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Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] - Education (3) - Nairaland

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2016 NAIRALAND JAMB TUTORIAL {the Physics Thread } / Nairaland 2016 Jamb Tutorial Classroon [use Of English Thread] / Nairaland 2016 Jamb Tutorial Classroom [chemistry Thread] (2) (3) (4)

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Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Geofavor(m): 10:25am On Sep 07, 2015
Ride on cool
Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by HomoSapiien: 10:28am On Sep 07, 2015
Sure bro, go on
Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 10:34am On Sep 07, 2015
Example one: determining the gradient of a straight line.

using the general equation form given earlier, the slope of a straight line can easily be determined.

1. find the gradient and the y-intercept of the equation: y=3x + 1

solution:

using the general form; y=mx + c,
compare the two equations:
y=3x + 1.........(1)
y=mx + c...............(2)

from the above comparism, it's clear that the gradient of the line is 3 and its y-intercept is 1

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Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 10:47am On Sep 07, 2015
Sometimes, the question may want us to also determine the x-intercept of some equations.

lets take another example:

1. find the x-intercept of y=3x + 1

to find the x-intercept, there's no need using the general form, this is what we're going to do:

soln:
since we've gotten the slope of the equation in the first example,
y=3x + 1

to find the x-intercept,
we equate y=0, i.e
when y=0,
0=3x + 1

collect like terms
-3x=1

divide both sides by -3

-3x/-3=1/-3

x= -1/3

therefore, the x-intercept is -1/3, i.e the point on the graph where the line crosses the x-axis.

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Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by angiography(m): 11:04am On Sep 07, 2015
Good day fellow student,
I am an undergraduate student of OAU who has now being offered medicine...hence leaving school for the meantime. I offer services in intensive tutoring in chemistry, biology, physics and English using JAMB syllabus with a guarantee of 280 and above (ceteris paribus). I stay in Abuja. Home tutoring service is acceptable too provided the cost of transport wld be added. You can reach me via akisamuela@gmail.com. I'm interested in seeing we young people achieving our goals.
Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 11:11am On Sep 07, 2015
Gradient of a straight line

The gradient of a straight line is the rate of change of y compared to x. For example, if the gradient is 3, then for any increase in x, y increases three times as much.

Example one:

1. find the gradients of the lines joining:
(a) A(-1, 2) and B(3, -2)
(b) C(0, -1) and D(4, 1)

The formula for calculating gradient/slope is: m=y2-y1/x2-x1

1 Like 1 Share

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Nobody: 11:12am On Sep 07, 2015
angiography:
Good day fellow student,
I am an undergraduate student of OAU who has now being offered medicine...hence leaving school for the meantime. I offer services in intensive tutoring in chemistry, biology, physics and English using JAMB syllabus with a guarantee of 280 and above (ceteris paribus). I stay in Abuja. Home tutoring service is acceptable too provided the cost of transport wld be added. You can reach me via akisamuela@gmail.com. I'm interested in seeing we young people achieving our goals.
Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Nobody: 11:14am On Sep 07, 2015
angiography:
Good day fellow student,
I am an undergraduate student of OAU who has now being offered medicine...hence leaving school for the meantime. I offer services in intensive tutoring in chemistry, biology, physics and English using JAMB syllabus with a guarantee of 280 and above (ceteris paribus). I stay in Abuja. Home tutoring service is acceptable too provided the cost of transport wld be added. You can reach me via akisamuela@gmail.com. I'm interested in seeing we young people achieving our goals.


If u want to help y not start from nairaland here, start assisting online, tutorial online wen pple see u work dey ll invite u for home tutoring or so

3 Likes

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 11:25am On Sep 07, 2015
Solution:
(a) using the form: (x1, y1) and (x2, y2)

we can point out from the first example, using the illustration above that:

x1= -1

y1= 2

x2= 3

y2= -2

now that we've written out our parameters, we start solving:
m=y2 -y1/x2- x1
m=(-2) - 2/3 - (-1)
m=-4/3 + 1
m=-4/4
m=-1

therefore, the gradient/slope is -1.

1 Like 1 Share

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 11:38am On Sep 07, 2015
solution to question 2

we're also going to follow the method used in solving the 1st question, i.e
x1= 0
y1= -1
x2= 4
y2=1
now, m=y2 - y1/x2 - x1
m= 1 -(-1)/4 - 0
m=1 + 1/4
m=2/4
m=1/2

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Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 11:40am On Sep 07, 2015
hope u guys r comprehending

1 Like

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by HomoSapiien: 11:42am On Sep 07, 2015
Orezy5:
hope u guys r comprehending

Yes sir, I've been noting everything down in my book.
Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 11:43am On Sep 07, 2015
HomoSapiien:



Yes sir, I've been noting everything down in my book.
lool. i'm nt sir o, i'm just a small boy

1 Like 1 Share

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by HomoSapiien: 11:50am On Sep 07, 2015
Orezy5:
lool. i'm nt sir o, i'm just a small boy

Lol.. It doesn't matter whether you are a boy - to me you are SIR.
Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 11:51am On Sep 07, 2015
HomoSapiien:


[i] Lol.. It doesn't matter whether you are a boy - to me you are SIR. [/b]
OWK SIR

1 Like

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by HomoSapiien: 11:54am On Sep 07, 2015
Orezy5:
OWK SIR
Lol. grin

1 Like 1 Share

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 12:06pm On Sep 07, 2015
DETERMINING THE EQUATION OF A LINE WITH A GIVEN GRADIENT(i.e the point-slope form)
some questions may come in this form. we may be asked to determine the equation of a line with a given gradient.

the formula for calculating this is:
y -y1 =m(x - x1)

example one: determine the equation of a straight line whose slope/gradient is -1/3 and that passes through the point (-3, 2).

solution
using the formula where (x1, y1)= (-3, 2) and m(the gradient)=-1/3,
y - y1=m(x - x1)
then y - 2=-1/3(x + 3)
cross multiply:
3(y - 2)= -1(x + 3)
open the brackets:
3y - 6= -x -3
bring all the similar variables to one side:
x + 3y= -3 + 6
x + 3y= 3
therefore the equation is x + 3y=3

SIMPLE. ISN'T IT?

