biolabee:



you know what guys lets slow down and teach him how to fish
johnpaul can u post what u have done so far
we tell u what uve done wrong
that way u learn the principles better

general biolabee and richiez, if you truly want to see my work on this question, u must know that your mathematical knowledge is more than mine, so plz however premature my ideas, solution, or logic may b to u, dont make me look like a mathematical fool, and to tell u the truth, i like this idea of posting my work, nd if not correct, u help me

for the question, this was what i did:
let the running speed be x
let the walking speed be y,
from the fist statement,
(8km/50min)x + (2km/50min)y=10km/50min..........eq 1
from the second statement
(4km/75min)x + (6km/75min)y=10km/75...........eq 2
now, the parameter(minute) need to b converted to hour to give the correct unit(km/hr).....in doing that.,

johnpaul1101: back to simultaneous, plz help me.....i've been trying this since this evening, but i keep getting confused:
a man travels 10km in 50mins if he runs for 8km and walks for 2km. If he runs 4km and walks 6km, his time is 1hour, 15min, find his running and walking speed.

well for the sake of others who wants to learn, here's the solution.

first of all, we must recall that Time = Distance/Speed .........(1)
lets convert unit of time to hours only
50mins = 50/60 =5/6 hrs
1hr 15mins =75mins = 75/60 = 5/4 hrs

let running speed = x
and walking speed = y

now applying eqn(1) we get
5/6 = 8/x + 2/y ......................(2)
5/6 = (8y + 2x)/xy
5xy = 6(8y + 2x)
5xy = 48y + 12x
x(5y-12) = 48y
x = 48y/(5y-12)................(3)
also;
5/4 = 6/x + 4/y ...............(4)
putting value of x obtained in eqn(3) into eqn(4) we get;
5/4 = 6(5y-12)/48y + 4/y
multiply thru by 'y'
5y =4[6(5y-12)/48 + 4]
5y= 5y/2 -6 +16
5y/2 = 10
y =20/5 = 4
from eqn(3)
x= 48y/(5y-12)
x= 48(4)/[5(4) -12)]
x= 192/8 =24
therefore running speed= x = 24km/hr
and walking speed = y = 4km/hr

Richiez:

well for the sake of others who wants to learn, here's the solution.

first of all, we must recall that Time = Distance/Speed .........(1)
lets convert unit of time to hours only
50mins = 50/60 =5/6 hrs
1hr 15mins =75mins = 75/60 = 5/4 hrs

let running speed = x
and walking speed = y

now applying eqn(1) we get
5/6 = 8/x + 2/y ......................(2)
5/6 = (8y + 2x)/xy
5xy = 6(8y + 2x)
5xy = 48y + 12x
x(5y-12) = 48y
x = 48y/(5y-12)................(3)
also;
5/4 = 6/x + 4/y ...............(4)
putting value of x obtained in eqn(3) into eqn(4) we get;
5/4 = 6(5y-12)/48y + 4/y
multiply thru by 'y'
5y =4[6(5y-12)/48 + 4]
5y= 5y/2 -6 +16
5y/2 = 10
y =20/5 = 4
from eqn(3)
x= 48y/(5y-12)
x= 48(4)/[5(4) -12)]
x= 192/8 =24
therefore running speed= x = 24km/hr
and walking speed = y = 4km/hr

but the answer here says 16km/hr and 6km/hr

johnpaul1101: for the question, this was what i did:
let the running speed be x
let the walking speed be y,
from the fist statement,
(8km/50min)x + (2km/50min)y=10km/50min..........eq 1
from the second statement
(4km/75min)x + (6km/75min)y=10km/75...........eq 2
now, the parameter(minute) need to b converted to hour to give the correct unit(km/hr).....in doing that.,

now you anticipated the solution very well, but i must tell you, knowing how to play with your distance, speed and time relationships give you an edge over how to generate your simultaneous equations.

1. distance = speed * time
2. speed = distance/time
3. time = distance/speed

now i did not choose the first and the second one because the time given was for distance covered after running and walking, and not for either running or walking alone.
for example it will be a big error to say 8km = running speed/(5/6)hrs note 5/6 hrs = 50mins
this is 50 mins is not for running alone, it was time taken after running and walking.

i chose the 3rd relationship because it keeps total time on one side of the eqn and the i can equate it to the addition of the time after running and the time after walking
for example 5/6 hrs = 8km/running speed + 2km/walking speed note; time=distance/speed

[quote author=Richiez]

nice try bro, although you made some some mistakes especially at the point where you took log of both sides. e.g it ought to be log(x+y)= log5 and not logx + logy = log5. but i must admit, you did a very good job.

