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Re: Nairaland Mathematics Clinic by Emdee590(m): 4:00pm On Mar 18, 2015 |
Let B=[1 3 5 3] i.e a 2*2 matrix , if f(x)=2x^2-4x 3 , what is f(B)? N/B the options are also in matrix form . Please solve it vividly for me thank you |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 4:51pm On Mar 18, 2015 |
Emdee590: OK check the ur f(x) d last part is it +3. or -3 ? |
Re: Nairaland Mathematics Clinic by Emdee590(m): 5:27pm On Mar 18, 2015 |
agentofchange1: It is +3 |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 11:05pm On Mar 18, 2015 |
Emdee590: OK here 1 Share
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Re: Nairaland Mathematics Clinic by Emdee590(m): 7:22am On Mar 19, 2015 |
So that is the solving and you Have just gotten it as the answer . How do I thank you and the gurus here ? Either way I remain loyal , thank you and may God almighty keep you fit so to continue solving our questions for us . That question above na 200 level question ooo and we for we class nobody fit get that question right ... Once again thank you |
Re: Nairaland Mathematics Clinic by Emdee590(m): 7:28am On Mar 19, 2015 |
Abeg make una no tire for me and ma questions ooo, because it is good that I know every steps involved in any solving as ma younger ones are looking unto me for solutions to some of their mathematical problems...una no know that kind thing again ? |
Re: Nairaland Mathematics Clinic by dejt4u(m): 7:36am On Mar 19, 2015 |
Emdee590:did you just say 200level?? Is ur school a university, Polytechnic, college of education or a monotechnic?? Your dept?? Is it science or social sciences? If it is social science, then no problem but if otherwise, wahala dey oo |
Re: Nairaland Mathematics Clinic by Nobody: 8:13am On Mar 19, 2015 |
dejt4u: LOL, I believe he is in social sciences! |
Re: Nairaland Mathematics Clinic by bolkay47(m): 9:13am On Mar 19, 2015 |
y=(sinx)^x^3 find dy/dx |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 9:52am On Mar 19, 2015 |
bolkay47: here take natural logarithm of both sides => lny=x^3 ln(sinx) differentiating implicitly we have , => y'/y = 3x^2 ln(sinx) + x^3cosx/sinx => y' =dy/dx =y(3x^2ln(sinx) +x^3cotx ) or dy/dx = x^2[ xcotx + 3ln(sinx) ] * (sinx)^(x^3) that's it man. |
Re: Nairaland Mathematics Clinic by bolkay47(m): 4:58pm On Mar 19, 2015 |
agentofchange1:exactly sir... That's what I get too |
Re: Nairaland Mathematics Clinic by Emdee590(m): 6:04pm On Mar 19, 2015 |
doubleDx:You are far away from the truth , it I federal uni ooo and maths department sef |
Re: Nairaland Mathematics Clinic by Nobody: 6:17pm On Mar 19, 2015 |
Emdee590: 200L Math department? Na wa oh, na hin that matrix problem dey give una wahala like that? But no offense, una really need to buckle-up be that oh! Is it that most of you were pushed to Math department when you applied to study something else? Anyways, na determination na hin be the key sha! |
Re: Nairaland Mathematics Clinic by dejt4u(m): 9:21pm On Mar 19, 2015 |
doubleDx:bros, it is very ridiculous to be teaching 200L mathematics student that concept.. Infact, it shld be in MTH101 and 3 by 3 matrix.. Your mates in other skuls are busy battling wit tensor analysis and the likes.. Not ur fault though, but the fault of those ppl that designed your school curriculum I'm very disappointed OP.. Emdee590: 1 Like |
Re: Nairaland Mathematics Clinic by Emdee590(m): 9:55pm On Mar 19, 2015 |
Well , una fit yarn anything shaa . The blaim is on both sides shaa |
Re: Nairaland Mathematics Clinic by Nobody: 11:54pm On Mar 19, 2015 |
yoji: At 0 year the car cost 1,750,000 (0, 1,750,000) In 3 years the car worth is 437,500 (3, 437,500) Slope: m=y2-y1/x2-x1 437500-1750000/3-0= -1312500/3=-437500 Line of equation using point slope form y-y1=m(x-x1) y-1750000=-437500(x-0) y-1750000=-437500x y=-437500x + 1750000 y=-437500(4.3) + 1750000 y=-131,250 |
Re: Nairaland Mathematics Clinic by Nobody: 2:56am On Mar 20, 2015 |
