Tutor:

New! Get regular updates about newly solved problems via algebra.com's RSS system.

---

Recent problems solved by 'mananth'

mananth answered: 12270 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 , 3300..3329 , 3330..3359 , 3360..3389 , 3390..3419 , 3420..3449 , 3450..3479 , 3480..3509 , 3510..3539 , 3540..3569 , 3570..3599 , 3600..3629 , 3630..3659 , 3660..3689 , 3690..3719 , 3720..3749 , 3750..3779 , 3780..3809 , 3810..3839 , 3840..3869 , 3870..3899 , 3900..3929 , 3930..3959 , 3960..3989 , 3990..4019 , 4020..4049 , 4050..4079 , 4080..4109 , 4110..4139 , 4140..4169 , 4170..4199 , 4200..4229 , 4230..4259 , 4260..4289 , 4290..4319 , 4320..4349 , 4350..4379 , 4380..4409 , 4410..4439 , 4440..4469 , 4470..4499 , 4500..4529 , 4530..4559 , 4560..4589 , 4590..4619 , 4620..4649 , 4650..4679 , 4680..4709 , 4710..4739 , 4740..4769 , 4770..4799 , 4800..4829 , 4830..4859 , 4860..4889 , 4890..4919 , 4920..4949 , 4950..4979 , 4980..5009 , 5010..5039 , 5040..5069 , 5070..5099 , 5100..5129 , 5130..5159 , 5160..5189 , 5190..5219 , 5220..5249 , 5250..5279 , 5280..5309 , 5310..5339 , 5340..5369 , 5370..5399 , 5400..5429 , 5430..5459 , 5460..5489 , 5490..5519 , 5520..5549 , 5550..5579 , 5580..5609 , 5610..5639 , 5640..5669 , 5670..5699 , 5700..5729 , 5730..5759 , 5760..5789 , 5790..5819 , 5820..5849 , 5850..5879 , 5880..5909 , 5910..5939 , 5940..5969 , 5970..5999 , 6000..6029 , 6030..6059 , 6060..6089 , 6090..6119 , 6120..6149 , 6150..6179 , 6180..6209 , 6210..6239 , 6240..6269 , 6270..6299 , 6300..6329 , 6330..6359 , 6360..6389 , 6390..6419 , 6420..6449 , 6450..6479 , 6480..6509 , 6510..6539 , 6540..6569 , 6570..6599 , 6600..6629 , 6630..6659 , 6660..6689 , 6690..6719 , 6720..6749 , 6750..6779 , 6780..6809 , 6810..6839 , 6840..6869 , 6870..6899 , 6900..6929 , 6930..6959 , 6960..6989 , 6990..7019 , 7020..7049 , 7050..7079 , 7080..7109 , 7110..7139 , 7140..7169 , 7170..7199 , 7200..7229 , 7230..7259 , 7260..7289 , 7290..7319 , 7320..7349 , 7350..7379 , 7380..7409 , 7410..7439 , 7440..7469 , 7470..7499 , 7500..7529 , 7530..7559 , 7560..7589 , 7590..7619 , 7620..7649 , 7650..7679 , 7680..7709 , 7710..7739 , 7740..7769 , 7770..7799 , 7800..7829 , 7830..7859 , 7860..7889 , 7890..7919 , 7920..7949 , 7950..7979 , 7980..8009 , 8010..8039 , 8040..8069 , 8070..8099 , 8100..8129 , 8130..8159 , 8160..8189 , 8190..8219 , 8220..8249 , 8250..8279 , 8280..8309 , 8310..8339 , 8340..8369 , 8370..8399 , 8400..8429 , 8430..8459 , 8460..8489 , 8490..8519 , 8520..8549 , 8550..8579 , 8580..8609 , 8610..8639 , 8640..8669 , 8670..8699 , 8700..8729 , 8730..8759 , 8760..8789 , 8790..8819 , 8820..8849 , 8850..8879 , 8880..8909 , 8910..8939 , 8940..8969 , 8970..8999 , 9000..9029 , 9030..9059 , 9060..9089 , 9090..9119 , 9120..9149 , 9150..9179 , 9180..9209 , 9210..9239 , 9240..9269 , 9270..9299 , 9300..9329 , 9330..9359 , 9360..9389 , 9390..9419 , 9420..9449 , 9450..9479 , 9480..9509 , 9510..9539 , 9540..9569 , 9570..9599 , 9600..9629 , 9630..9659 , 9660..9689 , 9690..9719 , 9720..9749 , 9750..9779 , 9780..9809 , 9810..9839 , 9840..9869 , 9870..9899 , 9900..9929 , 9930..9959 , 9960..9989 , 9990..10019 , 10020..10049 , 10050..10079 , 10080..10109 , 10110..10139 , 10140..10169 , 10170..10199 , 10200..10229 , 10230..10259 , 10260..10289 , 10290..10319 , 10320..10349 , 10350..10379 , 10380..10409 , 10410..10439 , 10440..10469 , 10470..10499 , 10500..10529 , 10530..10559 , 10560..10589 , 10590..10619 , 10620..10649 , 10650..10679 , 10680..10709 , 10710..10739 , 10740..10769 , 10770..10799 , 10800..10829 , 10830..10859 , 10860..10889 , 10890..10919 , 10920..10949 , 10950..10979 , 10980..11009 , 11010..11039 , 11040..11069 , 11070..11099 , 11100..11129 , 11130..11159 , 11160..11189 , 11190..11219 , 11220..11249 , 11250..11279 , 11280..11309 , 11310..11339 , 11340..11369 , 11370..11399 , 11400..11429 , 11430..11459 , 11460..11489 , 11490..11519 , 11520..11549 , 11550..11579 , 11580..11609 , 11610..11639 , 11640..11669 , 11670..11699 , 11700..11729 , 11730..11759 , 11760..11789 , 11790..11819 , 11820..11849 , 11850..11879 , 11880..11909 , 11910..11939 , 11940..11969 , 11970..11999 , 12000..12029 , 12030..12059 , 12060..12089 , 12090..12119 , 12120..12149 , 12150..12179 , 12180..12209 , 12210..12239 , 12240..12269 , 12270..12299, >>Next

