benji93:


let rate =r and volume =v
R is proportional to V
R1/V1 = R2/V2
at the point where it drops to 1%,the rate is (99/100*2500)/10
rate=247.5
but the volume at that point is 2500-2475=25
when the level drops to 10% volume=(10% of 2500)=250
247.5/25=R2/250 ,R1 =2475
R1 = Volume out/ total time
2475 = 250/total time
total time taken to reduce to 10% = 6.1 minutes

akpos4uall:


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.

Emodeee:
Prove the de morgan theorem for 3sets A,B and C (AUBUC)'=A'nB'nC'

mathematicians in the house, solve.

thankyouJesus:

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

benji93:

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.

akpos4uall:
Lemme also drop mine. Btw, its off head assumption, but its a lil bit realistic.
Assuming that the rate of outflow of water from a tank of capacity 2500 dm³ as a result of a hole at the base of the tank is directly proportional to the volume of water in it at each point in time, if it takes 10 hours for the water to drop from full capacity to 1%, find
A) how long it'll take to drop to 10%
B) the volume of water left after 1 hour
C) the volume of water left after 9 hours
In short in a tabular form, show the % of amount left for each of the first ten hours.
*the solution is in the thinking*

akpos4uall:


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



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

toobby:
Please help solve make n subject of the formular
Sum of ap.
S=n/2(2a+(n-1)d)

dejt4u:

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

Ubking0:
Pls I need a PDF of advanced engineering mathematics by H. k. Dass.....

frob0genius:

I do have it. How can I send it to you?
It's over 10mb

RobinHez:
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

benji93:

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

RobinHez:
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

Ubking0:
Pls I need a PDF of advanced engineering mathematics by H. k. Dass.....

Kendzyma:
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