here's another approach;

x + y = 5.......(1)
x^x + y^y =31.......(2)

from eqn(1), x=5-y, hence we substitute this value for x in eqn(2).
(5-y)^(5-y) + y^y = 31.......(3)
now, we have reduced the problem to what PLS AM NEW HERE HELP ME IN MATH

johnpaul1101:
but the answer here says 16km/hr and 6km/hr

Richiez:

well for the sake of others who wants to learn, here's the solution.

first of all, we must recall that Time = Distance/Speed .........(1)
lets convert unit of time to hours only
50mins = 50/60 =5/6 hrs
1hr 15mins =75mins = 75/60 = 5/4 hrs

let running speed = x
and walking speed = y

now applying eqn(1) we get
5/6 = 8/x + 2/y ......................(2)
5/6 = (8y + 2x)/xy
5xy = 6(8y + 2x)
5xy = 48y + 12x
x(5y-12) = 48y
x = 48y/(5y-12)................(3)
also;
5/4 = 6/x + 4/y ...............(4)
sorry i interchanged running distance with walking distance,the correct eqn ought to be
5/4 = 4/x + 6/y ..................(4)
putting value of x obtained in eqn(3) into eqn(4) we get;
5/4 = 4(5y-12)/48y + 6/y
multiply thru by 'y'
5y =4[4(5y-12)/48 + 6]
5y= 5y/3 -4 +24
10y/3 = 20
y =60/10 = 6
from eqn(3)
x= 48y/(5y-12)
x= 48(6)/[5(6) -12)]
x= 288/18 = 16
therefore running speed= x = 16km/hr
and walking speed = y = 6km/hr

this goes to show that even the minutest mistakes could cost alot in mathematics

SOBSISRAEL: now, we have reduced the problem to what PLS AM NEW HERE HELP ME IN MATH

you're most welcome. how do you want us to help you? please feel free to express yourself clearly

johnpaul, this is not to make fun of you but use the opportunity to learn

you used km and mins which are not the common units of speed and as richie showed u, you have to take steps and recheck
More questions but please put whAT YOU have done so far before we reply

i salute all ma 5* generals for here lo........abeg its this question i would like sum1 to solve oo.....i knw its gonna b too simple for ppl like Richiez,ositadima,biolabee and d likes........find §^2V/§y^2 wen V=tan`(y/x)........§ means partial differentiation,tan` means arc tan......tanks bros

kasbeats: i salute all ma 5* generals for here lo........abeg its this question i would like sum1 to solve oo.....i knw its gonna b too simple for ppl like Richiez,ositadima,biolabee and d likes........find §²V/§y² wen V=tan`(y/x)........§ means partial differentiation,tan` means arc tan......tanks bros

I want to believe your question is, find ∂²v/∂y² if v = tan^(-1) (y/x)? If it is then here is the solution.

To solve this we have to use the derivative of tan(-1)A or arcTanA which is 1/(A² + 1). Therefore, if A represents => y/x, the first derivative of the entire expression can be expressed using chain rule =>

∂[tan^(-1) (y/x)]/∂y =>

∂v/∂y = {1/[(y/x)² +1]}.∂(y/x)/∂y
∂v/∂y = {1/[(y/x)² +1]}.1/x
∂v/∂y = x/(y² + x²)

Having computed the first derivative of function wrt y, we can now differentiate the result wrt y for the second derivative of the given function => ∂²v/∂y².

x/(y² + x²)
Lets put t = y² + x², ∂t/∂y = 2y, ∂[x]/∂t = 0
Then the function becomes = x/t and using quotient rule its derivative wrt t =>

{t[ ∂(x)/∂t] - x( ∂t/∂y)}/t²

Subtituting t and the values of the other derivatives in the above formula, yields=>
= - x(2y)/t²) = -2yx/(y² + x²)²
Hence ∂[x/(y² + x²)]/∂y => -2yx/(y² + x²)²
Which is also the 2nd derivative of the implicit function =>

:. ∂²[arcTan(y/x)]/∂y² => -2yx/(y² + x²)²

I'm impressed

arctan(y/x)=v so, tan(v)=y/x I am going to manipulate with sin^2(v)+cos^2(v)=1

Now, dividing sin^2(v)+cos^2(y/x)=1 by cos^2(v)... u know sin^2(v)/cos^2(x)=tan^2(v) and 1/cos^2(v)=sec^2(v)
1+tan^2(v)=sec^2(v)

So, sec^2(v)=1+(y/x)^2
recalling arctan(y/x)=v
sec(arctan(y/x))=sec(v)
square both sides sec^2(arctan(y/x))=sec^2(v)=1+(y/x)^2
sec^2(arctan(y/x))=1+(y/x)^2

now, using chain rule and taking partial derivative of both sides with respect to x we have...