Money_Word_Problems/750078: A man has invested $8000, part of 10% and the remainder at 6% simple interest. how much has he invested at each rate if his annual income from these investments is $536 1 solutions Answer 456340 by mananth(12270)   on 2013-05-18 08:53:23 (Show Source): You can put this solution on YOUR website! Part I 10.00% per annum ------------- Amount invested =x Part II 6.00% per annum ------------ Amount invested = y 8000 Interest----- 536.00 Part I 10.00% per annum ---x Part II 6.00% per annum ---y Total investment x + 1 y= 8000 -------------1 Interest on both investments 10.00% x + 6.00% y= 536 Multiply by 100 10 x + 6 y= 53600.00 --------2 Multiply (1) by -10 we get -10 x -10 y= -80000.00 Add this to (2) 0 x -4 y= -26400 divide by -4 y = 6600 Part I 10.00% $ 1400 Part II 6.00% $ 6600 CHECK 1400 --------- 10.00% ------- 140.00 6600 ------------- 6.00% ------- 396.00 Total -------------------- 536.00 m.ananth@hotmail.ca

Equations/746325: I need to know which one of these equations has the same solution as the equation 2x+6=32. 2x=38 x-3=6 x+6=16 2(x-3)=16 2(x+3)=32 I know there is a difference between sloveing, solutions, and simplifying but I just don't know wtah it is? 1 solutions Answer 454229 by mananth(12270)   on 2013-05-08 08:50:22 (Show Source): You can put this solution on YOUR website! 2x+6=32. /2 x+3=16 Options 2(x+3)=32 is right /2 x+3=16

Polynomials-and-rational-expressions/746323: Factor c^2 - 4d^2 = 1 solutions Answer 454228 by mananth(12270)   on 2013-05-08 08:46:57 (Show Source): You can put this solution on YOUR website! use the identity

