## Thursday, 30 March 2017

### LOGARITHMS 7

Evaluate Log100 - log

_{ }0.001 - log_{ }0.0001+ log_{ }0.00001

__Solution__
= Log100 - log

_{ }0.001 - log_{ }0.0001+ log_{ }0.00001
= Log10

^{2}- log_{ }10^{-3}– log 10^{-4 }+ log 10^{-5}
= 2Log10 - (-3 log

_{ }10) – (-4log 10) + (-5log 10)
= (2x1) - (-3x1) - (-8x1) + (-5 x 1)

= 2 - (-3) – (-4) -5

= 2 + 3 + 4 - 5

= 9 - 5

**Hence Log100 - log**

_{ }0.001- log_{ }0.0001**+ log**

_{ }0.00001 = 4

__TRY THIS………………………__
Evaluate Log10,000 - log

_{ }0.001 - log_{ }0.00001+ log_{ }0.00001### FACTORIZING 1

Factorize
81-m

^{2}

__Solution__
We
use difference of two squares a

^{2}– b^{2}= (a - b)(a + b)
81-m

^{2}= 9^{2}-m^{2 }
= (9 - m)(9 + m)

**Hence 81 - m**

^{2}

^{ }**= (9 - m)(9 + m)**

__TRY THIS…………….__
Factorize
169 - c

^{2}### SIMPLE INTEREST 1

Asad
deposited the amount of 200,000/= in a bank for 3 years and got a profit of 72000/=.
Find the interest rate?

= R
~~6000~~ ~~6000~~

__Solution__
I
= 72000/=, P = 200,000/=, T = 3 years, R = ?

I =

__PRT__
100

72,000 =

__200,000 x R x 3__
100

72,000 =

__200,0__~~00~~ x R x 3
1~~00~~

72,000 = 2000 x R x 3

72,000 = 6000R

^{12}

__72,000__

__6000__

__Hence the interest rate was 12%__

__TRY THIS…………………..__
Sonali
deposited the amount of 300,000/= in a bank for 6 years and got a profit of 27,000/=.
Find the interest rate?

## Tuesday, 28 March 2017

### LOGARITHMS 6

If Log

_{a}x = 0.9, find Log_{a}(x^{2})

__Solution__
= Log

_{a}(x^{2})
= Log

_{a}x^{2}
= 2 x Log

_{a}x
= 2 x 0.9

= 1.8

__TRY THIS………..__
If Log

_{a}w= 3.2, find Log_{a}(w^{2})### LOGARITHMS 5

Evaluate Log100 - log

_{ }0.001 - log_{ }0.0001

__Solution__
= Log100 - log

_{ }0.001 - log_{ }0.0001
= Log10

^{2}- log_{ }10^{-3}– log 10^{-4}
= 2Log10 - (-3 log

_{ }10) – (-4log 10)
= (2x1) - (-3x1) - (-8x1)

=2 - (-3) – (-4)

= 2 + 3 + 4

=9

__Hence Log100 - log___{ }0.001- log_{ }0.0001 = 9

__TRY THIS………………………__
Evaluate Log10000 - log

_{ }0.000001 - log_{ }0.00000001### VARIATIONS 1

x
is inversely proportional to y. x=18 while y=5. Find y when x is 15.

__Solution__
x
⍺

__k__
y

x
=

__k__
y

18
=

__k__
5

k
= 18 x 5

k
= 90

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k
= 90, y= ?, x = 15.

x
=

__k__
y

15
=

__90__
y

y
x 15 = ~~y~~ (multiply by y on both
sides)

__90__x__15y__=

__90__

15
15

__Hence y = 6 .__

__TRY THIS…………….__
x
is inversely proportional to y. x=80 while y=4. Find y when x is 10.

### LOGARITHMS 4

If
log

_{9}(2x + 33)=2; find x.

__Solution__
Log

_{9}(2x + 33)=2
(2x
+ 33)=9

^{2 }^{ }

2x
+ 33=81

2x
=81 – 33

2x
= 48

__2x__=

__48__

2
2

x = 24

__Hence x = 24__

__TRY THIS………………__
If
log

_{7}(2t - 39)=2; find t### ARITHMETIC PROGRESSION 1

The first term of an AP is 41 and
the last term is 97.If the arithmetic progression consists of 20 terms,
calculate the sum of all the terms.

__Solution__
A

_{1}= 41, n=20, A_{n}= 97
S

_{n}=__n__(A_{1}+ A_{n})
2

S

_{20}=__20__(41 + 97)
2

S

_{20}= 10 x 138
S

_{20}= 1380*Hence the sum of all 20 terms is 1380.*

__TRYTHIS………………….__
The first term of an AP is 11 and
the last term is 53.If the arithmetic progression consists of 40 terms,
calculate the sum of all the terms.

