mathefaro:
1. A man and wife went to buy an article costing N400. the woman had 10% of the cost and the man 40% of the remainder. How much did they have altogether?

2. A bag contains 5 black, 4 white and x red marbles.
If the probability of picking a red marble is 2/5 find the value of x

3. A sector of a circle is bounded by two radii 7cm long and arc length 6cm. find the area of the sector.

4. Find the slope of the curve y = 2x^2 + 5x - 3 at (1,4)

1. #184


2. X=6

alakara:


1. #184


2. X=6

Yeah, you're correct. But it's better you show your full workings to the house. Moreover, where's the remaining 2 answers

mathefaro:
1. A man and wife went to buy an article costing N400. the woman had 10% of the cost and the man 40% of the remainder. How much did they have altogether?

2. A bag contains 5 black, 4 white and x red marbles.
If the probability of picking a red marble is 2/5 find the value of x

3. A sector of a circle is bounded by two radii 7cm long and arc length 6cm. find the area of the sector.

4. Find the slope of the curve of y = 2x^2 + 5x - 3 at (1,4)

---

(1) the woman had 10% of the cost.
10% of the cost is 10/100 × N400 = N40

the man had 40% of the remainder.
the remainder is N400- N40= N360
40% of the remainder is 40/100 × N360= N 144

the both had N184 altogether

---

(2) Total marble in the bag= 5+4+x= 9+x
probability of picking a red marble;

x/(9+x) =2/5
cross multiply to get;
5x= 18+2x
solve the linear equation to get;

x= 6

---

(3) radius of the circle: 7cm
length of the arc: 6cm
first obtain the angel of the arc in degree, then proceed to use;
area of a sector= pir²¢/360

2pir¢/360 =6

pi= 22/7

(2×22/7 ×7¢)/360=6

44¢/360=6

A¢=6 ( A=44/360)

¢= 6/A

¢= 49°

area of the sector is;

49×22/7×49/360= 20.96cm²

---

(4) Gradient of the slope of the curve of
y= 2x² + 5x -3 at (1,4)

NB: the (1,4) is the coordinate (x,y)

first differentiate the equation of the curve to obtain;

y' = 4x + 5 ______(1)

obtain the gradient of the slope of the curve by substituting for x in equation (1)

y' = 4(1) + 5 = 9

NB: y' is not y, it is the first derivative of y.

Equation of the slope;

(x-1)/(y-4) = 9

(this is synonymous to slope=(x₂-x₁)/(y₂-y₁) )


cross multiply to get;

x-1 = 9y - 36

9y - x = 35

---

you didn't show up in chemistry thread, please we need more tutors in the thread.

---

(1) Find the normal to the slope of the curve y= x² - 9 at x= 3.

---

(2) Find the geometric mean of the series 2, x, 25/2, hence find x.

---

(3) 6cosx + 9sinx = 10, find x

---

(4) find the integral of tanx

icecalm:

---

(3) radius of the circle: 7cm
length of the arc: 6cm
first obtain the angel of the arc in degree, then proceed to use;
area of a sector= 1/2 r²sin¢

2pir¢/360 =6

pi= 22/7

(2×22/7 ×7¢)/360=6

44¢/360=6

A¢=6 ( A=44/360)

¢= 6/A

¢= 49°

area of the sector is;

1/2×7²×sin49 = 49/2×sin49

= 18.49cm²

---

I'm sorry, this is not correct. Actually this formula you used is for getting area of the triangle inscribed in the sector. The area of a sector has a formula different from this... Check it out please. Put a mention on my handle in the chemistry thread so that I can lend a helping hand when it's needed.
Good morning to y'all

mathefaro:
I'm sorry, this is not correct. Actually this formula you used is for getting area of the triangle inscribed in the sector. The area of a sector has a formula different from this... Check it out please. Put a mention on my handle in the chemistry thread so that I can lend a helping hand when it's needed.
Good morning to y'all

the problem here is that, I couldn't remember what a sector really is,

oh, is the sector that bounded portion by the radii of the circle?

icecalm:
the problem here is that, I couldn't remember what a sector really is,

oh, is the sector that bounded portion by the radii of the circle?

exactly

mathefaro:
exactly

okay i have corrected it

Followin

https://www.nairaland.com/2717258/nairaland-2016-jamb-tutorial-classroon

Economic thread

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

bro..can we be studyin at my place in nwaniba? I need a readin partner

Please you guys should create a thread for physics...and also, is there any maths freak that lives in uyo here? Please holla...i really need you, thanks

Godzlove8:
bro..can we be studyin at my place in nwaniba? I need a readin partner

gimme ur number. Its me prophetudeme, did u apply in UI?

SaintHephzibah:
gimme ur number. Its me prophetudeme, did u apply in UI?

nah..i later change institution..the distance is too far from home

Godzlove8:
nah..i later change institution..the distance is too far from home

but u go dey board for hostel nah, so which school did u apply into?

