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# Recent problems solved by 'venugopalramana'

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 Money_Word_Problems/33747: Calculate the NPV(Net Present Value) for a project costing \$250,000 which will then provide \$25,000 annually for 25 years. The discount rate is 5%. What would the NPV be if the inflows were \$20.000 annually for only 20 years?1 solutions Answer 20163 by venugopalramana(3286)   on 2006-04-16 11:45:48 (Show Source): You can put this solution on YOUR website!Calculate the NPV(Net Present Value) for a project costing \$250,000 which will then provide \$25,000 annually for 25 years. The discount rate is 5%. What would the NPV be if the inflows were \$20.000 annually for only 20 years? TABULATE AS FOLLOWS IN XL WORK SHEET WHICH IS SELF EXPLANATORY.IF YOU GO TO FUNCTIONS-FINANCIALS AT THE END AND GIVE THE DISCOUNT RATE OF 0.05 AND NET INFLOW COLUMN AS THE DATA,YOU WILL GET NPV AT THE END.THE FORMULA IN XL IS. =NPV(0.05,D256:D281)WHERE 0.05 IS THE DISCOUNT RATE GIVEN AS 5% PER YEAR AND D256:D281 IS THE COLUMN WHERE NETINFLOW IS DEPICTED.IF YOU WANT TO KNOW MATHMATICAL FORMULA FOR THIS PLEASE COME BACK AND WE SHALL EXPLAIN YEAR INFLOW OUTFLOW NETINFLOW YEAR INFLOW OUTFLOW NETINFLOW 1 250000 -250000... 1 250000 -250000 2 25000 25000... 2 20000 20000 3 25000 25000... 3 20000 20000 4 25000 25000... 4 20000 20000 5 25000 25000... 5 20000 20000 6 25000 25000... 6 20000 20000 7 25000 25000... 7 20000 20000 8 25000 25000... 8 20000 20000 9 25000 25000... 9 20000 20000 10 25000 25000... 10 20000 20000 11 25000 25000... 11 20000 20000 12 25000 25000... 12 20000 20000 13 25000 25000... 13 20000 20000 14 25000 25000... 14 20000 20000 15 25000 25000... 15 20000 20000 16 25000 25000... 16 20000 20000 17 25000 25000... 17 20000 20000 18 25000 25000... 18 20000 20000 19 25000 25000... 19 20000 20000 20 25000 25000... 20 20000 20000 21 25000 25000... 21 20000 20000 22 25000 25000 23 25000 25000 24 25000 25000 25 25000 25000 26 25000 25000 NPV \$97,474.87 NPV-NEGATIVE (\$719.80)
 Miscellaneous_Word_Problems/32301: A Regualr polygon of n sides inscribed in a circle of radius r has 3r squared. What is the value of n?1 solutions Answer 20162 by venugopalramana(3286)   on 2006-04-16 11:26:59 (Show Source): You can put this solution on YOUR website!A Regualr polygon of n sides inscribed in a circle of radius r has 3r squared. What is the value of n? YOU MEAN AREA =3R^2...ASSUMING SO... THERE WILL BE N TRIAGLES WHEN N VERTICES ARE JOINED TO THE CENTRE OF THE CIRCUM CIRCLE. SO EACH RIANGLE HAS AREA OF 3R^2/N NOW AREA OF EACH TRIANGLE IS GIVEN BY 0.5*R^2*SIN(X)...WHERE X IS THE ANGLE SUBTENDED BY ONE SIDE OF REGULAR POLYGON AT CENTRE OF CIRCUMCIRCLE. HENCE 0.5*R^2*SIN(X)=3R^2/N SIN(X)=3/(N*0.5)= 6/N NOW IF THERE ARE N SIDES THEN EACH SIDE SUBTENDS 360/N ANGLE AT CENTRE .HENCE X=2PI/N...OR....N=2PI/X SIN(X)=6*X/2PI=3X/PI WE FIND THAT AT X=30...OR...PI/6,WE HAVE SIN(X)=SIN(30)=SIN(PI/6)=0.5 AND 3X/PI=(3*PI)/(6*PI)=0.5 HENCE X=PI/6 IS THE SOLUTION HENCE N=2PI/X=(2PI)/(PI/6)=2*6=12 HENCE THE REGULR POLYGON IS 12 SIDED
