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Mathematics Clinic / 2015 Cowbell Mathematics Champion, Akinkuowo Honoured By School. / Lead City University Clinic Welcomes First Ever Baby Since 10 Years Of Opening (2) (3) (4)
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Re: Nairaland Mathematics Clinic by bolkay47(m): 7:32pm On Mar 23, 2015 |
doubleDx:yes sir.. But I used lagragian multiplier and I got different answer and it satisfies the equation like yours. |
Re: Nairaland Mathematics Clinic by Nobody: 7:36pm On Mar 23, 2015 |
bolkay47: Yes, that's what I was thinking too! There are many answers that satisfies the equation....so, whichever works for you is okay! |
Re: Nairaland Mathematics Clinic by bolkay47(m): 9:25pm On Mar 23, 2015 |
doubleDx:exactly sir |
Re: Nairaland Mathematics Clinic by Emdee590(m): 2:56pm On Mar 24, 2015 |
Please fellas , can I post programming questions here . Pls make una help |
Re: Nairaland Mathematics Clinic by thankyouJesus(m): 3:50pm On Mar 24, 2015 |
Re: Nairaland Mathematics Clinic by thankyouJesus(m): 3:55pm On Mar 24, 2015 |
DrMaths:if i stands for complex number, then, i1001 = (i2)500(i). The answer is i. 24th of March, 2015. 15:55:32 1 Like |
Re: Nairaland Mathematics Clinic by Nobody: 4:07pm On Mar 24, 2015 |
agentofchange1:Will be expecting.. Thanks |
Re: Nairaland Mathematics Clinic by Nobody: 4:12pm On Mar 24, 2015 |
[color=#990000][/color]Gurus in the house ![]() Projection of a vector u=4i-3j+k on the line passing through the points A(2,3,-1) and B(-2,-4,3). Thanks ![]() |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:00pm On Mar 28, 2015 |
miniyi: sorry , didn't reply u all dix while . set x+1=3sinhx. instead that's it 1 Like |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 10:27pm On Mar 28, 2015 |
Laplacian: nice1 .prof. how about using determinant , ? which is better ?. |
Re: Nairaland Mathematics Clinic by Abdullahi4u7(m): 11:11am On Apr 01, 2015 |
Pls explain with an example the concept of arithmetic and geometric progressions. |
Re: Nairaland Mathematics Clinic by thankyouJesus(m): 12:52pm On Apr 01, 2015 |
Abdullahi4u7:Google should be the best bet but. . . . . . .SMH |
Re: Nairaland Mathematics Clinic by Youngsage: 2:57pm On Apr 01, 2015 |
![]() ![]() This thread still exists!!! wow... Good to find out. kudos y'all |
Re: Nairaland Mathematics Clinic by calculus3e(m): 7:06am On Apr 02, 2015 |
find the value of 5e^0.5 to 5significant figure |
Re: Nairaland Mathematics Clinic by calculus3e(m): 7:08am On Apr 02, 2015 |
integrate sinXcosX |
Re: Nairaland Mathematics Clinic by rhydex247(m): 9:15am On Apr 02, 2015 |
calculus3e:$sinxcosx dx using substitution method let u=cosx.... du=-sinxdx..... dx= -du/sinx $ sinx(u) -du/sinx -$udu.... -u^2/2 + C finally, we have -cos^2 x/2 + C 1 Like |
Re: Nairaland Mathematics Clinic by rhydex247(m): 9:18am On Apr 02, 2015 |
calculus3e:8.2436 |
Re: Nairaland Mathematics Clinic by jackpot(f): 5:58pm On Apr 04, 2015 |
@Maths generals, Please help meeeeeee A combat-ready soldier on a level ground, wants to toss a grenade from a height of 1.8 meters off the ground, to a stationary enemy target, on the edge of a cliff of height 50metres. The horizontal distance of the soldier from the foot of the cliff is 20metres. (a) If he intends tossing at an angle of 75degrees, what is the least initial velocity of his projection? (b)If he intends tossing with an initial velocity of 70m/s, what is the least angle of projection if his missile will hit the target at the shortest possible time? (take g=9.8m/s2, ignore spin, drag force, air resistance) Tags: agentofchange1, doubleDx, Laplacian, AlphaMaximus, STENON, AmazingAngel, Richiez, rhydex247, montty, CalculusFx, thankyouJesus, calculus3e, all maths generals, sergeants |
Re: Nairaland Mathematics Clinic by benji93: 10:47pm On Apr 05, 2015 |
rhydex247:you tried bro,but maybe you should try it this way sin(2x) =2sinxcosx (1/2)sin(2x) =sinxcosx dy = sinxcosx dx= (1/2)sin(2x) dx y=(-1/24)cos(2x) |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 11:06pm On Apr 05, 2015 |
