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 Geometry_proofs/33977: Given: Triangle ABC, AD=CB, and AC>DB Prove: M angle ADC> M angle DCB1 solutions Answer 21617 by venugopalramana(3286)   on 2006-04-29 05:49:02 (Show Source): You can put this solution on YOUR website!PLEASE BE CLEAR IN DESCRIPTION OR GIVE A SKETCH.WHERE IS D?THEN ONLY WE CAN HLP YOU
 Matrices-and-determiminant/33970: (1)A=[-3 2] B= [0 2] (find AB and BA if possible) [2 -2] [-2 4] (2)A=[-2 3] B=[ 3 ] (find AB if possible) [1 -2] [-2 ] [0 0] (3)A=[2 -2 4] B=[2 1 -3 0] [1 0 -1] [0 -2 1 -2] (find AB if possible) [2 1 3] [1 -1 0 2] I am having trouble with these problems i hope you can help me thank you Brittny1 solutions Answer 21616 by venugopalramana(3286)   on 2006-04-29 05:43:11 (Show Source): You can put this solution on YOUR website!THEY ARE ALL OF SAME MODEL .I WILL SHOW YOU ONE .YOU TRY OTHERS.COME BACK IF IN DIFFICULTY. (1)A=[-3 2] B= [0 2] (find AB and BA if possible) [2 -2] [-2 4] FIRST CHECK COMPATABILITY...THAT IS FIND THE ORDER OF A (M,N) = (2,2) MATRIX HERE. AND ORDER OF B (P,Q)= (2,2)..N SHOULD EQUAL P .THEN WE SAY THEY ARE COMPATABLE AND THEY CAN BE MULTIPLIED.OTHERWISE NOT.THE ANSWER WILL BE C (M,Q) MATRIX THEN .HERE WE HAVE N=2=P...SO THEY ARE COMPATABLE AND CAN BE MULTIPLIED.THE NSWER WILL BE M=2 AND Q=2...THAT IS C(2,2)MATRIX NEXT HOW DO WE FIND C =.... E11,E12 E21,E22 E11 IS OBTAINED BY MULTIPLYING EACH ELEMENT OF ROW 1 OF A WITH EACH ELEMENT OF COLUMN 1 OF B AND ADDING THEM UP.THAT IS E11=-3*0+2*-2=-4 E12 IS OBTAINED BY MULTIPLYING EACH ELEMENT OF ROW 1 OF A WITH EACH ELEMENT OF COLUMN 2 OF B AND ADDING THEM UP.THAT IS E12=-3*2+2*4=2 E21 IS OBTAINED BY MULTIPLYING EACH ELEMENT OF ROW 2 OF A WITH EACH ELEMENT OF COLUMN 1 OF B AND ADDING THEM UP.THAT IS E21=2*0+-2*-2=4 E22 IS OBTAINED BY MULTIPLYING EACH ELEMENT OF ROW 2 OF A WITH EACH ELEMENT OF COLUMN 2 OF B AND ADDING THEM UP.THAT IS E22=2*2+-2*4=-4 HENCE THE ANSWER IS C= -4,2 4,-4 (2)A=[-2 3] B=[ 3 ] (find AB if possible) [1 -2] [-2 ] [0 0] (3)A=[2 -2 4] B=[2 1 -3 0] [1 0 -1] [0 -2 1 -2] (find AB if possible) [2 1 3] [1 -1 0 2]
 Geometry_proofs/33947: Prove: If an isosceles triangle has an altitude from the vertex to the base, then the altitude bisects the vertex angle. Given: Triangle ABC is isoceles; Segment CD is the altitude to base Segment AB To Prove: Segment CD bisects angle C Plan: ?? Proof:?? I need to know what the plan would be and the proof, the statements and reasons.1 solutions Answer 21615 by venugopalramana(3286)   on 2006-04-29 05:23:54 (Show Source): You can put this solution on YOUR website!Prove: If an isosceles triangle has an altitude from the vertex to the base, then the altitude bisects the vertex angle. Given: Triangle ABC is isoceles; AC=CB Segment CD is the altitude to base Segment AB... WHERE D IS ON AB. HENCE ANGLE CDA=90=ANGLE CDB To Prove: Segment CD bisects angle C THAT IS ANGLE DCA=ANGLE DCB Plan: ??. WE SHALL TRY TO PROVE BOTH TRIANGLES CDA AND CDB TO BE CONGRUENT Proof:?? IN TRIANGLES CDA AND CDB WE HAVE AC=CB........GIVEN CD=CD......SAME COMMON SIDE ANGLE CDA=90=ANGLE CDB........GIVEN HENCE THE 2 TRIANGLES ARE CONGRUENT BY THEOREM ON CONGUENCE OF RIGHT TRIANGLES HENCE ANGLE DCA=ANGLE DCB