 Probability-and-statistics/31598: A box contains 2 fifty paise coins,5 twenty five paise coins and a certain fixed number(N<=2) of ten and five paise coins.If five coins are taken out of the box at random,then the probability that the total value of these 5 coins is less than one rupee and fifty paise is which of the following:- A.10*(N+2)/(N+7C5) B.1-(10*(N+2)/(N+7C5)) C.5(N+3)/(N+5C2) D.1-(5*(N+3)/(N+5C2))1 solutions Answer 20154 by venugopalramana(3286)   on 2006-04-16 07:54:23 (Show Source): You can put this solution on YOUR website!A box contains 2 fifty paise coins,5 twenty five paise coins and a certain fixed number(N<=2) of ten and five paise coins.If five coins are taken out of the box at random,then the probability that the total value of these 5 coins is less than one rupee and fifty paise is which of the following:- SINCE IT IS EASIER TO FIND THE NUMBER OF WAYS BY WHICH THE SUM OF 5 COINS CAN EQUAL OR EXCEED RS.1.50,LET US FIND IT OUT AND SUBTRACT THE RESULTANT PROBABILITY FROM ONE TO GET THE ANSWER WE HAVE IN ALL 2+5+N=N+7 COINS 5 CAN BE TAKEN IN ...(N+7)C5 WAYS CASE 1.....1*50+4*25=1.50....IN..(2C1)(5C4)=2*5=10 CASE 2.....2*50+3*25=1.75....IN..(2C2)(5C3)=1*10=10 CASE 3.....2*50+2*25+1*5 OR 10=1.55 OR 1.60...IN...(2C2)(5C2)(NC1)=1*10*N=10N TOTAL NUMBER .....................=10+10+10N=20+10N PROBABILITY OF GETTING 1.50 OR MORE =(20+10N)/{(N+7)C5} PROBABILITY OF GETTING LESS THAN 1.50 =1- [(20+10N)/{(N+7)C5}]
 Complex_Numbers/31805: 1.|3-5i| (3-5i)-(3+5i) (3-3)(-5i-5)i 0-10i For this problem i took the absolute value and made it in two seperate problems. After i did that i used foil, and seperated again problems. AFter that i solved what was in the parenthesis, then i 0-10i for the answer. I dont i did it right i asked my brother he said it was wrong i dont know if it is right.1 solutions Answer 20153 by venugopalramana(3286)   on 2006-04-16 06:49:43 (Show Source): You can put this solution on YOUR website!.|3-5i| ABSOLUTE VALUE OR MOD OF A COMPLEX NUMBER X+iY IS GIVEN BY SQUAREROOT OF (X^2+Y^2) AND IS SYMBOLISED AS |X+iY|=SQRT(X^2+Y^2)...HENCE WE GET |3=5i|=SQRT.(3^2+5^2)=SQRT(9+25)=SQRT(34)
 Distributive-associative-commutative-properties/31644: Show that every polynomial of degree 1,2 or 4 in Z2[t] has a root in Z2[t](t^4+t=1)1 solutions Answer 20152 by venugopalramana(3286)   on 2006-04-16 06:42:04 (Show Source): You can put this solution on YOUR website!I DONT THINK THE PROBLEM IS CORRECT IN Z2 F(T)=T^2+T+1...IS A SECOND DEGREE POLYNOMIAL...IT HAS NO ZERO IN REAL FIELD OF Z2, AND IS NOT REDUCIBLE.PLEASE CHECK BACK WITH YOUR BOOK/QUESTION SOURCE OR CLASS TEACHER
 Numeric_Fractions/33678: This question is from textbook numeric Fraction please ! could you help me in these problems 1) The denominator of a fraction is three more than the numerator. If the numerator is doubled and the denominator is increased by nine, the resulting fraction is equal to ½. What is the original fraction? 2) The denominator of a fraction is three more than numerator. If the numerator is tripled and the denominator is increased by 17, the resulting fraction is ½. What is the original fraction? 