2 Likes 1 Share

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by HomoSapiien: 12:15pm On Sep 07, 2015
^ No embarassed
But with time, i believe its going to be YES.

1 Like 1 Share

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 12:46pm On Sep 07, 2015
DETERMINING THE EQUATION OF A LINE WITH NO GIVEN GRADIENT(i.e the two-point form)

sometimes, jamb and other examination bodies can set funny questions. imagine them asking us to determine the equation of a line without giving us the gradient, it sounds ridiculous ryt?

don't worry, we've to show them that we are smarter. lol

the formula for calculating equations with no gradient given is: y -y1/y2 -y1=x- x1/x2 -x1

example:
find the equation of the straight line passing through the points (1, 4) and (-2, 6).
solution:
using the formula above,
where(x1, y1)= (1, 4) and (x2, y2)= (-2, 6),

y - 4/6 - 4= x - 1/-2 - 1

y - 4/2= x - 1/-3

cross multiply:
-3(y - 4)= 2(x - 1)
open the brackets:

-3y + 12= 2x - 2
collect all the variables to one side:
-3y - 2x= -2 - 12
-3y - 2x= -14

since minus(-) is common to both sides, multiply both sides by (-)

-(-3y - 2x)=-(-14)
3y + 2x=14
don't mind jamb, they're naughty, very naughty set of people.
NB: if the 1st answer we got(i.e -3y - 2x= -14) is the option u saw, go for it, bt if it's the other one, it's still the same thing oo.

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Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 12:57pm On Sep 07, 2015
pls oo, i wan go eat.
we're going to continue by 5pm.

1 Like 1 Share

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by thankyouJesus(m): 1:01pm On Sep 07, 2015
Watching from sideline
Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 1:52pm On Sep 07, 2015
thankyouJesus:
Watching from sideline
Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by markpeakson(m): 2:06pm On Sep 07, 2015
Orezy5:
DETERMINING THE EQUATION OF A LINE WITH NO GIVEN GRADIENT(i.e the two-point form)

sometimes, jamb and other examination bodies can set funny questions. imagine them asking us to determine the equation of a line without giving us the gradient, it sounds ridiculous ryt?

don't worry, we've to show them that we are smarter. lol

the formula for calculating equations with no gradient given is: y -y1/y2 -y1=x- x1/x2 -x1

example:
find the equation of the straight line passing through the points (1, 4) and (-2, 6).
solution:
using the formula above,
where(x1, y1)= (1, 4) and (x2, y2)= (-2, 6),

y - 4/6 - 4= x - 1/-2 - 1

y - 4/2= x - 1/-3

cross multiply:
-3(y - 4)= 2(x - 1)
open the brackets:

-3y + 12= 2x - 2
collect all the variables to one side:
-3y - 2x= -2 - 12
-3y - 2x= -14

since minus(-) is common to both sides, multiply both sides by (-)

-(-3y - 2x)=-(-14)
3y + 2x=14
don't mind jamb, they're naughty, very naughty set of people.
NB: if the 1st answer we got(i.e -3y - 2x= -14) is the option u saw, go for it, bt if it's the other one, it's still the same thing oo.
U ARE DOING A GREAT WORK HERE BRO.

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Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by markpeakson(m): 2:07pm On Sep 07, 2015
following

1 Like

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by HomoSapiien: 2:39pm On Sep 07, 2015
Orezy5:
pls oo, i wan go eat.
we're going to continue by 5pm.
Okay sir.

1 Like 1 Share

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 2:58pm On Sep 07, 2015
markpeakson:
U ARE DOING A GREAT WORK HERE BRO.
thanks for d encouragement boss!

1 Like

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Affable0709(m): 3:03pm On Sep 07, 2015
I'm totally interested. I'll be sitting for it 2016.

1 Like

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 4:15pm On Sep 07, 2015
Affable0709:
I'm totally interested. I'll be sitting for it 2016.
U ARE WELCOME BRO...

2 Likes

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 4:18pm On Sep 07, 2015
hmm.....bro umartins. i salute u o

1 Like

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 4:18pm On Sep 07, 2015
hmm.....bro umartins. i salute u o.
Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Umartins1(m): 4:34pm On Sep 07, 2015
Orezy5:
hmm.....bro umartins. i salute u o.

I'm Umartins1...... I no dey get una mentions.

I salute you too.

1 Like

Re: Nairaland 2016 Jamb Tutorial Classroon [mathematics Thread] by Orezy5(m): 5:15pm On Sep 07, 2015
Umartins1:

I'm Umartins1...... I no dey get una mentions.
I salute you too.
owk boss

1 Like

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