# See tutors' answers!

Algebra ->  Tutoring on algebra.com -> See tutors' answers!      Log On

 Tutoring Home For Students Tools for Tutors Our Tutors Register Recently Solved
 By Tutor
| By Problem Number |

Tutor:

# Recent problems solved by 'venugopalramana'

venugopalramana answered: 3288 problems
Jump to solutions: 0..29 , 30..59 , 60..89 , 90..119 , 120..149 , 150..179 , 180..209 , 210..239 , 240..269 , 270..299 , 300..329 , 330..359 , 360..389 , 390..419 , 420..449 , 450..479 , 480..509 , 510..539 , 540..569 , 570..599 , 600..629 , 630..659 , 660..689 , 690..719 , 720..749 , 750..779 , 780..809 , 810..839 , 840..869 , 870..899 , 900..929 , 930..959 , 960..989 , 990..1019 , 1020..1049 , 1050..1079 , 1080..1109 , 1110..1139 , 1140..1169 , 1170..1199 , 1200..1229 , 1230..1259 , 1260..1289 , 1290..1319 , 1320..1349 , 1350..1379 , 1380..1409 , 1410..1439 , 1440..1469 , 1470..1499 , 1500..1529 , 1530..1559 , 1560..1589 , 1590..1619 , 1620..1649 , 1650..1679 , 1680..1709 , 1710..1739 , 1740..1769 , 1770..1799 , 1800..1829 , 1830..1859 , 1860..1889 , 1890..1919 , 1920..1949 , 1950..1979 , 1980..2009 , 2010..2039 , 2040..2069 , 2070..2099 , 2100..2129 , 2130..2159 , 2160..2189 , 2190..2219 , 2220..2249 , 2250..2279 , 2280..2309 , 2310..2339 , 2340..2369 , 2370..2399 , 2400..2429 , 2430..2459 , 2460..2489 , 2490..2519 , 2520..2549 , 2550..2579 , 2580..2609 , 2610..2639 , 2640..2669 , 2670..2699 , 2700..2729 , 2730..2759 , 2760..2789 , 2790..2819 , 2820..2849 , 2850..2879 , 2880..2909 , 2910..2939 , 2940..2969 , 2970..2999 , 3000..3029 , 3030..3059 , 3060..3089 , 3090..3119 , 3120..3149 , 3150..3179 , 3180..3209 , 3210..3239 , 3240..3269 , 3270..3299, >>Next