3) Bob, Celia, Sam and Daniel contributed money to buy their boss a retirement gift. Sam and Daniel each gave the same amount of money. Celia gave \$12 more than Daniel. Bob gave half as much as Celia gave. If the four workers gave a total of \$81, how much did Sam give? From your faithfully student1 solutions Answer 20151 by venugopalramana(3286)   on 2006-04-16 06:30:18 (Show Source): You can put this solution on YOUR website!1) The denominator(DR) of a fraction is three more than the numerator(NR). If the numerator is doubled and the denominator is increased by nine, the resulting fraction is equal to ½. What is the original fraction? LET NR =N 3 MORE =N+3=DR...SO FRACTION IS N/(N+3) NR IS DOUBLED...WE GET 2N DR INCREASED BY 9...WE GET DR+9=N+3+9=N+12 NEW FRACTION=2N/(N+12)=1/2 2*2N=N+12 4N-N=12.....3N=12.....N=4 HENCE ORIGINAL FRACTION =N/(N+3)=4/7 2) The denominator of a fraction is three more than numerator. If the numerator is tripled and the denominator is increased by 17, the resulting fraction is ½. What is the original fraction? PROCEDING IN SAME WAY AS ABOVE ORIGINAL FRACTION =N/(N+3)...AND 3N/(N+3+17)=1/2 2*3N=N+20 6N-N=20 N=4 ORIGINAL FRACTION =4/7 3) Bob, Celia, Sam and Daniel contributed money to buy their boss a retirement gift. Sam and Daniel each gave the same amount of money. Celia gave \$12 more than Daniel. Bob gave half as much as Celia gave. If the four workers gave a total of \$81, how much did Sam give? LSET SAM GIVE =S DANIEL GAVE =S 12 MORE =S+12=CELIA GAVE HALF THIS =(S+12)/2=BOB GAVE TOTAL GIVEN =S+S+S+12+(S+12)/2=81 {2(3S+12)+S+12}=2*81=162 6S+24+S+12=162 7S=162-36=126 S=126/7=18 HENCE SAM GAVE 18 \$
 Polynomials-and-rational-expressions/33694: Find a polynomial equation of lowest degree with integer coefficients and has 0, 3, and plus/minus sq rt of 2 as zeros. Please, anyone help.1 solutions Answer 20128 by venugopalramana(3286)   on 2006-04-15 22:09:07 (Show Source): You can put this solution on YOUR website!Find a polynomial equation of lowest degree with integer coefficients and has 0, 3, and plus/minus sq rt of 2 as zeros. Please, anyone help. ROOTS ARE ...0,3,SQRT(2),-SQRT(2)...SO EQN IS (X-0)(X-3)(X-SQRT2)(X+SQRT2)=0 (X^2-3X)(X^2-2)=0 X^4-3X^3-2X^2+6X=0
 Graphs/33705: Give the coordinates of the vertex. 7) y = (x - 2)^2 A) ( -2, 0) B) (0, -2) C) (0, 2) D) ( 2, 0)1 solutions Answer 20124 by venugopalramana(3286)   on 2006-04-15 21:35:40 (Show Source): You can put this solution on YOUR website! y = (x - 2)^2 VERTEX IS WHERE Y BECOMES MAXIMUM OR MINIMUM...WE FIND THAT IT BECOMES MINIMUM HERE OF ZERO WHEN X=2...HENCE VERTEX IS X=2 AND Y=0...OR...(2,0)
 Quadratic_Equations/33709: 6) If we apply the quadratic formula and find that the value of b^2 - 4ac equals zero, what can we conclude about the solutions? A) The equation has no real number solutions. B) The equation has exactly one irrational solution. C) The equation has two different rational solutions D) The equation has exactly one rational solution.1 solutions Answer 20123 by venugopalramana(3286)   on 2006-04-15 21:33:02 (Show Source): You can put this solution on YOUR website! If we apply the quadratic formula and find that the value of b^2 - 4ac equals zero, what can we conclude about the solutions? A) The equation has no real number solutions. B) The equation has exactly one irrational solution. C) The equation has two different rational solutions D) The equation has exactly one rational solution. X=-B/2A...... The equation has exactly one rational solution. D IS THE ANSWER..