SaintHephzibah:
but u go dey board for hostel nah, so which school did u apply into?

unical

Godzlove8:
unical

if so then u must be bigneorapz

SaintHephzibah:
if so then u must be bigneorapz

no

Godzlove8:
bro..can we be studyin at my place in nwaniba? I need a readin partner

[email] sorry, too far. [/email]

What happened to this thread? Let's help out while we can shall we. I volunteer to take section IV - Calculus tutorial.

SECTION IV: CALCULUS
I. Differentiation:
(a) limit of a function;
(b) differentiation of explicit algebraic and simple trigonometric functions - sine, cosine and tangent.

2. Application of differentiation:
(a) rate of change
(b) maxima and minima

3. Integration:
(a) integration of explicit algebraic and simple trigonometric functions.
(a) area under the curve.

I call on our able moderators and all maths gurus on this forum to lend a helping hand in demystifying mathematics for the younger generation.

Geofavor
ProphetUdeme
Orezy5
agentofchange1
stlaibrowne
MathsChic
thankyouJesus
Madmathecian
ladokuntlad
masperano
olawalebabs
Richiez
Fynestboi
Lalasticlala
Seun

Mathemagician1:
What happened to this thread? Let's help out while we can shall we. I volunteer to take section IV - Calculus tutorial.

SECTION IV: CALCULUS
I. Differentiation:
(a) limit of a function;
(b) differentiation of explicit algebraic and simple trigonometric functions - sine, cosine and tangent.

2. Application of differentiation:
(a) rate of change
(b) maxima and minima

3. Integration:
(a) integration of explicit algebraic and simple trigonometric functions.
(a) area under the curve.

I call on our able moderators and all maths gurus on this forum to lend a helping hand in demystifying mathematics for the younger generation.

pls start, i'm following.

Mathemagician1:
What happened to this thread? Let's help out while we can shall we. I volunteer to take section IV - Calculus tutorial.

SECTION IV: CALCULUS
I. Differentiation:
(a) limit of a function;
(b) differentiation of explicit algebraic and simple trigonometric functions - sine, cosine and tangent.

2. Application of differentiation:
(a) rate of change
(b) maxima and minima

3. Integration:
(a) integration of explicit algebraic and simple trigonometric functions.
(a) area under the curve.

I call on our able moderators and all maths gurus on this forum to lend a helping hand in demystifying mathematics for the younger generation.

Geofavor
ProphetUdeme
Orezy5
agentofchange1
stlaibrowne
MathsChic
thankyouJesus
Madmathecian
ladokuntlad
masperano
olawalebabs
Richiez
Fynestboi
Lalasticlala
Seun

Present sir, I will try my best to contribute. I tutored some University of Ibadan aspirants on calculus, I will advice you skip it, all I have to do is copy and paste

thankyouJesus:
Present sir, I will try my best to contribute. I tutored some University of Ibadan aspirants on calculus, I will advice you skip it, all I have to do is copy and paste

Very well then, I'll take SECTION II - ALGEBRA. I'm going to start preparing my material right away. Let's keep this thread alive.

Cheers.

SECTION II: ALGEBRA
1. Polynomials:
(a) change of subject of formula
(b) factor and remainder theorems
(c) factorization of polynomials of degree not exceeding 3.
(d) multiplication and division of polynomials
(e) roots of polynomials not exceeding degree 3
(f) simultaneous equations including one linear, one quadratic
(g) graphs of polynomials of degree not greater than 3


2. Variation:
(a) direct
(b) inverse
(c) joint
(d) partial
(e) percentage increase and decrease.


3. Inequalities:
(a) analytical and graphical solutions of linear inequalities.
(b) quadratic inequalities with integral roots only.

4. Progression:
(a) nth term of a progression
(b) sum of A. P. and G. P.

5. Binary Operations:
(a) properties of closure, commutativity, associativity and distributivity.
(b) identity and inverse elements.

6. Matrices and Determinants:
(a) algebra of matrices not exceeding 3 x 3.
(b) determinants of matrices not exceeding 3 x 3.
(c) inverses of 2 x 2 matrices [excluding quadratic and higher degree equations].

Let me get my writing materials to learn from the bosses....