benji93: let's differentiate his answer & see if we could get back the integrand d(-cos^2 x/2) /dx = -(-2) cosxsinx /2 = cosxsinx = sinxcosx..............(as expected) now let's check yours dy/dx = -(-2)/24 sin2x = 1/12 sin2x = 2/12 * sinxcosx =1/6 sinxcosx =/= sinxcosx ??. so what can you say now ?. |
Re: Nairaland Mathematics Clinic by agentofchange1(m): 11:08pm On Apr 05, 2015 |
jackpot: OK , on bed now without writing stuffs , maybe 2maro b4 nw.. |
Re: Nairaland Mathematics Clinic by jackpot(f): 3:23pm On Apr 06, 2015 |
Re: Nairaland Mathematics Clinic by jackpot(f): 3:26pm On Apr 06, 2015 |
question number 2: An athlete can throw a javelin 60m from a standing position. If he can run 100m at constant velocity in 10s, how far could he hope to throw the javelin while running? tags: agentofchange1, doubleDx, Laplacian, rhydex247, STENON, AmazingAngel, etc. |
Re: Nairaland Mathematics Clinic by mdee1(m): 6:37am On Apr 07, 2015 |
pls any help will be appreciated, thanks
|
Re: Nairaland Mathematics Clinic by tohero(m): 2:37pm On Apr 07, 2015 |
jackpot: Using the formula v2 = u2+2gh Where v=0 , g=-9.8m/s2 h= 50 - 1.8 = 48.2m Vertical component of the initial velocity = u sinA Where A=75o. Then the formula bcoms 02 = u2sin275o - 2*9.8*48.2 That is 19.6*48.2 = u2sin275o Calculator should finish this. |
Re: Nairaland Mathematics Clinic by jackpot(f): 3:57pm On Apr 07, 2015 |
tohero:but Sir, does it mean that the distance (20m) of the cliff from the soldier doesn't matter in the solution? |
Re: Nairaland Mathematics Clinic by jackpot(f): 4:10pm On Apr 07, 2015 |
question 3 a military jet bomber flying at an altitude 600m in a horizontal straight line above a target with a speed of 360km/h drops off a bomb in order to hit the target with an initial velocity 0km/h. Suppose that at the time of drop, the horizontal distance of the target from the jet is 750m and the bomb's explosion radius is 300m, determine whether the target is hit by the bomb. If not, what is the range of the miss? |
Re: Nairaland Mathematics Clinic by jackpot(f): 4:25pm On Apr 07, 2015 |
Question four A poacher standing at a distance of 25m from a tree saw, aims and takes a shot with a catapult at a bird whose vertical distance from the ground is 30m. Suppose that (i) the catapult is held at a distance of 1.5m above ground level, (ii) the initial velocity of the shot was 40m/s, (iii) the bird was hit before it could make a move, then calculate (a) the duration of the shot (b) the angle of the shot |
Re: Nairaland Mathematics Clinic by benji93: 5:40pm On Apr 07, 2015 |
agentofchange1:sorry y = (-1/4)cos(2x) i didnt realize the typographical error |
Re: Nairaland Mathematics Clinic by tohero(m): 9:01pm On Apr 07, 2015 |
jackpot: Think about you throwing me a stone at a fixed distance. If you want the stone to hit me hard, you must throw it directly at me. If you threw it in a parabolic form in the sense that it went up and on coming down it hit me, believe me, it is not going to be painful as the previous. NB: -This distance between us doesn't determine how painful it will be to me. -what determines it is the height of where I am which will prompt the angle of elevation. I hope you get the analogy. |
Re: Nairaland Mathematics Clinic by bolkay47(m): 9:05pm On Apr 07, 2015 |
rhydex247:the answer can also be sin^2x/2 taking u=sinx dv=cosx and integrate by parts |
Re: Nairaland Mathematics Clinic by benji93: 11:47pm On Apr 07, 2015 |
jackpot: the catapult was held at 1.5m above the ground means it is 28.5m below the bird vertically. the poacher is 25m from the tree horizontally we have a right angled triangle with 28.5m as the opposite and 25m as the adjascent, and the path taken by the catapult is the hypotenuse x^2 = 25^2 + 28.5^2 x^2 =1437.25 x=37.911m duration of shot = 37.911/40=0.955s tan theta = 28.5/25 theta = tan inverse of 28.5/25=0.851m 1 Like |
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