 Linear-equations/33713: Determine the equation of each line (write in slope-intercept form): 1. The line through (-1, 3) with a slope 1/3. 2. The line through (3, 5) that is parallel to the x-axis. 3. The line through (0, 6) that is parallel to the line 3x + 5y = 15. 4. The line through (8, 0) that is perpendicular to the line x + y = 3. 1 solutions Answer 20122 by venugopalramana(3286)   on 2006-04-15 21:30:07 (Show Source): You can put this solution on YOUR website!Graphs/24460: find the slope,m,and the y intercept,b,of the line equation 2y=28 1 solutions Answer 13033 by venugopalramana(1088) About Me on 2006-01-15 10:53:09 (Show Source): SEE THE FOLLOWING WHICH IS SIMILAR TO YOUR PROBLEM AND DO GIVEN: · There is a line (L1) that passes through the points (8,-3) and (3,3/4). · There is another line (L2) with slope m=2/3 that intersects L1 at the point (-4,6). · A third line (L3) is parallel to L2 that passes through the (7,-13 1/2). · Yet another line (L4) is perpendicular to L3, and passes through the point (1/2,5 2/3). · The fifth line (L5) has the equation 2/5y-6/10x=24/5. Using whatever method, find the following: 2. The point of intersection of L1 and L3 3. The point of intersection of L1 and L4 4. The point of intersection of L1 and L5 5. The point of intersection of L2 and L3 6. The point of intersection of L2 and L4 7. The point of intersection of L2 and L5 8. The point of intersection of L3 and L4 9. The point of intersection of L3 and L5 10. The point of intersection of L4 and L5 PLEASE NOTE THE FOLLOWING FORMULAE FOR EQUATION OF A STRAIGHT LINE: slope(m)and intercept(c) form...y=mx+c point (x1,y1) and slope (m) form...y-y1=m(x-x1) two point (x1,y1)and(x2,y2)form..................... y-y1=((y2-y1)/(x2-x1))*(x-x1) standard linear form..ax+by+c=0..here by transforming we get by=-ax-c..or y=(-a/b)x+(-c/b)..comparing with slope intercept form we get ...slope = -a/b and intercept = -c/b ***************************************************** line (L1) that passes through the points (8,-3) and (3,3/4). eqn.of L1..y-(-3)=((3/4+3)/(3-8))*(x-8) y+3=((15/4(-5)))(x-8)=(-3/4)(x-8) or multiplying with 4 throughout 4y+12=-3x+24 3x+4y+12-24=0 3x+4y-12=0.........L1 There is another line (L2) with slope m=2/3 that intersects L1 at the point (-4,6). This means (-4,6)lies on both L1 and L2.(you can check the eqn.of L1 we got by substituting this point in equation of L1 and see whether it is satisfied).So eqn.of L2... y-6=(2/3)(x+4)..multiplying with 3 throughout.. 3y-18=2x+8 -2x+3y-26=0.........L2 A third line (L3) is parallel to L2 that passes through the (7,-13 1/2). lines are parallel mean their slopes are same . so we keep coefficients of x and y same for both parallel lines and change the constant term only.. eqn.of L2 from above is ...-2x+3y-26=0.........L2 hence L3,its parallel will be ...-2x+3y+k=0..now it passes through (7,-13 1/2)=(7,-13.5)......substituting in L3..we get k -2*7+3*(-13.5)+k=0...or k=14+40.5=54.5..hence eqn.of L3 is........-2x+3y+54.5=0......................L3 Yet another line (L4) is perpendicular to L3, and passes through the point (1/2,5 2/3). lines are perpendicular when the product of their slopes is equal to -1..so ,we interchange coefficients of x and y from the first line and insert a negative sign to one of them and then change the constant term. L3 is........