This year's jamb exam kicks off from the 27th NOT 29th.

You guys should go and reprint o. I have. My exam date is sooner than expected - 27th of this month. That's 7 days from now.

I no know why jamb carry me put with the first batch.

14*xP3=X+2P4 the * mean raise to power

Geofavor:
This year's jamb exam kicks off from the 27th NOT 29th.

You guys should go and reprint o. I have. My exam date is sooner than expected - 27th of this month. That's 7 days from now.

I no know why jamb carry me put with the first batch.

Lolz... i'll reprint soon.

lammy197:
14*xP3=X+2P4 the * mean raise to power

It'll be better you snap and upload the question

guys I just create a whatsapp group for all utme aspirants , we can ask eachoda question, share ideas also.. whatsapp me on 07066155123

Thats the question

lammy197:
Thats the question

But ²P₄ does not exist

ShyNeptune:
But ²P₄ does not exist

am also confused o

lammy197:
Thats the question

That question isn't right.

Geofavor is presently reading everything readable

Examination Town: Afuze in Edo State. Examination Centre Name: Diict Nigeria , Ase Secondary School Afuze, Edo State Examination Date: Friday, March 11, 2016 By 06:30AM PLS DO YOU KNOW THE ABOVE LOCATION

ShyNeptune:
Geofavor is presently reading everything readable

as in ehn, e no easy. Anyway anyhow, we go pass am sha.

Please mathlethes quick question

If some strips of stick of length 7cm 9cm and 27cm were cut independently from a tree find d shortest possible length of d stem

A 197cm
B 54cm
C 81cm
D 189cm

I have seen d solution to d Q but still can't comprehend d method applied so I brought here wit hopes of gettn a good explanation... Thanks in adv....d tutors here a doing a great job God bless u all

lekelistik:
Please mathlethes quick question

If some strips of stick of length 7cm 9cm and 27cm were cut independently from a tree find d shortest possible length of d stem

A 197cm
B 54cm
C 81cm
D 189cm

I have seen d solution to d Q but still can't comprehend d method applied so I brought here wit hopes of gettn a good explanation... Thanks in adv....d tutors here a doing a great job God bless u all

Shortest possible length = least common multiple. LCM ...
.
.
simply find the LCM of the the numbers which is your ans
.
.
189
.
My view though ... you can as well wait for the ogas them.

lekelistik:
Please mathlethes quick question

If some strips of stick of length 7cm 9cm and 27cm were cut independently from a tree find d shortest possible length of d stem

A 197cm
B 54cm
C 81cm
D 189cm

I have seen d solution to d Q but still can't comprehend d method applied so I brought here wit hopes of gettn a good explanation... Thanks in adv....d tutors here a doing a great job God bless u all

Just like stoicguy said..


Find the LCM of the numbers

3. 7 |9 |27
3. 7 |3 |9
3. 7 | 1| 3
7. 7 | 1| 1
. . .1|1 | 1



So it will be.. 3× 3× 3×7=189

If the sum of the first two terms of a G.P is 3, and the sum of the second and the third term is -6, find the sum of the first and the common ratio.

Thankyoujesus, mathefaro, shyneptune, oxytocin, orezy5 etc.

Geofavor:
If the sum of the first two terms of a G.P is 3, and the sum of the second and the third term is -6, find the sum of the first and the common ratio.

Thankyoujesus, mathefaro, shyneptune, oxytocin, orezy5 etc.

From Tn=ar^n-1
1st term=a
2nd term=ar
3rd term =ar²

a+ ar=3
ar+ar²=-6


a(1+r)=3..............(i)
ar(1+r)=-6............(ii)

Divide (ii) by (i)

r=-2

Substitute - 2 for r in...... (i)

a(1+(-2) =3
a(-1)=3
a =-3

Sum of first(a) and common ratio(r) =a +r=-3 + -2

=-5

Oxytocin:
From Tn=ar^n-1 1st term=a 2nd term=ar 3rd term =ar²
a+ ar=3 ar+ar²=-6

a(1+r)=3..............(i) ar(1+r)=-6............(ii)
Divide (ii) by (i)
r=-2
Substitute - 2 for r in...... (i)
a(1+(-2) =3 a(-1)=3 a =-3
Sum of first(a) and common ratio(r) =a +r=-3 + -2
=-5

thanks.

