Drone10:
Please, i need a help on this!

On No 1. What kind of equation is this? Who can help me out or maybe a guidelines

No 2. How do i approach this type of maths? Is it the normal BODMAS method approach?

1. First, expand the equation and then solve as a quadratic equation.

Clue: 2(x²+3)-2+4(6x-6)
2x²+6-2+24x-24
2x²+24x-20
Factorize and solve the quadratic function

2 and 3. Approach using BODMAS

Cheers

Can someone pls explain to me how the integral of 1/(y+1)(y-1) dy is equal to 1/2log(y-1/y+1)

Soneh:
Can someone pls explain to me how the integral of 1/(y+1)(y-1) dy is equal to 1/2log(y-1/y+1)

first split 1/(y+1)(y-1) into partial fractions den integrate.


I assume u know how to break it down to partial fractions

ladokuntlad:

first split 1/(y+1)(y-1) into partial fractions den integrate.

I assume u know how to break it down to partial fractions

tthanks

Soneh:
Can someone pls explain to me how the integral of 1/(y+1)(y-1) dy is equal to 1/2log(y-1/y+1)

SOLUTION.

$1/(Y+1)(Y-1)dy

DECOMPOSING INTO PARTIAL FRACTION

1=A/(Y+1) +B/(Y-1)

1=A(Y-1)+B(Y+1)

WHEN Y=1 , B=1/2

WHEN Y=-1 ,A=-1/2

THE INTEGRAL IS NOW OF THE FORM

$ {-1/2(Y+1) +1/2(Y-1)}dy

Upon integration we have

1/2(ln(Y-1))-1/2(ln(Y+1))+K

let K=lnA

ln(Y-1)^1/2-ln(Y+1)^1/2+lnA

finally, we have 1/2ln{(Y-1)/(Y+1)}A

OR

ln{A(Y-1)/(Y+1)}^1/2

DONE!
P.SERIES

Drone10:
Please, i need a help on this!

On No 1. What kind of equation is this? Who can help me out or maybe a guidelines

No 2. How do i approach this type of maths? Is it the normal BODMAS method approach?

The first question is not an equation but an expression and that's why you we're asked to simplify and not "solve". So all you need do is to expand brackets and collect like terms.
It becomes an equation only when the equation sign "=" is introduced and to solve that, just use any of the methods used for solving a quadratic equation as someone has stated earlier

Gurus, what's fibonacci series all about

Best thread ever

jessetom:
Gurus, what's fibonacci series all about

Fibonacci series is a series of numbers in which each Fibonacci number is the sum of the two preceding numbers. The simplest is the series 1, 1, 2, 3, 5, 8, 13, 21, 34.....

Mehn I miss these things

ElDeeVee:


Fibonacci series is a series of numbers in which each Fibonacci number is the sum of the two preceding numbers. The simplest is the series 1, 1, 2, 3, 5, 8, 13, 21, 34.....

Mehn I miss these things

Clap for yourself

frob0genius:


Clap for yourself

Clap for me

Where have you been sef

Still following since thread was opened *though From sidelines* can't find most of the old generals around any more.............. #remindsmeofmyundergraduatedays_niceworkpeople.

ElDeeVee:


Fibonacci series is a series of numbers in which each Fibonacci number is the sum of the two preceding numbers. The simplest is the series 1, 1, 2, 3, 5, 8, 13, 21, 34.....

Mehn I miss these things

Wow pretty cool. How about fibonacci spiral

jessetom:


Wow pretty cool. How about fibonacci spiral

Well, Fibonacci spiral is created by drawing circular arcs connecting the opposite corners of squares in Fibonacci series. Generally, the spiral starts from right bottom edge of the square and moves in an anti-clockwise direction till the end of the Fibonacci tiling.

Richiez:
We diagnose and solve math problems here
This thread is the meeting point for nairaland math gurus...I dare anyone to ask a question in mathematics without me having an answer to them, LETS START

A student needs at least 3 notebooks and 3 pencils.notebooks cost 60naira and pencils 36naira and the student has 360naira to spend.the student decides to spend as much as possible of his 360naira.
A)how many ways can he spend his money?
B)does any of the ways give him change?if so how much?