-2x+3y+54.5=0......................L3 hence L4,its perpendicular will be ..3x+2y+p=0...L4 this passes through (1/2,5 2/3)=(1/2,17/3).hence.. 3*1/2+2*17/3+p=0..or ..p= -77/6.so eqn.of L4 is 3x+2y-77/6=0..............L4 The fifth line (L5) has the equation 2/5y-6/10x=24/5. now to find a point of intersection means to find a point say P(x,y) which lies on both the lines ..that is, it satisfies both the equations..so we have to simply solve the 2 equations of the 2 lines for x and y to get their point of intersection.For example to find the point of intersection of L1 and L3 we have to solve for x and y the 2 equations.... 3x+4y-12=0.........L1....(1) and -2x+3y+40.5=0......L3.....(2) I TRUST YOU CAN CONTINUE FROM HERE TO GET THE ANSWERS.If you have any doubts or get into any difficulty ,please ask me.
 Graphs/33714: Write an equation in slope-intercept form of the line that passes through the given point and is parallel to the graph of each equation. y= -(two thirds)x+4; (1,-3)1 solutions Answer 20121 by venugopalramana(3286)   on 2006-04-15 21:29:25 (Show Source): You can put this solution on YOUR website!Graphs/24460: find the slope,m,and the y intercept,b,of the line equation 2y=28 1 solutions Answer 13033 by venugopalramana(1088) About Me on 2006-01-15 10:53:09 (Show Source): SEE THE FOLLOWING WHICH IS SIMILAR TO YOUR PROBLEM AND DO GIVEN: · There is a line (L1) that passes through the points (8,-3) and (3,3/4). · There is another line (L2) with slope m=2/3 that intersects L1 at the point (-4,6). · A third line (L3) is parallel to L2 that passes through the (7,-13 1/2). · Yet another line (L4) is perpendicular to L3, and passes through the point (1/2,5 2/3). · The fifth line (L5) has the equation 2/5y-6/10x=24/5. Using whatever method, find the following: 2. The point of intersection of L1 and L3 3. The point of intersection of L1 and L4 4. The point of intersection of L1 and L5 5. The point of intersection of L2 and L3 6. The point of intersection of L2 and L4 7. The point of intersection of L2 and L5 8. The point of intersection of L3 and L4 9. The point of intersection of L3 and L5 10. The point of intersection of L4 and L5 PLEASE NOTE THE FOLLOWING FORMULAE FOR EQUATION OF A STRAIGHT LINE: slope(m)and intercept(c) form...y=mx+c point (x1,y1) and slope (m) form...y-y1=m(x-x1) two point (x1,y1)and(x2,y2)form..................... y-y1=((y2-y1)/(x2-x1))*(x-x1) standard linear form..ax+by+c=0..here by transforming we get by=-ax-c..or y=(-a/b)x+(-c/b)..comparing with slope intercept form we get ...slope = -a/b and intercept = -c/b ***************************************************** line (L1) that passes through the points (8,-3) and (3,3/4). eqn.of L1..y-(-3)=((3/4+3)/(3-8))*(x-8) y+3=((15/4(-5)))(x-8)=(-3/4)(x-8) or multiplying with 4 throughout 4y+12=-3x+24 3x+4y+12-24=0 3x+4y-12=0.........L1 There is another line (L2) with slope m=2/3 that intersects L1 at the point (-4,6). This means (-4,6)lies on both L1 and L2.