 Trigonometry-basics/33860: cos theta = 5/12 and tan theta is negative, find the exact value for ctn theta.1 solutions Answer 21614 by venugopalramana(3286)   on 2006-04-29 05:14:21 (Show Source): You can put this solution on YOUR website!cos theta = 5/12 and tan theta is negative, find the exact value for ctn theta. USING ALL SILVER TEA CUPS,SINCE COS IS +VE AND TAN IS -VE,THETA IS IN IV Q.WHERE CTAN IS ALSO -VE. HENCE COMPLETING TRIANGLE 12^2=5^2+X^2 X=SQRT(144-25)=SQRT(119) HENCE COTAN(THETA)= 5/-SQRT(119)=-(5SQRT(119)/119)
 test/35133: We are working on combinations and permutaions which involve factorials most of the time. How many 4 letter combinations can be made from the letters f,g,h,i,j & k. No letters can repeat themselves and the letter do not have to form a word. How do you solve this?1 solutions Answer 21613 by venugopalramana(3286)   on 2006-04-29 05:08:30 (Show Source): You can put this solution on YOUR website!We are working on combinations and permutaions which involve factorials most of the time. How many 4 letter combinations can be made from the letters f,g,h,i,j & k. No letters can repeat themselves and the letter do not have to form a word. How do you solve this? COUNT TOTAL NUMBER OF LETTERS=6 TO MAKE 4 LETTER COMBINATIONS YOU SAID THAT IS NOT PROPER I THINK..THAT MEANS A TEAM OF 4 PLAYERS FROM 6 PERSONS.THAT MEANS ABCD..OR...BCDA...OR...CDBA..ETC...ARE ALL SAME TEAM.THE ARRANGEMENT IS NOT IMPORATANT.THEN ONLY IT IS CALLED COMBINATION.THEN THE ANSWER IS 6C4..THE FOMULA FOR NCR = N!/{(R!)*(N-R)!} SO WE GET 6!/2!*(4-2)!=6!/2!*4!=6*5*4*3*2*1/2*1*4*3*2*1=15... BUT I THINK YOU WANT ARRANGEMENTS .THEN ITS ANSWER IS 6P4..THE FORMULA FOR NPR=N!/(N-R)! 6P4=6!/(6-4)!=6!/2!=360
 Trigonometry-basics/33858: cos theta = 5/12 and tan theta is negative, find the exact valcue for sin theta.1 solutions Answer 21612 by venugopalramana(3286)   on 2006-04-29 04:59:59 (Show Source): You can put this solution on YOUR website!cos theta = 5/12 and tan theta is negative, find the exact valcue for sin theta. USING ALL SILVER TEA CUPS,SINCE COS IS +VE AND TAN IS -VE,THETA IS IN IV Q.WHERE SINE IS ALSO -VE. HENCE COMPLETING TRIANGLE 12^2=5^2+X^2 X=SQRT(144-25)=SQRT(119) HENCE SIN(THETA)= -SQRT(119)/12