 Linear_Algebra/31747: Let P (n) represent the statment 2+6+10+...+(4n-2)=2n^2 In the proof that P (n) is true for all integers n, n>1, what term must be added to both sides of P (k) to show P (k+1) follows form p(k)? (A) 4k+2 (B) P(k+1) (C) 4k+6 (D) 4k-2 Thank you for your assistance!1 solutions Answer 20613 by venugopalramana(3286)   on 2006-04-20 11:37:54 (Show Source): You can put this solution on YOUR website!Let P (n) represent the statment 2+6+10+...+(4n-2)=2n^2 In the proof that P (n) is true for all integers n, n>1, what term must be added to both sides of P (k) to show P (k+1) follows form p(k)? (A) 4k+2 (B) P(k+1) (C) 4k+6 (D) 4k-2 P(K)=(4K-2)...SO WE HAVE TO ADD THE NEXT TERM THAT IS P(K+1) TO BOTH SIDES TO PROVE THE EQUALITY BY INDUCTION METHOD. SO P(K+1)=4(K+1)-2=4K+4-2=4K+2 IS THE NUMBER WE SHOULD ADD TO BOTH SIDES ....A IS THE ANSWER.
 Quadratic_Equations/34339: Can you show me how to simplify this problem? 3x+9/x^2-x-12 * x^2+5x+6/24x+481 solutions Answer 20608 by venugopalramana(3286)   on 2006-04-20 10:55:22 (Show Source): You can put this solution on YOUR website!{(3x+9)/(x^2-x-12)} * {(x^2+5x+6)/(24x+48)} ={3(X+3)/(X^2-4X+3X-12)}*{X^2+3X+2X+6)/24(X+2)} ={3(X+3)/(X(X-4)+3(X-4))}*{(X(X+3)+2(X+3))/24(X+2)} ={3(X+3)/(X+3)(X-4)}{(X+3)(X+2)/24(X+2)} ={3/(X-4)}{(X+3)/24} =(X+3)/8(X-4)
 Trigonometry-basics/34341: Find the projection on the x-axis if AB = 10 and theta = 30 degrees.1 solutions Answer 20607 by venugopalramana(3286)   on 2006-04-20 10:46:02 (Show Source): You can put this solution on YOUR website!PROJECTION ON X AXIS OF AB = AB*COS(THETA)...WHERE THETA IS ANGLE BETWEEN X AXIA AND LINE AB. SO PROJECTION =10*COS(30)=10/2=5
 Quadratic_Equations/34352: (x^2+2x-15)/(6-2x) can you help simplify. I got -x-5/21 solutions Answer 20606 by venugopalramana(3286)   on 2006-04-20 10:43:30 (Show Source): You can put this solution on YOUR website!(x^2+2x-15)/(6-2x) =(X^2+5X-3X-15)/2(3-X)={X(X+5)-3(X+5)}/2(3-X)=(X+5)(X-3)/{(-2)(X-3) =-(X+5)/2 can you help simplify. I got -x-5/2...CORRECT..BUT PUT BRACKETS AS OTHERWISE IT MIGHT BE READ AS -X -(5/2)..WHERE AS IT IS ACTUALLY -(X+5)/2...OR...(-X-5)/2
 Trigonometry-basics/34345: express in terms of theta: csc (90 degrees + theta) sin (180 degrees + theta) - tan (270 degrees - theta). 1 solutions Answer 20604 by venugopalramana(3286)   on 2006-04-20 10:37:49 (Show Source): You can put this solution on YOUR website!express in terms of theta: csc (90 degrees + theta) sin (180 degrees + theta) - tan (270 degrees - theta). LET US USE A INSTEAD OF THETA.....USING ALL SILVER TEA CUPS FOR SIGN =-SEC(A)*(-SIN(A))*(COT(A))=(1/COS(A))(SIN(A))(COS(A)/SIN(A))=1
 Graphs/34246: Determine whether the following pairs of lines are parallel, perpendicular or neither. A) 2x + y = 1 and x-2y-1 b) 3x-y + 2= 0 and 3x + y = 0 C) y = 2x - 1 and 2x-y + 3=01 solutions Answer 20585 by venugopalramana(3286)   on 2006-04-20 03:29:21 (Show Source): You can put this solution on YOUR website!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.
 Trigonometry-basics/34256: find the eact value is 0 tan(x+y) if cscx=5/3 and cosy=5/13 I am aware that I have to use the sum idenity for the tangent function but I have no clue how to change csc and cos into a tangent idenity if that is even what I am supposed to do. I have tried a few times but I get different wrong answers each time. 1 solutions Answer 20584 by venugopalramana(3286)   on 2006-04-20 03:27:34 (Show Source): You can put this solution on YOUR website!tan(x+y)=(TAN(X)+TAN(Y))/(1-TAN(X)*TAN(Y)) if cscx=5/3=HYPOTENUSE/OPPOSITE SIDE....SO USING PYTHOGARUS ADJ.SIDE^2+OPP.SIDE^2=HYP.^2.....ADJ.^2+3^2=5^2...ADJ.^2=25-9=16 ADJ.SIDE=4....TAN(X)=OPP.SIDE/ADJ.SIDE=3/4 and cosy=5/13 =ADJ.SIDE/HYP. 5^2+OPP.SIDE^2=13^2 OPP.SIDE^2=169-25=144 OPP.SIDE=12 TAN(Y)=OPP.SIDE/ADJ.SIDE=12/5 TAN(X+Y)=(3/4 + 12/5 )/(1 - (3/4)(12/5)) =(3*5+12*4)/(4*5-3*12)=(15+48)/(20-36)=-63/16