(you can check the eqn.of L1 we got by substituting this point in equation of L1 and see whether it is satisfied).So eqn.of L2... y-6=(2/3)(x+4)..multiplying with 3 throughout.. 3y-18=2x+8 -2x+3y-26=0.........L2 A third line (L3) is parallel to L2 that passes through the (7,-13 1/2). lines are parallel mean their slopes are same . so we keep coefficients of x and y same for both parallel lines and change the constant term only.. eqn.of L2 from above is ...-2x+3y-26=0.........L2 hence L3,its parallel will be ...-2x+3y+k=0..now it passes through (7,-13 1/2)=(7,-13.5)......substituting in L3..we get k -2*7+3*(-13.5)+k=0...or k=14+40.5=54.5..hence eqn.of L3 is........-2x+3y+54.5=0......................L3 Yet another line (L4) is perpendicular to L3, and passes through the point (1/2,5 2/3). lines are perpendicular when the product of their slopes is equal to -1..so ,we interchange coefficients of x and y from the first line and insert a negative sign to one of them and then change the constant term. L3 is........-2x+3y+54.5=0......................L3 hence L4,its perpendicular will be ..3x+2y+p=0...L4 this passes through (1/2,5 2/3)=(1/2,17/3).hence.. 3*1/2+2*17/3+p=0..or ..p= -77/6.so eqn.of L4 is 3x+2y-77/6=0..............L4 The fifth line (L5) has the equation 2/5y-6/10x=24/5. now to find a point of intersection means to find a point say P(x,y) which lies on both the lines ..that is, it satisfies both the equations..so we have to simply solve the 2 equations of the 2 lines for x and y to get their point of intersection.For example to find the point of intersection of L1 and L3 we have to solve for x and y the 2 equations.... 3x+4y-12=0.........L1....(1) and -2x+3y+40.5=0......L3.....(2) I TRUST YOU CAN CONTINUE FROM HERE TO GET THE ANSWERS.If you have any doubts or get into any difficulty ,please ask me.
 Circles/33691: A regular Pentagon ABCDE is incribed in circle O. Chords EC and DB intersect at F (not diameter). Chord DB is Extended to G (outside circle), and Tangent GA is drawn. find measure of angle BDE find measure of angle BFC find measure of angle AGD1 solutions Answer 20107 by venugopalramana(3286)   on 2006-04-15 12:27:55 (Show Source): You can put this solution on YOUR website!A regular Pentagon ABCDE is incribed in circle O. Chords EC and DB intersect at F (not diameter). Chord DB is Extended to G (outside circle), and Tangent GA is drawn. find measure of angle BDE find measure of angle BFC find measure of angle AGD HOPE YOU HAVE A FIGURE OF THE ABOVE..ASSUMING THAT I AM GIVING THE SOLUTION. SUM OF ALL INTERIOR ANGLES IN A PENTAGON =(2*5-4)*90 =540 SINCE IT IS REGULAR PENTAGON EACH ANGLE =540/5=108 AS PER GIVEN DATA,A,B,E,D LIE ON A CIRCLE...SO ABED IS A CYCLIC QUADRILATERAL..HENCE SUM OF OPPOSITE ANGLES=180=ANGLE EAB+ANGLE BDE BUT ANGLE EAB=108 AS PROVED ABOVE ----------------------------------------- HENCE ANGLE BDE=180-108=72 ------------------------------------------- SIMILARLY FROM CYCLIC QUADRILATERAL ECBA,WE GET ANGLE ECB=72 NOW IN A REGULAR PENTAGON EACH SIDE SPANS 1/5 OF PERIMETER OF CIRCUM CIRCLE WITH O AS CENTRE.HENCE EACH SIDE SUBTENDS AN ANGLE OF 360/5=72 AT CENTRE AND HALF OF THAT THAT IS 36 AT THE CIRCUMFERENCE.. HENCE ANGLE ADB=ANGLE