 Polygons/33872: I have constructed a pentagon with a base of 529.2 ft. The other sides follow clockwise in this order: 57.6 ft, 248 ft, 323.4 ft, and 48.5 ft. How do I find the measure of a perpendicular drawn from the base to the opposite vertex? Base = 529.2 ft Vertex formed by sides measuring 248 ft and 323.4 ft.1 solutions Answer 21606 by venugopalramana(3286)   on 2006-04-29 01:17:20 (Show Source): You can put this solution on YOUR website!I have constructed a pentagon with a base of 529.2 ft.=AB SAY The other sides follow clockwise in this order: 57.6 ft=BC, 248 ft=CD, 323.4 ft=DE, and 48.5 ft=EA. How do I find the measure of a perpendicular drawn from..NO ..IT IS TO the base AB to..NO IT IS FROM the opposite vertex D? Base = 529.2 ft Vertex formed by sides measuring 248 ft and 323.4 ft. USING ABOVE NOMENCLATURE..... TAKE A COMPASS WITH SOME CONVENIENT RADIUS BIG ENOUGH SO THAT WHEN YOU DRAW AN ARC WITH THAT RADIUS KEEPING THE CENTRE AT D,THE ARC CUTS AB OR EXTENDED AB ON EITHER SIDE AT 2 POINTS ...SAY...P AND Q. NOW DRAW 2 ARCS WITH SAME RADIUS ,KEEPING THE CENTRE AT P AND Q..ON TO THE OTHER SIDE OF AB .LET THE 2 ARCS CUT AT R.JOIN DR.LET DR CUT AB AT Z.NOW DZ IS THE PERPENDICULAR FROM VERTEX...D TO BASE....AB.
 Rational-functions/33925: The radius of the circular top or bottom of a tin can with a surface area S and a height h is given by -h +√h^2 + .64s r =------------------------------- 2 What radius should be used to make a can with a height of 35 inches and a surface area of 11,025 square inches? A) 30 inches B) 15 inches C) 63 inches D) 28 inches Actual problem here: http://www.ccaurora.edu/mat090/Tests/test_chapter_8_files/image052.gif 1 solutions Answer 21602 by venugopalramana(3286)   on 2006-04-29 00:56:48 (Show Source): You can put this solution on YOUR website! X=28.................D IS THE ANSWER
 Volume/35408: i need to know what the volume measures and amount of space on an object is for a rectangular prism1 solutions Answer 21479 by venugopalramana(3286)   on 2006-04-28 12:17:11 (Show Source): You can put this solution on YOUR website!VOLUME OF A PRISM =BASE AREA *HEIGHT IF BASE IS RECTANGLE THEN VOLUME = L*B*H
 Volume/35409: i need to know the volume measure of an triangular prism is and the amount of space is1 solutions Answer 21477 by venugopalramana(3286)   on 2006-04-28 12:15:34 (Show Source): You can put this solution on YOUR website!VOLUME OF PRISM =BASE AREA *HEIGHT=TAKING BASE AS EQUILATERAL TRIANGLE VOLUME =A*A*H*SQRT(3)/4
 Matrices-and-determiminant/35276: factor the matrix A into a product of elementary matrices. A= [1 1] [2 1]1 solutions Answer 21475 by venugopalramana(3286)   on 2006-04-28 12:02:52 (Show Source): You can put this solution on YOUR website!A= [1 1] [2 1].... METHOD...WRITE IT IN NORMAL FORM.......1,0 .......................................0,1 STEP.1 USE....R21(-2)...THAT IS NEW ROW 2=OLD ROW 2 + (OLD ROW 1)*(-2)...WE GET 1,1 0,-1 STEP.2 USE R12(1).....THAT IS NEW ROW 1=OLD ROW 1 + OLD ROW 2 *1...WE GET.. 1,0 0,-1 STEP.3. USE C2(-1)....THATIS NEW COLUMN 2 =OLD COLUMN 2*(-1) 1,0 0,1 HENCE WE GOT I = R12(1)*R21(-2)*A*C2(-1) HENCE A = {R21(-2)}^(-1)*{R12(1)}^(-1)}*I*{C2(-1)}^(-1) A = {R21(-2)}^(-1)*{R12(1)}^(-1)}*{C2(-1)}^(-1) A = {R21(2)}*{R12(-1)}*{C2(-1)} NOW WE KNOW THAT I = 1,0 0,1 R21(2) = 1,0 2,1 R12(-1) = 1,-1 0,1 C2(-1) = 1,0 0,-1 HENCE A = |1,0|*|1,-1|*|1,.0| ..……|2,1|*|0,.1|*|0,-1| THUS A IS EXPRESSED AS PRODUCT OF ELEMENTARY MATRICES.YOU CAN MULTIPLY AND CHECK HOPE YOU ARE TAUGHT THE PROCEDURE AND YOU KNOW IT.OTHERWISE PLEASE COME BACK AND WE SHALL EXPLAIN
