Geofavor:
I've been under the weather - that's why i haven't been contributing much here. To read and study my textbooks sef na wahala.

Gear up bro..

Gifted18:
Bring on any maths problem and the Solution will be provided

Geofavor:
lol. Didn't you see the photo?

Gifted18:
I was unable to view it clearly. Write it out

Geofavor:
Orezy5 welldone.


I want to suggest that;

the syllabus be followed accordingly from the start.

One or two topics per week.(depending on how large a topic is)

pls do not hesitate to volunteer to teach a topic you are a genius at when it's time for that topic.


Jamb questions on the topic which anybody couldn't solve should be posted.
Then the topic's tutor or other students who have understood the topic properly can now do justice to the questions.


Cc
homosapiiens

affable0709

francis95

cee001

markpeakson

kunlexic

thankyoujesus

others.

Tobiee:
ProphetUdeme do u know any jamb lesson academy u can recommend?

Geofavor:

Pls solve this asap

ayokunlei:
prof -Umartins1 ,welldone for all the good work. Are you an undergraduate / or a graduate Sir?
Just interested in knowing you

Geofavor:
Orezy5 welldone.


I want to suggest that;

the syllabus be followed accordingly from the start.

One or two topics per week.(depending on how large a topic is)

pls do not hesitate to volunteer to teach a topic you are a genius at when it's time for that topic.


Jamb questions on the topic which anybody couldn't solve should be posted.
Then the topic's tutor or other students who have understood the topic properly can now do justice to the questions.


Cc
homosapiiens

affable0709

francis95

cee001

markpeakson

kunlexic

thankyoujesus

others.

Umartins1:


Hmmm... Let me say an undergraduate. But....


It's a long story.

Orezy5:
example three:

find the value of p, if the line which passes through (-1, -p) and (-2p, 2) is parallel to the line 2y + 8x - 17=0

solution
since both lines are parallel, they have the same slope/gradient.
let's find the gradient of 2y + 8x -17=0
as you all know that the general form of straight line equations is y=mx + c, let's re-arrange the equation to suit this arrangement:

2y + 8x - 17=0
2y= -8x + 17
make y the subject of the formula by dividing both sides by 2:

2y/2= -8x/2 + 17/2
y= -4x + 17/2
therefore, the gradient is -4.

now that we've gotten the value of the slope, let's find the value of p by using the gradient formula:
m=y2 - y1/x2 - x1

m=-4,
x1= -1,
x2= -2p,
y1= -p,
y2= 2.

-4= 2-(-p)/-2p -(-1)

-4= 2 + p/-2p + 1
cross multiply:
-4(-2p + 1)= 2 + p
open the bracket:

8p - 4= 2 + p
collect like terms:

8p - p= 2 + 4
7p= 6

divide both sides by 7:
7p/7 = 6/7
therefore, the value of p is= 6/7.

Orezy5:
example three:

find the value of p, if the line which passes through (-1, -p) and (-2p, 2) is parallel to the line 2y + 8x - 17=0

solution
since both lines are parallel, they have the same slope/gradient.
let's find the gradient of 2y + 8x -17=0
as you all know that the general form of straight line equations is y=mx + c, let's re-arrange the equation to suit this arrangement:

2y + 8x - 17=0
2y= -8x + 17
make y the subject of the formula by dividing both sides by 2:

2y/2= -8x/2 + 17/2
y= -4x + 17/2
therefore, the gradient is -4.

now that we've gotten the value of the slope, let's find the value of p by using the gradient formula:
m=y2 - y1/x2 - x1

m=-4,
x1= -1,
x2= -2p,
y1= -p,
y2= 2.

-4= 2-(-p)/-2p -(-1)

-4= 2 + p/-2p + 1
cross multiply:
-4(-2p + 1)= 2 + p
open the bracket:

8p - 4= 2 + p
collect like terms:

8p - p= 2 + 4
7p= 6

divide both sides by 7:
7p/7 = 6/7
therefore, the value of p is= 6/7.

Orestino:
nice one orezy... I would love to teach statistics though nice one orezy... I would love to teach statistics though nice one orezy... I would love to teach statistics though

Geofavor:
Orezy5 welldone.


I want to suggest that;

the syllabus be followed accordingly from the start.

One or two topics per week.(depending on how large a topic is)

pls do not hesitate to volunteer to teach a topic you are a genius at when it's time for that topic.


Jamb questions on the topic which anybody couldn't solve should be posted.
Then the topic's tutor or other students who have understood the topic properly can now do justice to the questions.


Cc
homosapiiens

affable0709

francis95

cee001

markpeakson

kunlexic

thankyoujesus

others.

Umartins1:



Consider triangles POQ and TOU,


POQ= TOU (alternate <s and PQ//TU)


QPO= TUO (alt <s and PQ//TU)


<POQ= <TOU (opposite angles)


Therefore, triangle POQ is similar to triangle TOU



Area of POQ / area of TOU = 3² / 5²


= 9/25


Area of triangle POQ = 9/25 X 50


= 18cm²

(B)

If you can sketch the diagram and apply the formular for finding the area of trapezium, that is 1/2 (a+b)h, you will arrive at 224cm²

...... But alternatively,



Let the height of triangle TOU= a


In TOU, 1/2 X 5a = 50


5a= 100


a= 20cm



Let the height of triangle POQ= b


In POQ, 1/2 X 3b = 18


3b= 36


b= 12cm




Therefore, the height of trapezium PQRS = h


h= a+b


H= 12cm+20cm


H= 32cm

Area of trapezium PQRS = area of parallelogram STQP + area of parallelogram RQPU + area of triangle POQ - area of triangle TOU



Area of parallelogram STQP= bXh


= 3X32


= 96cm²



Area of parallelogram RQPU is also 96cm²



Area of triangle POQ= 50cm²



Area of triangle TOU= 18cm²



Therefore, area of trapezium PQRS =



(96 + 96 + 50 - 18)cm²


= 224cm²

Geofavor:
Orezy5 well done bro.

Let's start using the syllabus by tomorrow. We don't have to waste too much time on a topic.

So, I'll begin with NUMBER BASE tomorrow.

Everyone should brush through the topic.(it's not hard). So whatever question you have will be attended to after/during llecture.

From the least difficult to the most, we can cover the syllabus if we start now.