DBC =36 NOW IN TRIANGLE CFB...WE HAVE... ANGLE FCB =ANGLE ECB =72..PROVED ABOVE. ANGLE FBC=ANGLE DBC=36......PROVED ABOVE -------------------------------------------- HENCE ANGLE BFC=180-72-36=72 -------------------------------- ABDE IS A CYCLIC QUADRILATERAL...AB IS A CHORD...AG IS A TANGENT TO CIRCUM CIRCLE..HENCE ANGLE BAG = ANGLE IN THE OPPOSITE SEGMENT =ANGLE ADB=36..PROVED ABOVE HENCE ANGLE EAG=ANGLE EAB + ANGLE BAG=108+36=144 IN QUADRILATERALDEAG...WE HAVE ANGLE AGD=360-ANGLE EDB-ANGLE EAG-ANGLE DEA =360-72-144-108=36 -------------------------- HENCE ANGLE AGD=36 -------------------------------
 logarithm/32682: For each of the following equations either prove that it is correct (by using the rules of logarithms and exponents) or else show that it is not correct (by finding numerical values for the variables that make the values of the two sides of the equation different). log (x/y) = log x/ log y log x - log y = log (x/y) log (2x) = 2 log x 2 log x = log (x^2) log x+1 / x+3 = log (x+1) - log (x+3) log (x square root x^2 + 1) = log x + 1/2 (x^2 +1) log (x^2 +1) = 2 log x + log 11 solutions Answer 20094 by venugopalramana(3286)   on 2006-04-15 08:39:33 (Show Source): You can put this solution on YOUR website!For each of the following equations either prove that it is correct (by using the rules of logarithms and exponents) or else show that it is not correct (by finding numerical values for the variables that make the values of the two sides of the equation different). log (x/y) = log x/ log y LHS....LOG(10/1)=LOG(10)=1...BUT RHS... LOG(10)/LOG(10)=1/0.....NOT CORRECT log x - log y = log (x/y) OK..AS PER FORMULA.. log (2x) = 2 log x LHS.....LOG(2*1)=LOG(2).....RHS....2*LOG(1)=2*0=0....NOT CORRECT 2 log x = log (x^2) CORRECT AS PER FORMULA LOG(X)^N=N*LOG(X) log {(x+1) / (x+3)} = log (x+1) - log (x+3)...I SUPPOSE THIS IS THE PROBLEM. CORRECT AS PER FORMULA...LOG(A/B)=LOG(A)-LOG(B) log {(x *square root (x^2 + 1)} = log x + 1/2 (x^2 +1) WRONG...LHS....LOG{1*SQRT(1^2+1)}=LOG{SQRT(2)} RHS......LOG(1)+(1/2)*(1+1)=0+1=1...WHICH IS NOT CORRECT. log (x^2 +1) = 2 log x + log 1 WRONG...LHS.....LOG(1^2+1)=LOG(2) RHS.....2*LOG(1)+LOG(1)=0+0=0.......WHICH IS NOT CORRECT...
 Quadratic_Equations/31497: Which of the below option's are right? If a+b+c=0,then 3a*x*x+2b*x+c=0 has A.atleast one root in [-1,0] B.atleast one root in [0,1] C.atleast one root in [-1,1] D.atleast one root in [0,2] 1 solutions Answer 20093 by venugopalramana(3286)   on 2006-04-15 08:06:14 (Show Source): You can put this solution on YOUR website!If a+b+c=0,then 3a*x*x+2b*x+c=0 has A.atleast one root in [-1,0] B.atleast one root in [0,1] C.atleast one root in [-1,1] D.atleast one root in [0,2] CONSIDER F(X)=AX^3+BX^2+CX... F(0)=0 F(1)=A+B+C=0...... GIVEN HENCE F(X) HASS 2 ZEROS AT EITHER END OF THE INTERVAL [0,1] DIFFERENTIATING F(X)... W.R.T.X,WE GET DF/DX=3AX^2+2BX+C....AS PER RTO;;ES THEOREM THIS MUST HAVE A ZERO IN THE INTERVAL [0,1]................SO...............B..IS CORRECT
