Word problem

A box contains 13 currency notes, which are either ₦50 or ₦20 notes.
The total value of the currency notes is ₦530. How many ₦50 notes are
in the box?

SOLUTION

Let the number of ₦50 notes be x and the number of ₦20 notes be y.

Number of notes = x + y = 13................(1).
Total value of the notes = 50x + 20y = 530................(2).
Multiply (1) by 20.
20x + 20y = 260................(3).
Subtract equation (3) from (2).
30x = 270.
X = 9.
Therefore,
9 + y = 13 ................(1).
Y= 13 - 9 = 4.
Hence, there are 9 ₦50 notes (and 4 ₦20 notes) in the box.

................................................
CHECK: 9 + 4 = 13, and (50 x 9) + (20 x 4) = 450 + 80 = 530.

How to convert recurring decimals to common fractions

Converting recurring decimals to common fractions.

This requires a special process. Let the recurring decimal be equal to
a letter of the alphabet. Multiply both sides by 10 raised to power n,
where n is the number of recurring digits in the recurring decimal.
Subtract the first equation from the second and you will get the
fractional value of the recurring decimal.

EXAMPLES

(a) 0.111................

Solution.

R = 0.11... .............(1).
10R = 1.11... ................(2).
Subtract equation (1) from (2).
9R = 1.
R = 1/9.

(b) 0.1515................

Solution

X = 0.1515... .............(1).
100X = 15.1515... .............(2).
Subtract (1) from (2).
99X = 15.
X = 15/99 = 5/33.

(c)0.285714285714................

Solution

t = 0.285714... .............(1).
1 000 000 t = 285 714. 285714... .............(2).
Subtract (1) from (2).
999 999 t = 285 714.
t = 285 714 / 999 999 = 2/7.

How to draw Maths diagrams with WPS office.

Even if you don't have a computer ,you can still draw Maths diagrams
with WPS Office on your Android smartphone . Download Photo Wonder and
the latest version of WPS office (which has "Sharing by Screenshots"
on its Writer) from Google Play. Then follow the steps in the picture
below ( which was created using the same procedure I am about to teach
you).

Solution to a tricky NECO SSCE 2017 question on Variation

Suppose y varies as the sum of two quantities, one of which varies
directly as x and the other inversely with x. When x = 2, y=4 and when
x = 5 , y = 7. Find the relationship between x and y.
















Solution.

Let the two quantities be a and b.
Then,
y= k (a + b ) ...........................................(1).
a= mx .......................................................(2)
b = n/x ......................................................(3).
K, m and n are the constants of variation. Using the same letter for
the 3 constants lead to a deadlock.
Substitute Equations 2 and 3 into Equation 1.

y = kmx + (kn /x) ..................................(4).

x =2 when y= 4 turns equation 4 into:

4 = 2mk + (kn/2).
8 = 4mk + kn ...................................(5).
x=5 when y =7 turns equation 4 into :
7 = 5mk + (kn /5).
35 = 25mk + kn .............................(6).
Solve equations 5 and 6 simultaneously.

km = 9/7 and kn = 20/7.
Hence, equation 4 becomes :
y = (9/7)x + [ (20/7) ÷ x ]
= (9x/7) + (20/7x)
= (9x² + 20)/(7x).
This is the required relationship.

Confirm that y =4 when x =2 and y =7 when x =5 by substituting 2 for x
and later 5 for x in the derived relationship.

⬛

Integration of inverse trig functions

Inverse trig functions are best integrated by parts. The following
trig identities are also important:
sin² x + cos² x = 1.
sec² x = tan² x + 1.
cosec² x = cot² x + 1.

1. ∫ Arc sin x dx.

Let u= arc sin x and dv = dx.
Then du= 1/(√1-x² ) dx and v=x.
∫ arc sin x dx = ∫ u dv = uv - ∫ v du = x arc sin x - ∫ x/(√1-x² )
dx= x arc sin x + √ (1-x²) + C.
NOTE: Since u= arc sin x implies x= sin u, ∫ x/(√1-x² ) dx = ∫(sin u)
/ √ (1- sin² u) . Cos u du = ∫[(sin u)/(cos u)]. Cos u du = ∫ sin u du
= - cos u = - √ (1- sin² u) = - √ (1 - x²).

------------------------------------------

2. ∫ arc cos x dx.

Let u = arc cos x and dv = dx.
Then du = -1 /√ (1- x²) dx and v = x.
Hence, ∫ arc cos x dx = ∫ u dv = uv - ∫ v du = x arc cos x - ∫ -x/√
(1-x²) dx = x arc cos x + ∫x/√(1-x²) dx = x arc cos x - √ (1-x²)
+ C.

---------------------------------------

3. ∫ Arc tan x dx.

Let u = arc tan x and dv = dx.
Then du= 1/(x² + 1) dx and v = x.
Therefore, ∫ arc tan x dx = ∫u dv = uv -∫ v du = x arc tan x - ∫ x/(x²
+ 1) dx = x arc tan x - (1/2) In (x² + 1) + C.

NOTE: ∫ x/(x² + 1) dx = ∫[(1/2) x 2x]/ (x² + 1) dx = (1/2) ∫ (2x)/(x²
+ 1) = (1/2) In (x² + 1) , since ∫f'(x)/f(x) dx = In f(x).

-----------------------------------------

4. Arc sec x.

Let u= arc sec x and dv = dx.
Then sec u = x ; dx/du = tan u sec u = sec u X √ (sec² u - 1) = x√(x²
- 1) ; du/dx = 1 /(dx/du) = 1/[x√(x² - 1) ] ; du = 1/[x√(x² - 1) ]
dx. V=
∫ dx = x.
Therefore, ∫ arc sec x dx = x arc sec x - ∫x . 1/[x√(x² - 1) ] dx =
x arc sec x - ∫ 1/√(x² - 1) dx = x arc sec x - In [x + √(x² - 1)] +
C.

NOTE: Since x = sec u, dx/du = tan u sec u ; dx= tan u sec u du.
Substituting, we have:
∫1/√ (x² - 1) dx = ∫ sec u du = ∫ sec u X [ (tan u + sec u)/ (tan u +
sec u)] du = ∫ (sec u tan u + sec² u)/ (tan u + sec u) du = In (tan u
+ sec u) = In [(x + √ (x² - 1)].
-----------------------------------------
5. ∫ arc cosec x dx.

Let u = arc cosec x and dv =dx.
Then du= -1 /[x√(x²-1)] dx and v = x.
Therefore, ∫ arc cosec x dx = x arc cosec x - ∫ x. -1 [x√(x²-1)] dx =
x arc cosec x + ∫ 1/√ (x²-1) dx = x arc cosec x + In [x + √(x²
- 1)] + C.

------------------------------------------
6. ∫ arc cot x dx.

Let u = arc cot x and dv = dx.
Then, du = -1/ (x² + 1) dx and v= x.
Therefore, ∫ arc cot x dx = x arc cot x - ∫ x . -1/(x² + 1) dx = x
arc cot x + ∫ x/(x² + 1) dx = x arc cot x + (1/2) In (x² + 1) + C.
(See the workings for the integral @ arc tan x above).

NOTE: u = arc cot x implies x = cot u = cos u/sin u. Then, using
Quotient rule, dx/du = -1 - cot² u = - 1 - x². Hence, du/dx = 1 / (-1
-x²) = -1 / (x² + 1);...; du = -1/(x² + 1) dx.

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