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Re: Nairaland Mathematics Clinic by Nobody: 12:20am On May 10, 2015 |
(1) solve | 3-x | ≥ 1 (2) solve x > x² (3) find the domain f(x)= 1/√1-x² (4) if f(x)= x² - 1 then f(secθ) - f(tanθ) =? (5) if f(x) = logx, then find f(2sinx) + f(cosx) at x=π/4 (6) period of tanx and cotx is ? (7) if f(x)= x+1/x, x≠0, then f(1/x) =? ( ![]() (9) lim (1+2x)^1/x as x→0 =? (10) evaluate lim 1-cos2x/x²(1+cos2x) as x→0 =? (11) lim (1+(4/x))^x as x→∞ =? (12) lim 2sinx-sin2x/x^3 as x→0 =? (13) evaluate lim x(π/2 - arc tanx) =? (14) derivative of x²/1+x² with respect to x² =? (15) if y=x^4 - 7x^3 + 3, then d^3 y/dx^3 at x=2 is? (16) d/dx (2sin²x + cos2x) is ? (17) differntiate with respect to x, the function 2arc tan√x (18) find d/dx f(x) where f(x) = cosh^-1 (2x) at x=2 (19) if x²/a² + logy²/b² =1, then dy/dx =? (20) differentiate x^sinx with respect to x (21) find ∫(sec²x + tan²x)dx =? (22) evaluate (x²-x+1/√x) dx (23) ∫ dx/√4-9x² =? (24) determine ∫ dx/4+x² (25) ∫ (logx)²/x dx =? (26) ∫ |
Re: Nairaland Mathematics Clinic by Nobody: 12:22am On May 10, 2015 |
sure men, where una dey? |
Re: Nairaland Mathematics Clinic by Nobody: 12:22am On May 10, 2015 |
.. |
Re: Nairaland Mathematics Clinic by akpos4uall(m): 12:54pm On May 10, 2015 |
benji93: Nice try. You are closer to solving this problem but not there yet. There is a flaw in your approach. It has to do with your rate of change Using your formula R1 = Volume out/ total time This can be rewritten as R1 = change in Volume/ change in time R1 = change in V/ change in t The above is simply average flow rate within two time frames. This is not what we need. We need to find the flow rate at any point in time t. We can achieve this by applying first principle of differentiation. Which is: as the change in time approaches zero, we will get the flow rate at that point in time R1 = change in V/ change in t As change in t approaches zero, the above equation becomes R1 = dV/dt There you have it. I guess you can continue from here. |
Re: Nairaland Mathematics Clinic by benji93: 6:00pm On May 10, 2015 |
akpos4uall:I see you really like differentiation, but i dont tink its necessary here,infact you can solve this problem without it if there is a solution,you were not given V as a function of t,neither were you given r as a function of t,so where are u going to apply this, i know i could have done sth with my average multiplying by two and subtracting the initial rate,but you dont even have one in this question,only if i take the initial rate as zero,tht i can have the final rate as average times 2(in that case u can continue), your use of differentiation will only come handy when it involves functions,but yours contained values. |
Re: Nairaland Mathematics Clinic by Admissnandjobs(m): 7:13pm On May 10, 2015 |
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Re: Nairaland Mathematics Clinic by Emodeee: 9:59pm On May 10, 2015 |
Prove the de morgan theorem for 3sets A,B and C (AUBUC)'=A'nB'nC' mathematicians in the house, solve. |
Re: Nairaland Mathematics Clinic by thankyouJesus(m): 12:16pm On May 11, 2015 |
Emodeee:let me use "ojoro" method to prove: let u = {1, 2, . . . .10} a = {1,2,3} b = {4,5,6} c = {7,8,9} AUBUC = {1,2,3. . . 9} (AUBUC)' = {10} A'nB'nC' = {10}. QED |
Re: Nairaland Mathematics Clinic by factorial1(m): 7:54pm On May 11, 2015 |
Prove and not "show" sir. thankyouJesus: |
Re: Nairaland Mathematics Clinic by akpos4uall(m): 12:16pm On May 12, 2015 |
