johnpaul1101: i believe that i am still a learner, i still think i'm entitled to trial, so if i may contribute, i would say: since this is bearing, after interpreting the question, i arrived at 574km(using pythagoras theorem). Please gurus, am i wrong?

you're very right, but you could also add the direction (angle) which can be gotten by; arctan (430/380)

Kindly help with dis...(1) the sum of the first n terms of a G.P is 127 and the sum of their reciprocal is 127/64.the first term is 1. Find n and the common ratio respectively...(2)find 3 numbers in G.P whose sum is 13 and whose product is -64.....(3) resolve into partial fraction 7x+2/(2x-3)(x+1)^2.......tank u and God bless as u asist me....

Hello Gurus, I need a help here.

1. The shopkeeper has 6 barrel...5 - wine...1 - beer with the volume: 60 64 72 76 80 and 124 litres. The first customer buys 2 wine barrels. The second customer buys twice as much wine as the first customer. What is the volume of the beer barrel?
a. 60 litres b. 72 litres c. 80 litres d. 124 litres

Fetus: Kindly help with dis...(1) the sum of the first n terms of a G.P is 127 and the sum of their reciprocal is 127/64.the first term is 1. Find n and the common ratio respectively...(2)find 3 numbers in G.P whose sum is 13 and whose product is -64.....(3) resolve into partial fraction 7x+2/(2x-3)(x+1)^2.......tank u and God bless as u asist me....

question (3)
7x+2/(2x-3)(x+1)^2=A/(2x-3)+B/(x+1)+C/(x+2)

multiply all numerators by (2x-3)(x+1)^2 LCM style.

=[Ax^2+2Ax+A+2Bx^2-Bx-3B+2Cx-3C]/(2x-3)(x+1)^2

equating coefficients

[x^2]: A+B=0
[x]: 2A-B+2C=7
[]: A-3B-3C=2

manipulating:
A=2
B=-1
C=1

7x+2/(2x-3)(x+1)^2=2/(2x-3)-1/(x+1)+1/(x+1)^2

question (2)

3 numbers a, ar, ar^2

product: a*ar*ar^2=(a^3)(r^3)=-64
sum: a+ar+ar^2=13

so, ar=-4
making a subject: a=-4/r. manipulating the sum equation we get
4r^2+17r+4...
(4r+1)(r+4)
r=-4 or -1/4
a=1 or 16

possible answers
a=1, ar=-4, ar^2=16
a=16, ar=-4, ar^2=1

like. 1, -4, 16. or 16, -4, 1


question (1)

formula: Sn=a(1-r^n)/(1-r)
a=1

sum of nth terms (1-r^n)/(1-r)=127
sum of reciprocal of nth terms [1-(1/r)^n]/[1-(1/r)]=127/64

manipulating sum of reciprocals we get

(r^n-1)r/(r-1)r^n=127/64
but (r^n-1)/(r-1)=127
so, 127r/r^n=127/64
or, r^n=64r

fix that in the sum of terms
(1-64r)/(1-r)=127
so, 1-64r=127-127r
or, 63r=126
r=2

since, r^n=64r
then 2^n=64*2
2^n=128
2^n=2^7
n=7

n=7
common ratio = 2

I pledge 2 always respect n fear maths...

Nice work Osita, how have you been man? It's been a while!

megatechgh: Hello Gurus, I need a help here.

1. The shopkeeper has 6 barrel...5 - wine...1 - beer with the volume: 60 64 72 76 80 and 124 litres. The first customer buys 2 wine barrels. The second customer buys twice as much wine as the first customer. What is the volume of the beer barrel?
a. 60 litres b. 72 litres c. 80 litres d. 124 litres

c. 80
guy there is no shortcut for this one.

*we have five wine barrels and one beer.
*first man buys 2 wine
*second guy buys 2 times the volume of wine as the first man.
*for this to work the sum of any 3 volumes must equal to 2times any 2 other volumes.
*once you get the right combination the 6th unused volume should be beer.

doubleDx: Nice work Osita, how have you been man? It's been a while!

I am cool man, ehmmm, how madam?

Generals Osita and Doubledy/dx where av u guys been?

Davidson267: I pledge 2 always respect n fear maths...