Signed-numbers/746310: If 2x+3y=24 and 2x-3y=12 then value of xy is_____________ 1 solutions Answer 454219 by mananth(12270)   on 2013-05-08 06:45:53 (Show Source): You can put this solution on YOUR website! 2.00 x + 3.00 y = 24.00 .............1 Total value 2.00 x -3.00 y = 12.00 .............2 Eliminate y multiply (1)by 1.00 Multiply (2) by 1.00 2.00 x 3.00 y = 24.00 2.00 x -3.00 y = 12.00 Add the two equations 4.00 x = 36.00 / 4.00 x = 9.00 plug value of x in (1) 2.00 x + 3.00 y = 24.00 18.00 + 3.00 y = 24.00 3.00 y = 24.00 -18.00 3.00 y = 6.00 y = 2.00 xy=9*2=18

Sequences-and-series/746281: If the third term and fifth term of a geometric progression are 225 and 5625 respectively, determine the sixth term. 1 solutions Answer 454217 by mananth(12270)   on 2013-05-08 06:43:25 (Show Source): You can put this solution on YOUR website! If the third term and fifth term of a geometric progression are 225 and 5625 respectively, determine the sixth term. Tn = ar^(n01) is the n th term of a GP T3= ar^(3-1) T3=ar^2 225=ar^2 Similarly t5=ar^4 5625=ar^4 divides ne equation by the other 225/5625= ar^2/ar^4 225/5625= 1/r^2 25/75= 1/r r=3 plug r 225=ar^2 225=a*9 a=225/9 a=25 t6=ar^5 t6=25*3^5 t6=25*243 t6=6075 6th term = 6075

Money_Word_Problems/746292: a total of $7000 is invested: part at 9% and the remainder at 11%. How much is invested at each rate if the annual interest is $700 1 solutions Answer 454216 by mananth(12270)   on 2013-05-08 06:35:02 (Show Source): You can put this solution on YOUR website! Part I 9.00% per annum ------------- Amount invested =x Part II 11.00% per annum ------------ Amount invested = y 7000 Interest----- 700.00 Part I 9.00% per annum ---x Part II 11.00% per annum ---y Total investment x + 1 y= 7000 -------------1 Interest on both investments 9.00% x + 11.00% y= 700 Multiply by 100 9 x + 11 y= 70000.00 --------2 Multiply (1) by -9 we get -9 x -9 y= -63000.00 Add this to (2) 0 x 2 y= 7000 divide by 2 y = 3500 Part I 9.00% $ 3500 Part II 11.00% $ 3500 CHECK 3500 --------- 9.00% ------- 315.00 3500 ------------- 11.00% ------- 385.00 Total -------------------- 700.00 m.ananth@hotmail.ca

Travel_Word_Problems/746167: Keisha rowed downstream for 5 miles and back upstream for 3 miles. She rowed an average of 3 times the rate of the current. Write an expression for her total time. This is on combining rational expressions with unlike denominators. 1 solutions Answer 454174 by mananth(12270)   on 2013-05-07 21:08:57 (Show Source): You can put this solution on YOUR website! Keisha rowed downstream for 5 miles and back upstream for 3 miles. She rowed an average of 3 times the rate of the current. Write an expression for her total time. let current speed be r rowing speed = 3r speed up stream = 3r-r=2r speed down stream = 3r+r=>4r time down stream = d/r => 5/4r time upstream = 3/2r Total time

Money_Word_Problems/746164: Josee invests a total of $15 000.00 in two different investments. The first amount is put into a long-term account that pays interest at a rate of 6.5% per year. The second amount is put into a short-term account earning interest at a rate of 5% per year. Josee's investments earn a total of $885 in interest in one year. How much money did Josee put into each investment? 1 solutions Answer 454172 by mananth(12270)   on 2013-05-07 21:04:15 (Show Source): You can put this solution on YOUR website! Part I 6.50% per annum ---x Part II 5.00% per annum ---y Total investment x + 1 y= 15000 -------------1 Interest on both investments 6.50% x + 5.00% y= 885 Multiply by 100 6.5 x + 5 y= 88500.00 --------2 Multiply (1) by -6.5 we get -6.5 x -6.5 y= -97500.00 Add this to (2) 0 x -1.5 y= -9000 divide by -1.5 y = 6000 Part I 6.50% $ 9000 Part II 5.00% $ 6000 CHECK 9000 --------- 6.50% ------- 585.00 6000 ------------- 5.00% ------- 300.00 Total -------------------- 885.00 m.ananth@hotmail.ca