benji93: This problem involves differential equation We were given an initial value as well as another value after a certain amount of time Here is the problem once again akpos4uall: The flow rate can be represented as dV/dt This is proportional to the volume V i.e. dV/dt = kV Where k is the constant of proportionality dV/dt = kV dV/V = kdt Integrate to get InV = kt + c Where c is constant of integration Applying the given conditions When t = 0, V = 2500 & When t = 10, V = 25 The equation becomes InV = In2500 - 0.1tIn100 InV = In2500 - In(10)0.2t Take exponential of both sides to get V = 2500*(10)-0.2t The above equation is the volume-time relationship. Using V = 250, Solve for t to get t = 5 Hence it'll take 5hrs for the water to reduce to 10% When t = 1, V = 1577.4 which is 63.10% When t = 9, V = 39.622 which is 1.585% In a table like form, here is corresponding % left for the first ten 10hrs Time(hrs). % left 1 63.10 2 39.81 3 25.12 4 15.85 5 10.00 6 6.310 7 3.981 8 2.512 9 1.585 10 1.000 I stand to be corrected 1 Like |
Re: Nairaland Mathematics Clinic by akpos4uall(m): 12:21pm On May 12, 2015 |
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Re: Nairaland Mathematics Clinic by benji93: 4:59pm On May 12, 2015 |
akpos4uall:oh ok,you are right,in that light? |
Re: Nairaland Mathematics Clinic by benji93: 5:10pm On May 12, 2015 |
do you know that limiting h to 0 dy/dx = f(x)-f(x-h)/h, where h is change in x is valid, i mean instead of the usual dy/dx=f(x+h)-f(x)/h as h approaches O and is change in x, we can use the that above. I stand to be corrected guys. |
Re: Nairaland Mathematics Clinic by benji93: 11:36pm On May 12, 2015 |
[quote author=factorial1 post=33641165]Prove and not "show" sir. [/quotI] We can use logic gate to prove this,using 3 inputs there are 8 possibilities,but only 7 can be true when at least one input is true(A U B U C),but when this result is barred(considering all others except it), there is only one possibility(no input is right)(A U B U C)', but the possibility that all three inputs are right at the same time is one and the possibility that all three inputs are wrong at the same time is also one,the desired is just one possibility since it cannot be the former it has to be the latter. A'nB'nC' = (AUBUC)' I would have loved to use the truth table of logic gates to prove this but i am unable to,there are 16 truth tables in all 8 for multiplication of the inputs and 8 for the Addition of the inputs. |
Re: Nairaland Mathematics Clinic by frob0genius(f): 10:52am On May 15, 2015 |
Let A=(12)(35) and B=(13)(56) be in S6.
I'm confused here, should A and B be in S6 or just B?
#product of disjoint cycle
Thanks in advance |
Re: Nairaland Mathematics Clinic by Olarewajub: 6:08pm On May 20, 2015 |
Please someöne should help with this statistic questiön. In 200 tosses of a coin, 115 heads and 85 tails where observed, test the hypothesis that the coin is fair using significant level of 0.05% 1 Like |
Re: Nairaland Mathematics Clinic by Soneh(m): 3:03pm On May 24, 2015 |
pls house come to my rescue : 1.express the scalar product of two vectors in term of their covariant and contravariant components. 2.write a condition for the point A,B and c to be collinear. (I) show that(a ×b)+(b×c)+(c×a)=0 3.let f(x)=x³+2x²_kx+5. k is defined to be one third of the rank of the matrix below A=|3 -4| .. ..|1 -1| compute f(A) @benji93, akpos4uall,thankyoujesus,kendzyma et al please i need your help on this |
Re: Nairaland Mathematics Clinic by toobby(m): 10:17am On May 25, 2015 |
Please help solve make n subject of the formular
Sum of ap.