Respect? YES but fear? HAHAHAHA

ositadima1:

c. 80
guy there is no shortcut for this one.

*we have five wine barrels and one beer.
*first man buys 2 wine
*second guy buys 2 times the volume of wine as the first man.
*for this to work the sum of any 3 volumes must equal to 2times any 2 other volumes.
*once you get the right combination the 6th unused volume should be beer.

Please check the question very well, there are answer options from a - d....i was asked to choose the correct one.

megatechgh:
Please check the question very well, there are answer options from a - d....i was asked to choose the correct one.

lol, well I pick 80, that's c

ositadima1: lol, well I pick 80, that's c

well Gen. Osita u'r right.
Reason; 60+72=132 twice of this is 132*2 =264
3 other combinations that yields same amount are; 64+76+124=264
the remaining volume(i.e vol of beer) is 80

Goodday all generals and gurus, please i need help with this circle geometry question:An equilateral triangle of side 20cm is inscribed in a circle. Calculate the distance of a side of the triangle from the centre of the circle?
After doing what i did, i got 8cm, but the answer here is 5.77cm, please how do i go about it?

Calculate the area of triangle abc. Ab is 9cm, bc is 5cm and angle acb is 50degree.


Maths guru, pls help. Thanks.

jhydebaba: Calculate the area of triangle abc. Ab is 9cm, bc is 5cm and angle acb is 50degree.


Maths guru, pls help. Thanks.

i'm not a guru but an amateur, so after solving this your question, i got 14.5cm. Please gurus, i would like to know if i'm wrong or right. Thanks

johnpaul1101: Goodday all generals and gurus, please i need help with this circle geometry question:An equilateral triangle of side 20cm is inscribed in a circle. Calculate the distance of a side of the triangle from the centre of the circle?
After doing what i did, i got 8cm, but the answer here is 5.77cm, please how do i go about it?

I don't know how visible my drawings.

[a/sin(30)]=[20/sin(120)]
a=20*sin(30)/sin(120)

sin(30)=b/a
a=b/sin(30)

b/sin(30)=20*sin(30)/sin(120)
b=20*[sin(30)]^2/sin(120)
b=5.77cm

johnpaul1101:
i'm not a guru but an amateur, so after solving this your question, i got 14.5cm. Please gurus, i would like to know if i'm wrong or right. Thanks

pls show workings

jhydebaba: pls show workings

area of triangle=1/2*side a*side b*sin(angle c)

Sumbody pls help me wit dis simple factorisation showin steps
x3 -3x2-10x+24=0

Soloblincus: Sumbody pls help me wit dis simple factorisation showin steps x3 -3x2-10x+24=0

Different methods can be used to solve this question, but let us apply factor theorem and long division method.
f(x) =x³ -3x² -10x+24
by factor theorem, if f(a) =0, then (x-a) is a factor.
Now let a=2 f(2)= (2)^3 -3(2)^2 -10(2) + 24
f(2) = 8-12-20+24=0 therefore (x-2) is a factor!
using the long division, x3-3x2-10x+24 divided by (x-2), the quotient is x2-x-12
hence, (x-2) and (x2-x-12) are factors.
x3-3x2-10x+24 = (x-2)(x2-x-12)
but x2-x-12 is a quadratic equation that can further be splitted into (x+3)(x-4)
thus; x3-3x2-10x+24=(x-2)(x+3)(x-4)

Richiez: Different methods can be used to solve this question, but let us apply factor theorem and long division method.
f(x) =x³ -3x² -10x+24
by factor theorem, if f(a) =0, then (x-a) is a factor.
Now let a=2 f(2)= (2)^3 -3(2)^2 -10(2) + 24
f(2) = 8-12-20+24=0 therefore (x-2) is a factor!
using the long division, x3-3x2-10x+24 divided by (x-2), the quotient is x2-x-12
hence, (x-2) and (x2-x-12) are factors.
x3-3x2-10x+24 = (x-2)(x2-x-12)
but x2-x-12 is a quadratic equation that can further be splitted into (x+3)(x-4)
thus; x3-3x2-10x+24=(x-2)(x+3)(x-4)

dis ans came a bit late because i jst wrote d exam i wuz preparin 4 and a quest to find engen values usin faddeev larierrier's mtd degenerated to a similar equation. Findin d roots were all i needed 2 get d values & complete mrks bt i could not. Thanks all d same... Cheers

First things first.... Una to mush!! Una bi confam Bosses!! Ah see brains dey kash faya!! Salute to u all, and specially to Man @Richlez! God go make una bigger!