dejt4u: Exactly! I was surprised myself bruv! How can a University be teaching 200 L Math students High School & UTME Mathematics? I wonder for some schools sha! Smdh... |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 3:16pm On Mar 20, 2015 |
Emdee590: its OK bro , to God be glory , we learn daily , put more effort , practice daily with determination , don't always rush to post questions here or give some1 else without you actually trying exhaustively & extensively to obtain valid solution . only after which still didn't get it , then you could seek for help from others that's how you learn / develop/ improve. your mathematical prowess . 1 Like |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 3:18pm On Mar 20, 2015 |
bolkay47: nice. |
Re: Nairaland Mathematics Clinic by Geofavor(m): 10:12am On Mar 22, 2015 |
Emdee590:haa! ![]() I answered exactly this type of question in this recently concluded jamb. Just a little change in the numerical values. 200L maths? Wow. Jamb is crazy!!! This question was one of the questions that took my time that i couldn't finish answering all the tough questions. It took me 2mins. I made sure i got the answer though ![]() |
Re: Nairaland Mathematics Clinic by Emdee590(m): 10:58am On Mar 22, 2015 |
agentofchange1:Thank you sir |
Re: Nairaland Mathematics Clinic by Nobody: 1:07pm On Mar 22, 2015 |
pls I need the solution to an integral.. Integral of (x+3)/((x^2+2x+10)^1/2) thanks |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 3:51pm On Mar 22, 2015 |
Emdee590: UWC. sir. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 6:27pm On Mar 22, 2015 |
miniyi: OK +++++++++++++solution +++++++++++ let's consider content of the denominator x2 +2x +10 complete the square we. get x2 +2x +1-1+10 =(x+1)2 +9 we now have $(x+3)dx /√((x+1)2 +32) set x+1=3tan¢ .........(*) dx=3sec2 ¢ d¢ thus we have $ (x+3)/√(9(tan2¢ +1) * 3sec2¢ d¢ simplifies to $(x+3)sec¢d¢ recall from (*), x+1=3tan¢ , => x= 3tan¢-1 => $(3tan¢ +2)sec¢ d¢ => 3$sin¢ d¢ /cos2 ¢ +2$sec¢ d¢ now computing using substitution , we get 3sec¢ +2In(sec¢ + tan¢ ) + k where sec¢ & tan¢ = √((x+1)2 +9 )/3 & (x+1)/3. respectively. deduce & reduce further that's it man , hope u get ? 1 Like |
Re: Nairaland Mathematics Clinic by akpos4uall(m): 6:36pm On Mar 22, 2015 |
benji93: benji93: Using the center of the figure as O, the point where the two circles meet as A & the points where the smaller circle touches the horizontal & vertical axis as C & B respectively, OC = x, OB = y OA = r = 6 + x = 4 + y => y - x = 2 AC = x + r BC2 = x2 + y2 AB2 = r2 + y2 Using circle theorem, AB2 + BC2 = AC2 r2 + y2 + x2 + y2 = (x + r)2 r2 + 2y2 + x2 = x2 + 2xr + r2 2y2 = 2xr Substitute 2 + x & 6 + x for y & r respectively into the above equation 2(2 + x)2 = 2x(6 + x) (2 + x)2 = 6x + x2 4 + 4x + x2 = 6x + x2 4 = 6x - 4x x = 2 => y = 4, r = 8 The area of the larger square = 4r2 = 4 * 8 * 8 256 1 Like
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Re: Nairaland Mathematics Clinic by Nobody: 12:44am On Mar 23, 2015 |
agentofchange1:thanks |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 3:41pm On Mar 23, 2015 |
[quote author=miniyi post=31903394] thanks [/quote Don't mention its fun solving but not fun typing .. its well. |
Re: Nairaland Mathematics Clinic by bolkay47(m): 5:12pm On Mar 23, 2015 |
Try this :::: 500=20x+22y. Find x and y |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 5:28pm On Mar 23, 2015 |
miniyi:oo my bad, it just dawn on me , there was a mistake in the substitution i will post the correction later . chai .. ![]() |
Re: Nairaland Mathematics Clinic by bolkay47(m): 5:42pm On Mar 23, 2015 |
agentofchange1:I tried it. Got 1/6ln{(x-3)/(x+4)}+c |
Re: Nairaland Mathematics Clinic by DrMaths(m): 7:06pm On Mar 23, 2015 |
Hello Guys, Forget about mah moniker ![]() Plz help me with dis Z = ¡^1001 Av tried all mathematical tricks i kw, but i stil got stooked! |
Re: Nairaland Mathematics Clinic by Nobody: 7:14pm On Mar 23, 2015 |
bolkay47: From LHS 500 => 20(14) + 22(10) Since LHS = RHS Comparing LHS & RHS yields => 20(14) + 22(10) = 20x + 22y 14 + 10 = x + y thus => x = 14 and y = 10 1 Like |
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