 Equations/31546: i am trying to find out how you would find the super compaonent of a full package of \$85000. I thought you would do: xy=z x=unknown y=.09 z=85000 so what is x? and how do you do it and an algebra formula.1 solutions Answer 20092 by venugopalramana(3286)   on 2006-04-15 07:58:41 (Show Source): You can put this solution on YOUR website!XY=Z....................I Z=85000....................II Y=0.09.............................III SUBSTITUTING II AND III IN I X*0.09=85000 X=85000/9=9444.44
 Probability-and-statistics/31453: An elevator containing 5 people can stop at any one of 7 floors. What is the probability that no 2 people get off on the same floor? (Assume each floor is equally like for each person)1 solutions Answer 20091 by venugopalramana(3286)   on 2006-04-15 07:53:13 (Show Source): You can put this solution on YOUR website!An elevator containing 5 people can stop at any one of 7 floors. What is the probability that no 2 people get off on the same floor? (Assume each floor is equally like for each person) LET US DEVELOP ALOGIC LIKE THIS THE FIRST PERSON CAN GET DOWN AT ANY ONE OF THE 7 FLOORS..IN....7 WAYS. THE SECOND PERSON CAN GET DOWN AT ANY FLOOR OTHER THAN THE ABOVE THAT IS IN 6 WAYS LIKE THAT IF WE CONTINUE WE GET THE ANSWER AS 7*6*5*4*3=2520 WAYS
 Triangles/33655: I have a right triangle with one known angle in addition to the right angle and the length of the side (b) between these two angles. The length of side (b) is five. How do I find the length of side (a)? Side (c) is also unknown. 1 solutions Answer 20089 by venugopalramana(3286)   on 2006-04-15 05:11:55 (Show Source): You can put this solution on YOUR website!I have a right triangle with one known angle in addition to the right angle and the length of the side (b) between these two angles. The length of side (b) is five. How do I find the length of side (a)? Side (c) is also unknown. LET THE TRIANGLE BE ABC b=AC=5 ANGLE BAC=90 ANGLE ACB IS KNOWN= C SAY YOU REFER TABLES OR A CALCULATOR TO FIND COS(C) AND TAN(C) TO DETERMINE SIDES BC=a AND c=AB USING THE FOLLOWING EQUATIONS COS(C)=AC/BC=5/a....OR.......a=5/COS(C)...KNOWING COS(C)FROM CALCULATOR / TABLE YOU CAN FIND a...IF ANGLE C=60 SAY....THEN COS(C)=COS(60)=0.5....a=5/0.5=10 TAN(C)=AB/AC=c/5....OR.......c=5*TAN(C)...KNOWING TAN(C)FROM CALCULATOR / TABLE YOU CAN FIND c...IF ANGLE C=60 SAY....THEN TAN(C)=TAN(60)=SQRT(3)=1.732....c=5*1.732=8.66
 logarithm/33674: log(10x)^2=2+2logx This one is false right? Thank you for helping me.1 solutions Answer 20088 by venugopalramana(3286)   on 2006-04-15 04:54:10 (Show Source): You can put this solution on YOUR website!log(10x)^2=2+2logx This one is false right? Thank you for helping me. LHS=2LOG(10X)=2{LOG(10)+LOG(X)}=2+2LOG(X)
 logarithm/33675: ln(1/x)+lnx=0 Is this one true or false? Thank you for your help.1 solutions Answer 20087 by venugopalramana(3286)   on 2006-04-15 04:51:52 (Show Source): You can put this solution on YOUR website!ln(1/x)+lnx=0 LHS=LN(X*1/X)=LN(1)=0
 logarithm/33676: log(2x)=2logx I believe this one is true. Am I right? Thank you.1 solutions Answer 20086 by venugopalramana(3286)   on 2006-04-15 04:48:18 (Show Source): You can put this solution on YOUR website!log(2x)=2logx I believe this one is true. Am I right? Thank you. NO THE FORMULA IS LOG(A*B)=LOG(A)+LOG(B) SO......LOG(2X)=LOG(2)+LOG(X)