Linear_Equations_And_Systems_Word_Problems/746161: kaitlin will rent car for weekend. she has two plans. the first plan has intial fee of $55 and cost of .7 cent per each mile driven. the 2nd has cost of .8 cent for each mile driven. how many miles would she need to drive for the two plan to cost the same? 1 solutions Answer 454171 by mananth(12270)   on 2013-05-07 20:55:06 (Show Source): You can put this solution on YOUR website! let she drive x miles on both plans C1= 55+0.0.07x C2=0.08x C1=C2 55+0.07x=0.08x 55=0.08x-0.07x 55=0.01x x=5500 miles

Age_Word_Problems/746162: IN 8 years time I will be twice the age I was 3 years ago. How old am I now? How will my equation look like? 1 solutions Answer 454170 by mananth(12270)   on 2013-05-07 20:50:34 (Show Source): You can put this solution on YOUR website! I am x years now 3 years ago I was x-3 years 8 years hence I will be x+ 8 years x+8=2(x-3) x+8=2x-6 x=14

Linear-systems/745343: x= adult tickets y= children tickets 5x + 3y = 134 x + y = 32 whats true about this equation? a.) More adults attended the reunion than children b.) six more children attended that adults c.) nineteen children attended the reunion d.) three more adults attended than children i didnt know how i would solve this? 1 solutions Answer 453812 by mananth(12270)   on 2013-05-05 20:57:35 (Show Source): You can put this solution on YOUR website! 5x + 3y = 134 x + y = 32. multiply by -3 -3x-3y=-96 add to the first equation we get 2x=38 x=19 if x=17 plug x in x+y=32 y=13 Adults = 19 children = 13 now you may find the correct statement

Sequences-and-series/745077: Please help If the number of terms of an Arithmetic sequence is 15 and the value of it's middle term is 10 then find the sum of it's terms is there any rule that i can solve it with?? Thanks 1 solutions Answer 453708 by mananth(12270)   on 2013-05-05 09:57:35 (Show Source): You can put this solution on YOUR website! The nth term of an AP is given by the formula Tn = a+(n-1)d t15 = a+14d The middle term is the 8th term T8=a+7d 10=a+7d multiply by 2 20=2(a+7d) 20=2a+14d Sum of n terms = Sn = n/2[2a+(n-1)d] s15 = 15/2[2a+(15-1)d] S15= 15/2[2a+14d] but 2a+14d=20 S15=15/2( 20) S15=150 sum of 15 terms = 150

Numbers_Word_Problems/745072: the sum of the digits of a three digit number is 14. The hundreds digit being 4 times the units digit. If 594 is subtracted from the number, the order of hundreds tens digits will be reversed. Find the number. 1 solutions Answer 453694 by mananth(12270)   on 2013-05-05 08:37:15 (Show Source): You can put this solution on YOUR website! the sum of the digits of a three digit number is 14. The hundreds digit being 4 times the units digit. If 594 is subtracted from the number, the order of hundreds tens digits will be reversed. Find the number. let the number be xyz x+y+z=14 x=4z Therefore 4z+y+z=14 y+5z=14..........................(1) 594 subtracted from the number 100x+10y+z-594 = 100z+10y+x 99x-99z=594 /99 x-z=6 But x=4z 4z-z=6 3z=6 z=2 if z=2 then x= 4z=>4*2=8 x+y+z=14 8+y+2=14 10+y=14 y=14-10 y=4 The number is xyz 842

�t�e�s�t/745017: Solve each system by substitution or elimination. 2x+5y=-7 3x-2y=-18 studying for a test so the steps will be helpful 1 solutions Answer 453680 by mananth(12270)   on 2013-05-05 06:11:17 (Show Source): You can put this solution on YOUR website! 2.00 x + 5.00 y = -7.00 .............1 Total value 3.00 x -2.00 y = -18.00 .............2 Eliminate y multiply (1)by 2.00 Multiply (2) by 5.00 4.00 x 10.00 y = -14.00 15.00 x -10.00 y = -90.00 Add the two equations 19.00 x = -104.00 / 19.00 x = -5.47 plug value of x in (1) 2.00 x + 5.00 y = -7.00 -10.95 + 5.00 y = -7.00 5.00 y = -7.00 + 10.95 5.00 y = 3.95 y = 0.79 x= -5.47 y= 0.79 m.ananth@hotmail.ca