∂[sec^2(arctan(y/x))]/∂x=∂[1+(y/x)^2]/∂x

∂[arctan(y/x))]/∂x*2[sec^2(arctan(y/x))]*tan[(arctan(y/x)]*(-(y/x)^2)=-2*y^2/x^3

now, tan[(arctan(y/x)]=y/x
sec^2(arctan(y/x)=1+(y/x)^2

substituting back

∂[arctan(y/x))]/∂x*2*[1+(y/x)^2]*(y/x)*(-(y/x)^2)=-2*y^2/x^3

∂[arctan(y/x))]/∂x*[1+(y/x)^2]*-2*y^3/x^3=-2*y^2/x^3

∂[arctan(y/x))]/∂x*[1+(y/x)^2]=y

∂[arctan(y/x))]/∂x=y/[1+(y/x)^2]

second partial derivative

∂^2[arctan(y/x))]/(∂x)^2=∂[y/[1+(y/x)^2]/∂x using chain rule again...
p
∂^2[arctan(y/x))]/(∂x)^2=-y[1+(y/x)^2]^(-2)*-2*y^2/x^3

∂^2[arctan(y/x))]/(∂x)^2=2xy(x^2+y^2)^2

DoubleDx there is no minus in the final solution, check again

kasbeats: i salute all ma 5* generals for here lo........abeg its this question i would like sum1 to solve oo.....i knw its gonna b too simple for ppl like Richiez,ositadima,biolabee and d likes........find §^2V/§y^2 wen V=tan`(y/x)........§ means partial differentiation,tan` means arc tan......tanks bros

@Osita, I think you didn't get the question properly, it says we should differentiate wrt to y and not x. You did a good job but you solved yours wrt x. That explains why your first derivative differs from mine.

There is no need removing your workings. It's perfect if we are to differentiate with respect to x. That could help some folks, just indicate that you are differentiating wrt x then check the one wrt y. 1heart bro.

I NEED YOUR ASSISTANCE GUYS?!! You dont know how happy i am to find this thread, feeling so relaxed now! Please help me solve the following questions on Matrix.

The Matrix of units of products produced by Daddy K cream factory in three shifts is given:
(2 3 7)
( X-Y 1 3)
(x+Y 3 5)
You are required to determine:
(a) Rhe values of x and y, if Matrix A is Symmetric.
(b) The gross revenue vector (AP) for each product, if the price $P per unit of the product is given by the vector P = (45)
(30)
(55)

please i need the solution before tomorrow!!! Pleaseeeeeeee!!!

Seriously! You guys are GURUS!!!

doubleDx:

@Osita, I think you didn't get the question properly, it says we should differentiate wrt to y and not x. You did a good job but you solved yours wrt x. That explains why your first derivative differs from mine.

There is no need to removing your workings. It's perfect if we are to differentiate with respect to x. That could help some folks, just indicate that you are differentiating wrt x then check the one wrt y. 1heart bro.

ha haha, my bad...

SpicyMimi: I NEED YOUR ASSISTANCE GUYS?!! You dont know how happy i am to find this thread, feeling so relaxed now! Please help me solve the following questions on Matrix.

The Matrix of units of products produced by Daddy K cream factory in three shifts is given:
(2 3 7)
( X-Y 1 3)
(x+Y 3 5)
You are required to determine:
(a) Rhe values of x and y, if Matrix A is Symmetric.
(b) The gross revenue vector (AP) for each product, if the price $P per unit of the product is given by the vector P = (45)
(30)
(55)

please i need the solution before tomorrow!!! Pleaseeeeeeee!!!

You want us to do your homework for you, while u dey chill out abi?

SpicyMimi: I NEED YOUR ASSISTANCE GUYS?!! You dont know how happy i am to find this thread, feeling so relaxed now! Please help me solve the following questions on Matrix.

The Matrix of units of products produced by Daddy K cream factory in three shifts is given:
(2 3 7)
( X-Y 1 3)
(x+Y 3 5)
You are required to determine:
(a) Rhe values of x and y, if Matrix A is Symmetric.
(b) The gross revenue vector (AP) for each product, if the price $P per unit of the product is given by the vector P = (45)
(30)
(55)

please i need the solution before tomorrow!!! Pleaseeeeeeee!!!

Questiob 1 a

[ 2 3 7]
[(x - y) 1 3]
[(x+y) 3 5]

A matrix A is said to be symmetric if A equals A transpose. i.e A => A^T which means that if we interchange rows and columns of A, the matrix remains unchanged.