S=n/2(2a+(n-1)d) |
Re: Nairaland Mathematics Clinic by dejt4u(m): 3:03pm On May 25, 2015 |
toobby: make n the subject of the relation, you are gonna get; n2d + n(2a-d) - 2s = 0.. From here, solve the above quadratic equation using quadratic formula method; your answer should be something like this: n = [(d-2a) +/- SQRT(4(a2+2ds) - d(4a-d))] / 2d |
Re: Nairaland Mathematics Clinic by RobinHez(m): 8:01am On May 26, 2015 |
pls help..
A contractor agreeing to finish a work in 150 days, employed 75men
each working 8 hrs daily. after 90 days, only 2/7 of the work was
completed. Increasing the number of men by each working now for
10hrs daily, the work can be completed in time
A. 120 men
B. 100 men
C. 75 men
D. 150 men |
Re: Nairaland Mathematics Clinic by toobby(m): 8:32am On May 26, 2015 |
dejt4u:Tanx bro |
Re: Nairaland Mathematics Clinic by Bolaji16(m): 10:32am On May 26, 2015 |
someone should pls help with the solution
|
Re: Nairaland Mathematics Clinic by frob0genius(f): 12:11am On May 27, 2015 |
@ Bolaji16 NB: opened to corrections
|
Re: Nairaland Mathematics Clinic by Ubking0(m): 2:49am On May 27, 2015 |
Pls I need a PDF of advanced engineering mathematics by H. k. Dass..... |
Re: Nairaland Mathematics Clinic by frob0genius(f): 9:57am On May 27, 2015 |
Ubking0:I do have it. How can I send it to you? It's over 10mb |
Re: Nairaland Mathematics Clinic by Nobody: 10:14am On May 27, 2015 |
Gurus! |
Re: Nairaland Mathematics Clinic by Soneh(m): 1:53pm On May 27, 2015 |
frob0genius:please i am also in need of it, you can send it to this |
Re: Nairaland Mathematics Clinic by benji93: 6:17pm On May 27, 2015 |
RobinHez:i think the answer is not in, let the number of men = n let time = t in this case n is inversely proportional to t, n*t=k1 k1 = 75 * 720 = 54000 with this relationship it will take 90 men 600 hrs(60 days at 10hrs/day) to do the same work. but k1 is directly proportional to amount of work done k1=wk2 but since only 2/7 of the work was done, inorder to do the remaining work(5/7) which is a multiple of the first, we would need to increase k1, k1 for 5/7 of work = ((54000*(5/7))/2/7)=135000 k1=n*t t=60*10 k1=n*600 135000=n*600 n=225 men, note that i converted the days to hours. i stand to be corrected |
Re: Nairaland Mathematics Clinic by RobinHez(m): 6:34pm On May 27, 2015 |
benji93:wow! it was actually a Pume question from Delsu... i just hope other people are going to come up with answers too ![]() |
Re: Nairaland Mathematics Clinic by Kendzyma(m): 6:59pm On May 27, 2015 |
RobinHez:let N=num of men required H=hours used by each men D=number of days used W=number of work done. D relationship btwn d four is N=KW/DH............eq 1 WHERE K Is the proportionality constant from d first statement.... N=150 W=2a/7 D=90 H=8 k=? After insertin d values to equation 1,u get ur K= 756,000/29a..[/b]Now to d sec statement [b]N=? K=756,000/29a w=1-2a/7=5a/7 H=10 D=150-90=60 insertin dis values to equation 1,u get N=100men |
Re: Nairaland Mathematics Clinic by Drniyi4u(m): 8:29pm On May 27, 2015 |
Ubking0:į'vę gāţ įţ too |
Re: Nairaland Mathematics Clinic by RobinHez(m): 8:56pm On May 27, 2015 |
Kendzyma:ok thanks! |
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