Quadratic_Equations/745041: I have this problem: A roof shingle is dropped from a rooftop that is 100 feet above the ground. The height y (in feet) of the dropped roof shingle is given by the function y=-16t^2+100 where t is the time (in second)s since the shingle is dropped. Graph the function. I have worked out that if I solve for t, t=2.5. But how do I proceed in graphing? My teacher doesn't explain anything and this problem is on my review sheet. We didn't ever see word problems like this in the chapter. Any help you can give me would be very much appreciated. 1 solutions Answer 453678 by mananth(12270)   on 2013-05-05 05:40:44 (Show Source): You can put this solution on YOUR website! y=-16t^2+100 since co-efficient of x^2 is negative the parabola opens downwards find the two values of x for which y=0 t= +2.5 and -2.5 the two points are (-2.5,0) and (2.5,0) when t=0, y=100 you get another point (0,100) this is the vertex of the parabola. take t=+/-1, y1=84 (+/-1,84) take t=+/-2,y2=68(+/-2,36) The parabola is symmetric about the axis of symmetry. Plot the points and trace a symmetric curve.

Linear_Equations_And_Systems_Word_Problems/744845: x and y are two numbers where x is greater than y. the sum of the two numbers is 20. twice the larger number is equal to 3 times the smaller one. (a) Write down a [pair of simultaneous equations in x and y. (b) Solve the simultaneous equations in (a), to find the two numbers. 1 solutions Answer 453582 by mananth(12270)   on 2013-05-04 09:07:43 (Show Source): You can put this solution on YOUR website! x+y=20>>>>>>>>>>>>>>>>>>>>(1) 2x=3y 2x-3y=0>>>>>>>>>>>>>>>>>>(2) multiply first equation by 3 3x+3y=60 add to (2) 5x=60 x=12 substitute x=12 in equation (1) x+y=20 12+y=20 y=20-12 x=12 y=8 x=12 & y =8

Linear_Equations_And_Systems_Word_Problems/744818: A boy in the age of 20 yrs bought a box and he started dropping 250 rupees on his every birthday and his sister use to take back 50 rupees from the box on her every birthday .. Boy dies at an age of 60 years.. When opened the box then found just 500 rupees in the box .. How just 500 Rs. ??? 1 solutions Answer 453579 by mananth(12270)   on 2013-05-04 08:02:47 (Show Source): You can put this solution on YOUR website! The boy's birthday falls on feb 29th.

Trigonometry-basics/744772: Solve the following: 0 less then or equal to x less then 2pi A) 4cos x + sqrt 3= 2cos x B) 2sin^2 x - 5 sin x= -2 Thanks again for all the help! 1 solutions Answer 453551 by mananth(12270)   on 2013-05-03 23:05:55 (Show Source): You can put this solution on YOUR website! A) 4cos x + sqrt 3= 2cos x 4cos x - 2cos x = -sqrt(3) 2cosx=-sqrt(3) cosx= -sqrt(3)/2 x= 7pi/6 B) 2sin^2 x - 5 sin x= -2 2sin^2 x - 5 sin x +2=0 2sin^2 x - 4 sin x -sinx+2=0 2sinx(sinx-2)-1(sinx-2)=0 (sinx-2)(2sinx-1)=0 sinx = 2 or sinx =1/2 between 0 & 2pi find x

logarithm/744756: Solve the following equations a) log(base 5) x+2 = 1- log(base 5) x-2 b) log(base 5)x^2-4 = 1 Can you please help me out with this question? Thanks so much in advance:) Can you show all the steps because it will help me understand more thanks 1 solutions Answer 453544 by mananth(12270)   on 2013-05-03 22:19:49 (Show Source): You can put this solution on YOUR website! a) log(base 5) x+2 = 1- log(base 5) x-2 log(base 5) x+2 +log(base 5) x-2=1 log(base 5) ((x+2)(x-2)) =log(base5)5 x^2-4=5 x^2=9 x= +/- 3 ignore negative x=3 b) log(base 5)x^2-4 = 1 log(base 5)x^2-4 = log (base5)5 x^2-4=5 x^2=9 x= +/-3