 test/34327: when working alone, Gail can finish her paper route in 6hrs. When Mike substitutes for her, he can complete the job in 9 hrs. If they work together, how long should it take them ? I dont quite get what does "paper route" mean? what is that?1 solutions Answer 20583 by venugopalramana(3286)   on 2006-04-20 03:15:41 (Show Source): You can put this solution on YOUR website!SEE THE FOLLOWING AND TRY Pump A can fill a tank in 6 hours. Pump A and B together can fill the same tank in 3.5 hours. How long will it take to pump B to fill the tank alone? Can you show me how to do this problem? Thank you. 1 solutions Answer 20208 by venugopalramana(1495) About Me on 2006-04-17 01:14:18 (Show Source): Pump A can fill a tank in 6 hours. A FILLS IN 1 HOUR..............1/6 TANK Pump A and B together can fill the same tank in 3.5 hours. A&B FILL IN 1 HOUR..........1/3.5 TANK=10/35=2/7 TANK SO IN 1 HOUR B FILLS 2/7 - 1/6 = (2*6-1*7)/(6*7)=5/42...TANK SO PUMP B ALONE WILL TAKE 1/(5/42) HRS TO FILL THE TANK =42/5=8.4 HRS. How long will it take to pump B to fill the tank alone? Can you show me how to do this problem? Thank you.
 Rational-functions/34335: rationalize the denominator, then simplify, if possible sqrt (x)-2/sqrt (x)-51 solutions Answer 20582 by venugopalramana(3286)   on 2006-04-20 03:12:14 (Show Source): You can put this solution on YOUR website!rationalize the denominator, then simplify, if possible (sqrt (x)-2)(SQRT(X)+5)/(SQRT(X)+5)(sqrt (x)-5) =(X+5SQRT(X)-2SQRT(X)-10)/(5-5^2) =-(X+3SQRT(X)-10)/20
 Linear_Algebra/34258: Problem: Is it possible for the columns of a 4x3 matrix to be linearly dependent? if so, give an example and demonstrate the dependence. If not, prove it. I understand what linearly dependent means, but I don't know what kind of a matrix we should use to prove it without coming up with a clear example with numbers. Please help. 1 solutions Answer 20579 by venugopalramana(3286)   on 2006-04-20 00:54:43 (Show Source): You can put this solution on YOUR website!Problem: Is it possible for the columns of a 4x3 matrix to be linearly dependent? YES ,WE CAN ALWAYS WRITE TO SUIT THE REQUIREMENT..FOR..EXAMPLE...LET THE MATRIX A BE EQUAL TO 1,2,3 2,4,6 3,6,9 4,8,12 THE COLUMNS ARE C1...........1,2,3,4 C2...........2,4,6,8 C3...........3,6,9,12 WE FIND EACH ELEMENT OF C2=2*EACH CORRESPONDING ELEMENT OF C1 WE FIND EACH ELEMENT OF C3=3*EACH CORRESPONDING ELEMENT OF C1 SO THE COLUMNS ARE LINEARLY DEPENDENT.... ANOTHER WAY TO MAKE SUCH MATRIX IS SAY A IS............. 1,2,3 4,5,9 6,7,13 8,9,17 THE COLUMNS ARE C1...1,4,6,8 C2...2,5,7,9 C3...3,9,13,17 WE FIND EACH ELEMENT OF C3=SUM OF EACH CORRESPONDING ELEMENT OF C1 AND C2 if so, give an example and demonstrate the dependence. If not, prove it. I understand what linearly dependent means, but I don't know what kind of a matrix we should use to prove it without coming up with a clear example with numbers. Please help. IN GENERAL JUST MAKE NUMBERS IN ONE COLUMN TO BE SUM OR DIFFERENCE OF ELEMENTS IN OTHER 2 COLUMNS...SAY..A..IS... X,Y,X-Y P,Q,P-Q T,U,T-U E,F,E-F...ETC...
 Geometry_Word_Problems/34303: Write and equation and solve the following problem. The area or a rectangle is 130 square feet. Find the dimensions of the rectangle if the length is 3 feet greater than the width. I know A=LW so: 130 = L+3 = w Did I get the formula correct1 solutions Answer 20575 by venugopalramana(3286)   on 2006-04-20 00:31:48 (Show Source): You can put this solution on YOUR website!SEE MY COMMENTS BELOW The area or a rectangle is 130 square feet. Find the dimensions of the rectangle if the length is 3 feet greater than the width. I know A=LW....OK so: 130 = L+3 = w....NO....GO STEP BY STEP....LENGTH IS MORE THAN WIDTH...SO L=W+3...OK...NOW AREA =LW=(W+3)W=130 SO WE GET W^2+3W=130 W^3+3W-130=0 W^2+13W-10W-130=0 W(W+13)-10(W+13)=0 (W-10)(W+13)=0 W-10=0....OR...W=10 SO L=W+3=10+3=13 Did I get the formula correct
 Graphs/34309: Geometry: Floor plans for a building have the four corners of a room located at the points (2,3), (11,6), (-3,18), and (8,21). Determine whether the side through the points (2,3) and (11,6) is perpendicular ot the side through the points (2,3) and (-3,18).1 solutions Answer 20574 by venugopalramana(3286)   on 2006-04-20 00:02:17 (Show Source): You can put this solution on YOUR website!Geometry: Floor plans for a building have the four corners of a room located at the points (2,3)...A , (11,6)....B , (-3,18)....C , and (8,21)...D . Determine whether the side through the points (2,3)....A and (11,6).....B is perpendicular ot the side through the points (2,3).....A and (-3,18)...........C. SLOPE OF AB = (6-3)/(11-2)=3/9=1/3 SLOPE OF AC = (18-3)/(-3-2)=15/-5=-3 PRODUCT OF SLOPES OF AB AND AC =(1/3)(-3)=-1.... HENCE THE 2 LINES ARE PERPENDICULAR.