Factorize x3+2x2-5x-6=0

parkparky:
Factorize x3+2x2-5x-6=0

Try x= -1



(-1)³ + 2(-1)² - 5(-1) -6 = 0



-1+2+5-6=0



0=0



Therefore x=-1



x+1=0




Divide the polynomial by x+1



You will get x²+x-6



Solve x²+x-6=0



x²+3x-2x-6= 0



x(x+3)-2(x+3)= 0



(x-2)(x+3) = 0



x=2 or x= -3



Therefore, x=-1, 2 or -3

Umartins1:



Try x= -1



(-1)³ + 2(-1)² - 5(-1) -6 = 0



-1+2+5-6=0



0=0



Therefore x=-1



x+1=0




Divide the polynomial by x+1



You will get x²+x-6



Solve x²+x-6=0



x²+3x-2x-6= 0



x(x+3)-2(x+3)= 0



(x-2)(x+3) = 0



x=2 or x= -3



Therefore, x=-1, 2 or -3

can I ask, what was the basis of your guessing that x = -1? Or was it just a trial and error with random numbers?

Umartins1:



Try x= -1



(-1)³ + 2(-1)² - 5(-1) -6 = 0



-1+2+5-6=0



0=0



Therefore x=-1



x+1=0




Divide the polynomial by x+1



You will get x²+x-6



Solve x²+x-6=0



x²+3x-2x-6= 0



x(x+3)-2(x+3)= 0



(x-2)(x+3) = 0



x=2 or x= -3



Therefore, x=-1, 2 or -3

must we use x=-1 in all cases

I guess I need to chip in my quota here,

parkparky:
must we use x=-1 in all cases

Geofavor:

can I ask, what was the basis of your guessing that x = -1? Or was it just a trial and error with random numbers?

The "-1" is not just a random number, we are expected to try out all the factors of the constant polynomial which is "-6" in this case. Apparently, the numbers we can use here (factors of -6) include ±1, ±2, ±3 and ±6. And according to factors theorem, if after trying x = -2 and it gives you zero i.e. f(-2) =0, it means x +2 is a factor of the polynomial. Then we can get the other factors by using the long division and subsequently factoring whatever quotient we get.

Mind you, this is the traditional (omoluabi) method. Jambites ain't advised to try this in the hall. Just cross check your options(expand the brackets) one after the other to see the one that works. It saves you done time and stress.
**winks**

mathefaro:
I guess I need to chip in my quota here, The "-1" is not just a random number, we are expected to try out all the factors of the constant polynomial which is "-6" in this case. Apparently, the numbers we can use here (factors of -6) include ±1, ±2, ±3 and ±6. And according to factors theorem, if after trying x = -2 and it gives you zero i.e. f(-2) =0, it means x +2 is a factor of the polynomial. Then we can get the other factors by using the long division and subsequently factoring whatever quotient we get.

Mind you, this is the traditional (omoluabi) method. Jambites ain't advised to try this in the hall. Just cross check your options(expand the brackets) one after the other to see the one that works. It saves you done time and stress.
**winks**

you're right. I saw a question like this one during the exam. Me wey no get time for nonsense, . . . I skip am sharpally.

Geofavor:

you're right. I saw a question like this one during the exam. Me wey no get time for nonsense, . . I skip am sharpally.

Hahahaha

Simplify 3^5n/9^1-n X 27^n+1

parkparky:
Simplify 3^5n/9^1-n X 27^n+1

3^{10n + 1}

Geofavor:

3^{10n + 1}

show working biko

parkparky:
show working biko

3⁵ⁿ/(9^{1 - n}) x 27^{n + 1}

3⁵ⁿ/(3^{2(1 - n)}) x 3^{3(n + 1)}

3⁵ⁿ/(3^{2 - 2n}) x 3^{3n + 3}

3^{5n - (2 - 2n)} x 3^{3n + 3}

3^{5n - (2 - 2n) + (3n + 3)}

3^{5n - 2 + 2n + 3n + 3}

3^{10n + 1}

...

293!!! maths 98... thank God

Orestino:
293!!! maths 98... thank God

which kind question came out abegooo

Orestino:
293!!! maths 98... thank God

98 in mathematics?
Not even my mathematics teacher can score that.
Geofavor come and see something oh