Go down

thankyouJesus:
[b]Good evening all, as per my obligation, today will be "our" first class. "We" will be considering differentiation.
Note 1:
1. dy/dx can also be represented as y'.
2. d²y/dx² can also be represented as y", till infinity.
3. d/dx means "we" are differentiating with respect to x, d/du means we are differentiating with respect to . . . . . . . . . . . .very good, u.
4. wrt means with respect to.
DIFFERENTIATION.
Differentiation means rate of change.
General formula:
if y = axⁿ,
a is the co-efficient of x.
n is the power of x.
Therefore, dy/dx = power "times" co-effecient of x "times" the power of x "minus" 1.
Mathematically,
y = axⁿ,
dy/dx = nax^{n - 1}.
-----------------------------
Example 1. Find y' [it also means dy/dx] if y = 2x.
Solution:
the co-efficient of x is 2.
The power of x is 1.
Therefore,
y = 2x
dy/dx = (1)(2)x^1-1
dy/dx = 2x⁰
dy/dx = 2(1)
because x⁰ is 1.
dy/dx = 2.
-----------------------------.
Example 2: find y' if y = 3x⁴.
Solution:
y = 3x⁴
y = 12x³
-----------------------------.
Example 3: find y' if y = x
.
.
.
.
.
.
.
y' = 1.
Very good, because,
y = x
the co-efficient of x is 1.
The power of x is 1.
Therefore,
y = x
y' = (1)(1)x^1-1
y' = 1x⁰
y' = 1(1)
y' = 1.
-----------------------------.
Example 4: find y' if y = 2.
Solution:
y = 2,
the question can also be changed into
y = 2(1)
it can be further changed into
y = 2(x⁰)
since,
y = 2x⁰
y' = (0)(2)x^0-1
y' = 0x^-1
y' = 0.
Note 2: The derivative of a constant is 0.
i.e
d/dx (7) = 0
d/dx (10) = 0, but,
d/dx (7x) = 7
d/dx (10x) = 10.
-----------------------------.
Summary:
1. Differentiation means rate of change.
2. If y = axⁿ, then dy/dx = nax^n-1.
3. The differentiation of constant values is 0. E.g,
y = 6,
dy/dx = 0,
y = 1000000000
dy/dx = 0.
-----------------------------.
Assignment: find dy/dx if y =
1. 20x
2. -50x³
3. -25x^-2
4. 30/x², [Hint: 1/x² is x^-2.

Any question?

Good night class.
23rd March, 2015.
21:35:45[/b]

This is a direct copy, I need to be sure people still visit this nairaland for knowledge to avoid the math gurus wasting their time. Just 2 or more comments will do.
www.nairaland.com/modifypost?redirect=%2Fthankyoujesus%2Fposts%2F16%2331931741&post=31931741

SECTION II - POLYNOMIALS

(A) CHANGE OF SUBJECT OF FORMULA

A formula is an equation containing two of more variables and it describes how the variables are related.

Example 1: The formula I = PTR/100 is used for finding simple interest where P = Principal, R = Rate and T = Time. Observe that the same letter can stand for different quantities in different formula.
To find the Principal (P) from I = PTR/100

I = PTR/100
Multiply both sides by 100
100I = PTR
Divide both sides by TR
P = 100I/TR

We can also find Time (T) by applying the same principles.

I = PTR/100
Multiply both sides by 100
100I = PTR
Divide both sides by PR
T = 100I/PR

CLASS WORK:
(1) Make R the subject of formula if I = PTR/100

Change of formula to be continued shortly.…

Mathemagician1:
SECTION II - POLYNOMIALS

(A) CHANGE OF SUBJECT OF FORMULA

A formula is an equation containing two of more variables and it describes how the variables are related.

Example 1: The formula I = PTR/100 is used for finding simple interest where P = Principal, R = Rate and T = Time. Observe that the same letter can stand for different quantities in different formula.
To find the Principal (P) from I = PTR/100

I = PTR/100
Multiply both sides by 100
100I = PTR
Divide both sides by TR
P = 100I/TR

We can also find Time (T) by applying the same principles.

I = PTR/100
Multiply both sides by 100
100I = PTR
Divide both sides by PR
T = 100I/PR

CLASS WORK:
(1) Make R the subject of formula if I = PTR/100

Change of formula to be continued shortly.…

Am not sure if we can cope with two topics, let me wait for you to round up.

thankyouJesus:
This is a direct copy, I need to be sure people still visit this nairaland for knowledge to avoid the math gurus wasting their time. Just 2 or more comments will do.

Sir, which application do you use that makes your math look so orderly? I don't like the idea of write power as(^) and square root as sqrt. I want people to see the real thing.

Mathemagician1:


Sir, which application do you use that makes your math so orderly? I don't like the idea of write power as(^) and square root as sqrt. I want people to see the real thing.

you can also do so without an app, sir. Let's say you wanna type 2^2, Just type like this, 2[sup]2[sup]....

for 2base2, type 2[sub]2[sub] ... But remember to put a backslash in the second parenthesis in each case.

Mathemagician1:


Sir, which application do you use that makes your math look so orderly? I don't like the idea of write power as(^) and square root as sqrt. I want people to see the real thing.

Sir ke, me na small boy o. For power, use [sup]the power stuff[/su p]without the space. Eg 2x^(y-4) =2x[sup](y-4)[/su p] remember, without the space.

Mathemagician1:
SECTION II - POLYNOMIALS

(A) CHANGE OF SUBJECT OF FORMULA

A formula is an equation containing two of more variables and it describes how the variables are related.

Example 1: The formula I = PTR/100 is used for finding simple interest where P = Principal, R = Rate and T = Time. Observe that the same letter can stand for different quantities in different formula.
To find the Principal (P) from I = PTR/100

I = PTR/100
Multiply both sides by 100
100I = PTR
Divide both sides by TR
P = 100I/TR

We can also find Time (T) by applying the same principles.

I = PTR/100
Multiply both sides by 100
100I = PTR
Divide both sides by PR
T = 100I/PR

CLASS WORK:
(1) Make R the subject of formula if I = PTR/100

Change of formula to be continued shortly.…

I x 100 = pTR
R = I x 100
..............pT