:. If the above matrix is A, then A transpose will be =>

[2 x-y x+y]
[3 1 3]
[7 3 5]

Looking at the two matrices, A and A^T. The number on the first row second column of A^T is given by the expession (x - y) which equals 3 in the matrix A.

Hence x - y = 3 ......(1)

Also, the number on the first row third column of A^T is given by the expression (x + y) which equals 7 in the matrix A.

:. x + y = 7 .....(2)

Adding equation (1) and (2) yields=>

x - y = 3
x + y = 7
2x = 10
x = 5

subtitute x in equation (1)

5 - y = 3
y = 5 - 3
y = 2

:. x = 5 and y = 2.

doubleDx:

Questiob 1 a

[ 2 3 7]
[(x - y) 1 3]
[(x+y) 3 5]

A matrix A is said to be symmetric if A equals K transpose. i.e A => A^T which means that if we interchange rows and columns of A, the matrix remains
unchanged.

:. If the above matrix is A, then K transpose will be =>

[2 x-y x+y]
[3 1 3]
[7 3 5]

Looking at the two matrices, A and A^T. The number on the first row second column of A^T is given by the expession (x - y) which equals 3 in the matrix A.

Hence x - y = 3 ......(1)

Also, the number on the first row third column of A^T is given by the expression (x + y) which equals 7 in the matrix A.

:. x + y = 7 .....(2)

Adding equation (1) and (2) yields=>

x - y = 3
x + y = 7
2x = 10
x = 5

subtitute x in equation (1)

5 - y = 3
y = 5 - 3
y = 2

:. x = 5 and y = 2.

See babe I wan chike..., u don quickm answer finish, finish it up joo...

ositadima1:

See babe I wan chike..., u don quickm answer finish, finish it up joo...

Lol bro, I'm on mobile, typing and walking help her finish the b part. Carry on bruv, I'm with you all the way

doubleDx:

Lol bro, I'm typing and walking help her finish the b part. Carry on bruv, I'm with you all the way

brother me dey walka too, branch some where and end what you started.

ositadima1:

brother me dey walka too, branch some where and end what you started.

Lol, what I started? When did we start doing that on this thread? That's quite funny! If that's the case then I think only the op should answer questions on here. Na wa oh

Well, I'm helping just like you are doing so when I get home I'll help her finish the b part.

doubleDx:

Lol, what I started? When did we start doing that on this thread? That's quite funny! If that's the case then I think only the op should answer questions on here. Na wa oh

Well, I'm helping just like you are doing so when I get home I'll help her finish the b part.

Yes sir!

ositadima1:

See babe I wan chike..., u don quickm answer finish, finish it up joo...

Oh, all the questions you've been solving here is so you can gain popularity and get chics online? Well, some of us are married and we do this to help people and not for some e-chics

doubleDx:

Oh, all the questions you've been solving here is so you can gain popularity and get chics online? Well, some of us are married and we do this to help people and not for some e-chics

lol, what did I do to you this Sunday

ositadima1:

You want us to do your homework for you, while u dey chill out abi?

hahahahahahahaha, u na bad guy

ositadima1:

lol, what did I do to you this Sunday

No beef bruv. I've got no beef with you!

But the way you addressed my post was uncalled for. It's not compulsory to help folks. That others are not willing to help doesn't mean we are the best on here, I studied Engineering and not BEd Mathematics.

I'm just giving the best I can out of my free will. If I decide not to log on to nairaland again, I'll leave and never come back. Nothing matters to me here because nairaland doesn't pay me. If you are quoting my post next time, make sure you write something reasonable. Gosh! We are not kids anymore bruv!

ositadima1:

You want us to do your homework for you, while u dey chill out abi?

hahhahahhahahahhahahahahhahahahahhah:...haba nah! U r sooooooooooo hilariousssssssssss!!!! Actually i aint o! I got loads of assignments on my table...so have been so busy surfing d internet! Chill ke? Nothn like dat o!

God bless You Doubledx! You ve just made ma day, cause right now, am the most happppiiiieeeeeeeesssst girl!!!! Words cant describe how excited and grateful i am right now! God Bless you and your family abundantly in Jesus name.Amen!!!

Let it go...bro

doubleDx:

Lol, what I started? When did we start doing that on this thread? That's quite funny! If that's the case then I think only the op should answer questions on here. Na wa oh

Well, I'm helping just like you are doing so when I get home I'll help her finish the b part.

Thank you!!! Thank You!!! Thank you!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! Thank You!!! ThaNk You!!! Thank You!!! And Thank You!!! God am sooooo happpyyyy!!!! And sooooo Grateful!!!!!!

Doubledx Ride on, you're a star
@Ositadima1 common show dis chic what you got!