Travel_Word_Problems/744771: Hello! I need help on this question. I understand all the speed x time equals distance and so, but i don't quite get this kinds of questions. A man walks to work at 3 1/2 kmph and is one minute late. If he walked at 3 1/3 kmph he would have been three minutes late. What is the distance to his office? Thanks :) 1 solutions Answer 453541 by mananth(12270)   on 2013-05-03 22:07:17 (Show Source): You can put this solution on YOUR website! let the right time be t speed = 3 1/2 kmph time taken = t+(1/60) hours speed =3 1/3 kmph time taken = t + (3/60) hours in both cases distance is the same d=rt 3 1/2 *( t+(1/60) )= distance 3 1/3 (t + (3/60))= distance 3 1/2 *( t+(1/60) )= 3 1/3 (t + (3/60)) 7/2((60t+1)/60) = 10/3((60t+3)/60) cross multiply 21(60t+1)/60 = 20(60t+3)/60 21(60t+1)=20(60t+3) 1260t+21=1200t+60 60t=60-21 60t=39 t= 39/60 t= 39 minutes With speed of 3 1/2 he takes one minute more => 39+1=40 minutes d=rt d= (7/2) *(40/60) d= 7/3 =2.33 km

Volume/744763: a cylindrical tennis ball container can contain maximum three ball stacked on one another.the top and bottom ball also touch the lid and the base of the container respectively if the volume of a tennis ball is 240 cm,then what is the volume of the container? 1 solutions Answer 453539 by mananth(12270)   on 2013-05-03 21:51:27 (Show Source): You can put this solution on YOUR website! V ball = 240 V = 4/3 * pi * r^3 240 = 4/3 * pi * r^3 (240*3)/(4 * pi) = r^3 r^3=57.29 r=3.85 cm radius of the ball diameter = 7.70cm Since the balls are touching each other and 3 balls can be place in the cylinder the height of cylinder = 3 * diameter =>23.10cm radius of cylinder = radius of ball So cylinder volume = pi * r^2*h r=3.85, h=23.10 substitute to get the answer 1075 cm^3 ( rounded off)

Expressions-with-variables/744551: I am having difficulty completing this question. Any help is appreciated. Thanks in advance. Substitution Method: x=2-y y=3-2x 1 solutions Answer 453353 by mananth(12270)   on 2013-05-03 02:05:55 (Show Source): You can put this solution on YOUR website! x=2-y y=3-2x y=3-2(2-y) y=3-4+2y y=-1+2y -2y -2y -y=-1 /-1 y=1 plug y in any equation to get x

Linear-equations/744534: whats the equation for a line passing through these two points:(4,0) and (6,-8) written in slope intercept form? 1 solutions Answer 453347 by mananth(12270)   on 2013-05-03 01:01:56 (Show Source): You can put this solution on YOUR website! x1 y1 x2 y2 4 0 6 -8 The general equation of the line when co-ordinates of two points are given is -2 y- 0 = 8 x 32 -2 y= 8 x+ 32 / -2 y= -4 x+ -16

Quadratic_Equations/744535: Solve 1 solutions Answer 453345 by mananth(12270)   on 2013-05-03 00:59:23 (Show Source): You can put this solution on YOUR website! (t+5)^2=12 t^2+10t+25=12 t^2+10t+13=0 Use the quadratic formula a= 1 , b= 10 , c= 13 b^2-4ac= 100 + -52 b^2-4ac= 48 x1=( -10 + 6.93 )/ 2 x1= -1.54 x2=( -10 -6.93 ) / 2 x2= -8.46 (-1.536 , -8.46 ) m.ananth@hotmail.ca

Radicals/744470: Need guidance with division. 2/2*sqrt(2)-1 Do I divide by the denominator? Thank you! 1 solutions Answer 453340 by mananth(12270)   on 2013-05-02 22:12:03 (Show Source): You can put this solution on YOUR website! multiply & divide by the conjugate (2*sqrt(2)+1)