 Volume/35333: I need as quick as possible for HW: FInd the volume of a regular octahedron with sides equalling fifty square root of three cubic units? I don't know what to do please help me1 solutions Answer 21472 by venugopalramana(3286)   on 2006-04-28 08:16:46 (Show Source): You can put this solution on YOUR website!I need as quick as possible for HW: FInd the volume of a regular octahedron with sides equalling fifty square root of three cubic units? I don't know what to do please help me FORMULA IS VOLUME =SQRT(2)*S^3/3={SQRT(2)/3}*{52*SQRT(3)}^3=344418
 Trigonometry-basics/35340: This question is from textbook Algebra 1 In a right triangle one side is 8 ft and the other side is 6 ft. What are the measures of angle A and B if C is 90 degrees? What is the hypotenuse? 1 solutions Answer 21471 by venugopalramana(3286)   on 2006-04-28 08:05:33 (Show Source): You can put this solution on YOUR website!In a right triangle one side is 8 ft and the other side is 6 ft. What are the measures of angle A and B if C is 90 degrees? What is the hypotenuse? HYPOTENUSE IS THE SIDE OPPOSITE RIGHT ANGLE OR 90 DEGREE ANGLE WHICH IS C HERE.SO AB IS THE HYPOTENUSE.AS PER PYTHOGARUS THEOREM AC^2=AB^2+BC^2=8^2+6^2=64+36=100 AC=10
 Polygons/35365: What is the foumula to find the sum of interior angles of any polygon?1 solutions Answer 21470 by venugopalramana(3286)   on 2006-04-28 08:02:40 (Show Source): You can put this solution on YOUR website!sum of interior angles of any polygon OF N SIDES=(2*N-4)*90 DEGREES
 Polygons/35366: What is the formula to find the sum of interior angles of any polygon?1 solutions Answer 21469 by venugopalramana(3286)   on 2006-04-28 07:55:02 (Show Source): You can put this solution on YOUR website!sum of interior angles of any polygon WITH N SIDES =(2*N-4)*90 DEGREES EX....N=4...QUADRILATERAL.......(2*4-4)*90=4*90=360
 Trigonometry-basics/35367: how do you solve: is triangle DEF, d=5 e=6 f=3 then COS E=???? i have the right set up i believe but i cant get the right answer helpp!!!1 solutions Answer 21468 by venugopalramana(3286)   on 2006-04-28 07:53:01 (Show Source): You can put this solution on YOUR website!is triangle DEF, d=5 e=6 f=3 then COS E=???? i have the right set up i believe but i cant get the right answer helpp!!! FORMULA IS e^2=d^2+f^2+2df*COS(E) 6^2=5^2+3^2+2*5*3*COS(E) COS(E)=(36-25-9)/(2*5*3)=1/6