2..a car repair workshop uses large numbers of filter,costing 300naira each,and plugs costing 400naira each.The workshop owner allows 30,000naira for these parts and needs at least twice as many filters as plug.there must also be at least 50 filters and 20 plugs..
A)what is the largest number of parts that can be bought,and in what?
B)if it is decided to buy as many plugs as conditions allow,how many of each part can be bought?





Pls show d workings

A student needs at least 3 notebooks and 3 pencils.notebooks cost 60naira and pencils 36naira and the student has 360naira to spend.the student decides to spend as much as possible of his 360naira.
A)how many ways can he spend his money?
B)does any of the ways give him change?if so how much?

2..a car repair workshop uses large numbers of filter,costing 300naira each,and plugs costing 400naira each.The workshop owner allows 30,000naira for these parts and needs at least twice as many filters as plug.there must also be at least 50 filters and 20 plugs..
A)what is the largest number of parts that can be bought,and in what?
B)if it is decided to buy as many plugs as conditions allow,how many of each part can be bought?





Pls show d workings

gnaseez:
A student needs at least 3 notebooks and 3 pencils.notebooks cost 60naira and pencils 36naira and the student has 360naira to spend.the student decides to spend as much as possible of his 360naira.
A)how many ways can he spend his money?
B)does any of the ways give him change?if so how much?

2..a car repair workshop uses large numbers of filter,costing 300naira each,and plugs costing 400naira each.The workshop owner allows 30,000naira for these parts and needs at least twice as many filters as plug.there must also be at least 50 filters and 20 plugs..
A)what is the largest number of parts that can be bought,and in what?
B)if it is decided to buy as many plugs as conditions allow,how many of each part can be bought?





Pls show d workings

Omo, e get as e be ooo. Na the interpretation dey give me problem pass. If only someone can help me interpret.

gnaseez:
A student needs at least 3 notebooks and 3 pencils.notebooks cost 60naira and pencils 36naira and the student has 360naira to spend.the student decides to spend as much as possible of his 360naira.
A)how many ways can he spend his money?

He sincerely has only one way since he must buy AT LEAST 3 of each.

gnaseez:
A student needs at least 3 notebooks and 3 pencils.notebooks cost 60naira and pencils 36naira and the student has 360naira to spend.the student decides to spend as much as possible of his 360naira.
A)how many ways can he spend his money?
B)does any of the ways give him change?if so how much?

Pls show d workings

A. 60*3+36*5=360 first possibility
60*4+36*3=336 second possibilty.
Thus A= 2ways

B. yes there is a change of 24naira

gnaseez:

2..a car repair workshop uses large numbers of filter,costing 300naira each,and plugs costing 400naira each.The workshop owner allows 30,000naira for these parts and needs at least twice as many filters as plug.there must also be at least 50 filters and 20 plugs..
A)what is the largest number of parts that can be bought,and in what?
B)if it is decided to buy as many plugs as conditions allow,how many of each part can be bought?

Pls show d workings

Don't really understand what you mean by "largest number of parts" but first let us look at the conditions
it says at least two filters to one plug i.e at least 2:1
but there is an initial condition of 50filters and 20plugs
let us get the value for that first

50*300(price for filter) + 20*400(price for plug)= 15,000+8,000=N23,000

Now we are left with 7,000, we have to permutate it in such a way that
300x+ 400y=7,000 where x= number of filter and y=number of plug


possible outcomes are

300*2+400*16=7,000 ......(1)
300*10+400*10=7,000 ......(2)
300*14+400*7=7,000 ......(3)
300*22+400*1=7,000 ......(4)

observing the four possibilities, we have eqn 4 as the highest number of parts


Thus, largest numbers of part that can be bought is
(50+22) filters and (20+1) plugs
implies 72filters and 21plugs
A= 72filters


B. First, let us add up our possibilities to the initial values given

eqn(1)..... (50+2)filters and (20+16) plugs =52:36=13:9 not our answer
eqn(2)..... (50+10)filters and (20+10) plugs =60:30=2:1 our answer
eqn(3)..... (50+14)filters and (20+7) plugs =64:27=also our answer
eqn(4)..... (50+22)filters and (20+1) plugs =72:21=24:7 also our answer


looking through,
eqn(1) didn't satisfy the condition filter must b at least twice plug

all other equations satisfy the conditions given but our aim is to find the one that has the highest number of plugs which makes eqn(2) the right choice.
Thus B. maximum of 30plugs+60filters=90numbers of parts

Richiez:
We diagnose and solve math problems here
This thread is the meeting point for nairaland math gurus...I dare anyone to ask a question in mathematics without me having an answer to them, LETS START

1)There are 5 more girls than boys in a class.if 2 boys join the class,the ratio of girls to boys will be 5:4.
Find the:
a)number of girls in the class
b)total number of pupils in the class
c)probability of selecting a boy as class captain.