Equations/744471: how do I find the equation of a line that passes through 1,2 and is perpendicular to y equals negative three fourths X plus two and three fourths 1 solutions Answer 453334 by mananth(12270)   on 2013-05-02 22:01:04 (Show Source): You can put this solution on YOUR website! 1 y = - 3/ 4 x + 11/4 Divide by 1 y = - 3/ 4 x + 11/4 Compare this equation with y=mx+b, m= slope & b= y intercept slope m = - 3/4 The slope of a line perpendicular to the above line will be the negative reciprocal 4/3 Because m1*m2 =-1 The slope of the required line will be4/3 m=4/3 ,point ( 1 , 2 ) Find b by plugging the values of m & the point in y=mx+b 2 = 4/ 3 + b b= 2/3 m= 4/3 The required equation is y = 4/ 3 x + 2/ 3 m.ananth@hotmail.ca

test/744498: Can you help me solve this equation? y^2+3y=36-2y Please help!! Thank you! 1 solutions Answer 453331 by mananth(12270)   on 2013-05-02 21:51:40 (Show Source): You can put this solution on YOUR website! +2y +2y -36-36 y=-9 OR 4

logarithm/744484: 1. Solve the following equations a) log(base 5)x+2 + log(base 5) x-2 - log(base 5)4 = log(base 5)3 b) log(base 5)x^2-4 - log(base 5)4 = log(base 5)3 c) log(base 5)x+2 = 1- log(base 5) x-2 d) log(base 5) (x^2-4) =1 Can you please help me out? Thanks so much in advance:) Can you please show all the steps it would really help me understand:) 1 solutions Answer 453329 by mananth(12270)   on 2013-05-02 21:47:55 (Show Source): You can put this solution on YOUR website! log(base 5)x+2 + log(base 5) x-2 - log(base 5)4 = log(base 5)3 log(base 5)((x+2)(x-2)) - log(base 5)4 = log(base 5)3 log(base 5)(((x+2)(x-2))/4) = log(base 5)3 common base x=+/-4 therefore x=4

logarithm/744488: 1. Solve the following equations a) 5^(2x) + 5^(x) - 6 = 0 (hint let y = 5^(x) b)9^(x)-5(3^(x))+6 = 0 ( hint let y = 3^(x) Can you please help me out? Thanks so much in advance:) Can you please show all the steps it would really help me understand:) 1 solutions Answer 453327 by mananth(12270)   on 2013-05-02 21:38:44 (Show Source): You can put this solution on YOUR website! 5^(2x) + 5^(x) - 6 = 0 let y = 5^(x) (y+3)(y-2)=0 y=-3 OR y=2 5^x=2 log 5^x=log2 xlog5 = log 2 x=log2/log5

Sequences-and-series/744441: Hello, this isn't an arithmetic sequence question, but a geometric one. I need to find the rate of the following geometric sequence, 9 (nine), 3√3 (three square root of 3),3(three),... and the next three terms of the sequence. How can I do it? 1 solutions Answer 453325 by mananth(12270)   on 2013-05-02 21:25:49 (Show Source): You can put this solution on YOUR website! r= => a=9, r= sqrt(1/3) you find the 5th and 6th in the same way

Money_Word_Problems/744222: David Saxon invested $25000 in two different corporate bonds for 1 year.One bond pays a 4.5% simple interest rate. and the other pays 6% simple interest rate. The Total annual interest David received from both bonds was $1380. Find the amount he invested for each bond. For 1 year,the simple interest on a specific corporate bond is found by multiplying the amount invested by the simple interest rate. 1 solutions Answer 453226 by mananth(12270)   on 2013-05-02 09:33:31 (Show Source): You can put this solution on YOUR website! Part I 4.50% per annum ------------- Amount invested =x Part II 6.00% per annum ------------ Amount invested = y 25000 Interest----- 1380.00 Part I 4.50% per annum ---x Part II 6.00% per annum ---y Total investment x + 1 y= 25000 -------------1 Interest on both investments 4.50% x + 6.00% y= 1380 Multiply by 100 4.5 x + 6 y= 138000.00 --------2 Multiply (1) by -4.5 we get -4.5 x -4.5 y= -112500.00 Add this to (2) 0 x 1.5 y= 25500 divide by 1.5 y = 17000 Part I 4.50% $ 8000 Part II 6.00% $ 17000 CHECK 8000 --------- 4.50% ------- 360.00 17000 ------------- 6.00% ------- 1020.00 Total -------------------- 1380.00 m.ananth@hotmail.ca