 Quadratic-relations-and-conic-sections/35337: 1. graph x-3= -1/8(y+2)^2. Write the coordinates of the vertex and the focus and the equation of the directrix. 2.Find all solution to each system of equations algerbaiclly. x^2 + y^2 =25 3x^2 + 2y^2 = 59 3. Write the standard equations of the conic section with the given characteristics hyperbola with foci at (5,-7) and (14,0) and vertices at (12,0) and (4,0)1 solutions Answer 21466 by venugopalramana(3286)   on 2006-04-28 02:05:58 (Show Source): You can put this solution on YOUR website! 1. graph x-3= -I/8 (y=2)^2. Write the coordinates of the vertex and > the > > > focus and the equation of the directrix. X-3=-(1/8)(Y-2)^2 (Y-2)^2=-8(X-3) COMPARING WITH STD.EQN.OF PARABOLA (Y-K)^2=4A(X-H)^2...WE GET VERTEX=(H,K)=(3,2) A=-2 FOCUS =(H+A,K)=(3-2,2)=(1,2) DIRECTIX IS ... X-H+A=0...THAT IS X-3-2=0...OR...X=5 GRAPH IS GIVEN BELOW > > > 2.Find all solution to each system of equations algerbaiclly. > > > x^2 + y^2 =25 SINCE SUM OF TWO SQUARES = 25 ,AND SQUARES ARE ALWYS POSITIVE ,WE GET THE SOLUTION SETS AS X^2<=25 AND Y^2<=25....THAT IS 1.0<=|X|<=5 AND 0<=|Y|<=5 > > > 3x^2 + 2y^2 = 59...YOU CAN DO THE SAME WAY AS ABOVE 3X^2<=59 AND 2Y^2<=59..ETC....... > > > 3. Write the standard equations of the conic section with the given > > > characteristics > > > hyperbola with foci at (5,-7) and (14,0) and vertices at (12,0) and > > > (4,0) STANDARD EQN.OF HYPERBOLA IS (X-H)^2/A^2 -(Y-K)^2/B^2 =1...WHERE CENTRE IS (H,K)...WE ARE GIVEN VERTICES ARE AT (12,0) AND (4,0)...SO CENTRE IS AT (12+4)/0,(0+0)/2=(8,0) HENCE H=8....K=0 TRANSVERSE AXIS IS Y=K=0 LENGTH OF TRANVERSE AXIS =2A=16 A=8 CONJUGATE AXIS IS X=H=8 PLEASE CHECK DATA....CENTRE WHICH IS MID POINT OF FOCI AND ALSO MIDPOINT OF VERTICES ARE NOT TALLYING
 Geometric_formulas/35222: Assume that the only area formula you know is the formula for the area of a triangle. Explain how to derive the formula for the area of the trapezoid1 solutions Answer 21418 by venugopalramana(3286)   on 2006-04-27 12:33:39 (Show Source): You can put this solution on YOUR website!Assume that the only area formula you know is the formula for the area of a triangle. Explain how to derive the formula for the area of the trapezoid LET ABCD BE THE TRAPEZOID WITH AB PARALLEL TO CD.SEE DIAGRAM BELOW. FROM C AND D DRAW CF AND DE PERPENDICULARS TO AB MEETING AB AT F AND E JOIN EC NOW AREA OF TRAPEZIUM =AREA OF TRIANGLE ADE+AREA OF TRIANGLE DEC+AREA OF TRIANGLE EFC+TRIANGLE CFB....ALL ARE RIGHT ANGLED TRIANGLES HENCE AREA =BASE*HT/2..HT.FOR ALL TRIANGLES =H=DE=CF=DISTANCE BETWEN PARALLEL SIDES OF TRAPEZIUM AREA=(1/2)*H{AE+DC+EF+FB)=(1/2)*H(AE+EF+FB+DC)=(1/2)*H(AB+DC) AREA OF TRAPEZIUM =(1/2)(SUM OF PARALLEL SIDES )(PERPENDICULAR DISTANCE BETWEEN THE 2 PARALLEL SIDES)
 Geometric_formulas/35223: If each dimension of a box is quadrupled, how are the surface area and the volume affected?1 solutions Answer 21416 by venugopalramana(3286)   on 2006-04-27 11:57:41 (Show Source): You can put this solution on YOUR website!If each dimension of a box is quadrupled, how are the surface area and the volume affected? SINCE AREA IS PROPORTIONAL TO SQUARE OF THE SIDE...AREA WILL INCREASE BY 4^2=16 TIMES SINCE VOLUME IS PROPORTIONAL TO CUBE OF THE SIDE...VOLUME WILL INCREASE BY 4^3=64 TIMES