2)A number is written as 37 in base x. Twice the number is written as 75 in base x. Find the value of x.

Anyone should kindly help solve this with workings...thanks

Azeola:


1)There are 5 more girls than boys in a class.if 2 boys join the class,the ratio of girls to boys will be 5:4.
Find the:
a)number of girls in the class
b)total number of pupils in the class
c)probability of selecting a boy as class captain.

Let number of boys be X and Girls be Y respectively

Then from information given, we have
y=x+5

now if there is an increment of 2 boys we will have a ratio of 5:4 thus interpreting this, we arrive at

yx+2)=5:4

recall that y=x+5

so

(x+5)x+2)=5:4

implies
(x+5)/(x+2)=5/4

4(x+5)=5(x+2) expanding gives

4x+20=5x+10
x=10 thus y=10+5=15

i.e Number of boys are 10 and girls 15

a). Number of girls is 15
b). Total number of pupil in class is 25
c). Probability of selecting a boy as captain is 10/25=2/5

Azeola:


2)A number is written as 37 in base x. Twice the number is written as 75 in base x. Find the value of x.

First we need to convert both numbers in base X to base Ten. i.e

37_x=(3x+7)₁₀

also
75_x=(7x+5)₁₀

Interpreting the question, we have

2*((3x+7)₁₀))=(7x+5)₁₀

2(3x+7)=7x+5
6x+14=7x+5
x=9
Thus the base value is 9.

Azeola:


1)There are 5 more girls than boys in a class.if 2 boys join the class,the ratio of girls to boys will be 5:4.
Find the:
a)number of girls in the class
b)total number of pupils in the class
c)probability of selecting a boy as class captain.

2)A number is written as 37 in base x. Twice the number is written as 75 in base x. Find the value of x.

Anyone should kindly help solve this with workings...thanks

1)solved

ladokuntlad:



Let number of boys be X and Girls be Y respectively

Then from information given, we have
y=x+5

now if there is an increment of 2 boys we will have a ratio of 5:4 thus interpreting this, we arrive at

yx+2)=5:4

recall that y=x+5

so

(x+5)x+2)=5:4

implies
(x+5)/(x+2)=5/4

4(x+5)=5(x+2) expanding gives

4x+20=5x+10
x=10 thus y=10+5=15

i.e Number of boys are 10 and girls 15

a). Number of girls is 15
b). Total number of pupil in class is 25
c). Probability of selecting a boy as captain is 10/25=2/5

Good one bro.seen my mistake.calculated 4 no of persons in class after d extra 2 boys joined

@ ladokuntlad and Joe82834 thanks guys....really appreciate the prompt response,would b posting some others soon,hope u guys won't get tired. Thanks once again.

Azeola:
@ ladokuntlad and Joe82834 thanks guys....really appreciate the prompt response,would b posting some others soon,hope u guys won't get tired. Thanks once again.

Not atall.


Bring it on anytime

Find the values of x, y and z in the systems of equation
.
2x + y - z = 8 ------(1)

x² - y² + 2z² = 14 ------(2)

3x³ + 4y³ + z³ = 195 -------(3)



greetings all great men here.. it's been a while.

Joe82834:


Good one bro.seen my mistake.calculated 4 no of persons in class after d extra 2 boys joined

I think ur own z more correct.....bcus d question said total numba of all d pupil(I.e including d additional 2 boy)......just my view...stand to be corrected

Richiez:
We diagnose and solve math problems here
This thread is the meeting point for nairaland math gurus...I dare anyone to ask a question in mathematics without me having an answer to them, LETS START

***

wow great one op, i really learn a lot from this


i rep funaab