles BAD and ADC,angle BAD = ACD (90 degrees)and angle D is common in bothSo by AA, triangles BAD and ACD are similarSo triangle BAD is similar to triangle BCA and ACD so triangle BCA is similar to triangle ACD too.
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VeritasPrepKarishma wrote:Bunuel wrote:rvinodhini wrote:Hi I need a quick clarification on the concept of perpendicular bisector.With a perpendicular bisector, the bisector always crosses the line segment at right anglesIf any line cuts another line at 90 then it should be a perpendicular bisector right - i.e it divided the line segment into equal halves at 90 ?So here BC should be the perpendicular bisector and the AC=CD=3 right ?Please let me know what am missing here.I do understand the explanations in the other thread mentioned,but can someone clarify as to why AC is not the perpendicular bisector ? A perpendicular bisector is a line which cuts a line segment into two equal parts at 90°. So AC to be a perpendicular bisector of BD it must not only cut it at 90° (which it does) but also cut it into two equal parts. Now, in order AC to cut BD into two equal parts right triangle ABD must be isosceles, which, as it turns out after some math, it is not.Complete solution:In triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?A. 3 B. 15/4C. 5 D. 16/3 E. 20/3Attachment:splittingtriangle.jpgImportant property: perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle.Thus, the perpendicular AC divides right triangle ABD into two similar triangles ACB and DCA (which are also similar to big triangle ABD). Now, in these three triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles marked with red and blue on the diagram). So, \(\frac{CD}{AC}=\frac{AC}{BC}\) --& \(\frac{CD}{4}=\frac{4}{3}\) --& \(CD=\frac{16}{3}\).Answer: D.For more on this subject please check Triangles chapter of Math Book: Hope it helps.This part is clear:
triangles the ratio of the corresponding sides will be equal (corresponding sides are the sides opposite the same angles)However, how does one determine which angles are equal? Except 90 degree angles of both triangles, i could not seem to follow how exactly other angles became equal?Both smaller triangles are similar to the large triangle. So they are similar to each other too.In triangles BAD and BCA, Angle BAD = BCA (90 degrees)and angle B is common in bothSo by AA, triangles BAD and BCA are similarSimilarly, in triangles BAD and ADC,angle BAD = ACD (90 degrees)and angle D is common in bothSo by AA, triangles BAD and ACD are similarSo triangle BAD is similar to triangle BCA and ACD so triangle BCA is similar to triangle ACD too.Hi Karishma,Thanks for explaining this. I understand how 3 triangles are similar to each other. However how do we determine which side is similar to which in order to set up the ratio. for example, if we take two smaller triangles - Triangle ABC and ADC....I think side AB is corresponding to AD, BC is corresponding to CD and AC is common. Is that correct? if yes, how do I find which side is corresponding in the smaller triangle with respect to the bigger triangle?Many thanks in advance.
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gmatkiller88 wrote:Thanks for explaining this. I understand how 3 triangles are similar to each other. However how do we determine which side is similar to which in order to set up the ratio. for example, if we take two smaller triangles - Triangle ABC and ADC....I think side AB is corresponding to AD, BC is corresponding to CD and AC is common. Is that correct? if yes, how do I find which side is corresponding in the smaller triangle with respect to the bigger triangle?Many thanks in advance.There is a very simple method of finding out the corresponding sides in similar triangles. Say, you have two triangles ABC and DEF. You find by AA that the triangles are similar. All you have to do is name the triangles the way the angles are equal. Say angle A = angle E, angle B = angle D and and hence angle C = angle F. Then we write:triangle ABC is similar to triangle EDF.Now you have the corresponding sides. That is, AB/ED = BC/DF = AC/EFIn the question above,triangles BAD and BCA are similarSo BA/BC = AD/CA = BD/BAtriangles BAD and ACD are similarSo BA/AC = AD/CD = BD/ADtriangles BCA is similar to triangle ACDSo BC/AC = CA/CD = BA/AD
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enigma123 wrote:Attachment:The attachment Triangle.jpg is no longer availableIn triangle ABC, if BC = 3 and AC = 4, then what is the length of segment CD?A. 3 B. 15/4C. 5 D. 16/3 E. 20/3such questions where there is a 90 degree angle in
a triangle can always be solved easily by drawing a circle .as we draw circle and extend AC to point E we will get a triangle BED , exact copy of triangle BAD --Eq1. as shown in the attached image , Triangle ABC and Triangle ECD are similar as all angles are equal. so
\(\frac{EC}{BC}
\frac{CD}{AC} = \frac{ED}{AB}\) 3x=4y --& \(x=\frac{4}{3}*y\)y=4 as shown in eq1so x= \(\frac{16}{3}\)
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