I like math and am very good at it too, but sadly enough, am still at a very low level in academics par say. O'Level no bi anything man!

God will grant me my academic desires IJN




By the way can somebody explain the law of large numbers with simple comprehensible examples for me. Kudos!

Na gurus really dey 4 dis thread... Abeg make una ride on jor.. Maths happen 2 be my best subject...

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 value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3.why not solve this equation instead of asumption aproach...what of If y tends to be a decimal in anoda problem?.

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 value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3

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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 value of y that must be inserted in eqn (3) to obtain 31.

a thorough look at eqns (1) & (2) shows that the values of x & y must be positive integers within the range 0<x<5 and 0<y<5
clearly, y=3
i.e (5-3)^(5-3) + 3^3
2^2 + 27
4+27=31
now that we have established that y=3, we shall substitute the value of y in eqn(1) to get x
recall that x+y=5
therefore; x+3=5 or x=5-3=2
hence, x=2 when y=3

this is a wow!!!

Hello guru pls help me with this:
Q1 use inverse method and gaussia elimination to solve the follow.
X+2y+3z=5
3x-y+2z =8
4x-6y-4z=-2

Q2 solve the equetion:
1. 2*4x+1=8
2. 3y+1=9y-2
3.10x=0.000001

Q3. To accumulate a fund for his daughter's higher education, a man invests a sum of #1000 on the daughter's first birthday and an equal amount on each of her subsequent's birthdays if the interest is compounded semi annually at the rate of 6% per annum, find the amount accumulated jst after the investment has been made on the 18th birthday?
Q4. If #1000 000 was deposited in a bank on 1st April 2013, what would it amount on 1st April 2033 if the compound interest at 3% per annum was allowed.
Q5. Demand and supplu equation are 2P2+Q2=11 and p+2q=7. Find the equilibriu price and quantity when p stand for price and Q for quantity.

Nice work gurus. General Richiez, Osita & others, I salute ona wella! I have been a little busy lately, will find time to contribute when I'm free. Take care y'all. 1heart!

Q3. To accumulate a fund for his daughter's higher education, a man invests a sum of #1000 on the daughter's first birthday and an equal amount on each of her subsequent's birthdays if the interest is compounded semi annually at the rate of 6% per annum, find the amount accumulated jst after the investment has been made on the 18th birthday?

Sorry I have checked in in a while, you guys were going too quantum for me


Semiannual compounding means 3% every 6 months
At the 18 birthday means there are 17 years since the investment started on the 1st year hence we will be looking at 33 periods (2n -2) Where n = 17 because this looks at the end of the 17th year

However because the payments are on an annual nature, the interest rate will be annualized to get an effective annual rate

Eff rate r _eff = 1.03 ^ 2 - 1 = 6.09%

This will be applied for 18 - 1 = 17 years
Formulas

Future Value factor = ((1 + r_eff )^N - 1)/r_eff

Formula Future Value = Equal payment * Future Value factor

FV factor = 1.0609^17-1/0/0609 = 28.4385106

FV = 1000 * 28.4385106 = 28,438.51

Q4. If #1000 000 was deposited in a bank on 1st April 2013, what would it amount on 1st April 2033 if the compound interest at 3% per annum was allowed.

4. Similar principle
Assuming a depoSit is made on 1st April 2033 and the investment is liquidated
n = 20 YEARS
FV factor = (1.03 ^ 20 - 1)/0/.03 = 26.87037449

if you assume you are collecting what you have as at that date without that day investment, you subtract one

FV factor will then be 25.87037449

So multipliy by 1,000,000 to get the FV

Q5. Demand and supplu equation are 2P2+Q2=11 and p+2q=7. Find the equilibriu price and quantity when p stand for price and Q for quantity.

At Equilibrum Supply = Demand
Solve using simultaneous equation
=
p = 7 - 2q

Substitute into equation one to yield 9q2 - 56q + 87 = 0

p = 1 q = 3

Alternate soln p = 0.56 q = 3.22