### EXPONENTIALS 1

Simplify the following by writing in
power form:

__w__

^{90 }

w

^{40 }

__Solution__
= w

^{90 - 40 }
=w

^{50}

__TRY THIS……………….__
Simplify the following by writing in
power form:

__h__

^{59 }

h

^{21 }### LOGARITHMS 3

If Log

_{a}x = 0.2, find Log_{a}(^{1}/x^{2})

__Solution__
= Log

_{a}(^{1}/x^{2})
= Log

_{a}x^{-2}
= -2 x Log

_{a}x
= -2 x 0.2

= -0.4

__TRY THIS………..__
If Log

_{a}w= 1.6, find Log_{a}(^{1}/w^{2})## Saturday, 25 March 2017

### ARITHMETIC 1

Evaluate 396

^{2}– 386^{2}

__Solution__
We apply difference of two squares:
a

^{2}- b^{2}= (a+b)(a-b)
396

^{2}– 386^{2 }= (396 + 386)( 396 – 386)
=
(782)( 10)

=
7820 [only add a zero at 782]

396

^{2}– 386^{2 }= 13820

__TRY THIS…………….__
Evaluate 508

^{2}– 408^{2}### LOGARITHMS 2

If Log

_{a}x = 0.9, find Log_{a}(^{1}/x)

__Solution__
= Log

_{a}(^{1}/x)
= Log

_{a}x^{-1}
= -1 x Log

_{a}x
= -1 x 0.9

= -0.9

__TRY THIS………..__
If Log

_{a}w= 3.6, find Log_{a}(^{1}/w)### SLOPE OR GRADIENT 1

Find the slope of a line which
passes through (-1, -8) and (6,-9)

__Solution__
x

_{1 }= -1, y_{1 }=-8, x_{2 }= 6, y_{2 }= -9
m =

__y2 –y1___{ }x2 – x1

m =

__-9 –(-8)__
6 –(-1)

m =

__-9 + 8__
6
+ 1

m =

__- 1__
7

__Hence the slope is -1/7__

__TRY THIS………………………__
Find the slope of a line which
passes through (-11, -2) and (9,-12).

### BINARIES-1

If p*k = 4pk + p – 2k; find 3*2.

__Solution__
p*k = (4 x p x k) + p –
(2xk) [rewriting the given expression more clearly]

3*2 = (4 x 3 x 2) + 7 –
(2x2) [substituting

**7**for p and**2**for k]
3*2 = 24 + 7 – 4

3*2 = 31 – 4 [we add
first before subtracting]

3*2 = 27.

__Hence (3*2) = 27.__

__TRY THIS........__
If p*k = 2p - k – 5pk; find (-3*4).

### QUADRATICS-1

Find
the maximum value of the quadratic equation:5-6t-8t

^{2}.

__Solution__
a=-8,
b=-6, c=5

Maximum
=

__4ac-b__^{2}
4a

Maximum
=

__(4 x -8 x 5) - (-6)__^{2}
4(-8)

Maximum
=

__(-160)- 36__
32

Maximum=

__-196__since (-160)- 36=-196
32

Maximum
=

__-98__=__-49__
16 8

__Hence maximum value is -49/8__**TRY THIS………………………………..**

__NECTA 2003 QN. 10c__
Find
the maximum value of the quadratic equation:3+30t-5t

^{2}.### POLYGONS 2

A regular polygon has 62 sides. Find the total degrees of
that polygon.

__Solution__
n = 62

Total degrees = (n - 2)180

^{0}
=
(62 – 2)180

^{0}
=
60 x 180

^{0}
=
10800

^{0}
Total degrees = 10800

^{0}

__TRY THIS……………………….__
A regular polygon has 53 sides. Find the total angles of
that polygon.

### LOGARITHMS 1

Evaluate Log10000 + log

_{ }0.001 - log_{ }0.00000001

__Solution__
= Log10000 + log

_{ }0.001 - log_{ }0.00000001
= Log10

^{4}+ log_{ }10^{-3}– log 10^{-8}
= 4Log10 + (-3 log

_{ }10) – (-8log 10)
= (4x1) + (-3x1) - (-8x1)

=4 + (-3) – (-8)

= 4 – 3 + 8

=9

**Hence Log10000 + log**

_{ }0.001- log_{ }0.00000001 = 9

__TRY THIS………………………__
Evaluate Log100 + log

_{ }0.0000001 - log_{ }0.00000001### MID-POINT 1

Find the midpoint of a line from (23,
12) to (5, 20)

__Solution__
x1=23,
x2=5, y1=12,y2=20.

Mid point = (

__x1 + x2__,__y1 + y2__)
2 2

Mid point = (2

__3 + 5__,__12 + 20__)
2 2

Mid point = (

__28__,__32__)
2 2

Hence
midpoint = (14, 16)

__TRY THIS……………………..__
Find the midpoint of a line from (-6,
14) to (6, -4)

## Friday, 24 March 2017

### PIE CHARTS 1

In a survey of 60 people, 28 people said
their favourite sport was football. What angle in a pie chart would this represent?

__Solution__

We make a fraction of 28 ot of 60, then
multiply by 360

^{0}.
=

__28__x 360^{0}.
60

= ~~360~~

__28__x^{0}_{6}_{1}

=
28 x 6

=
168

^{0}.

__Hence the angle is 168__

^{0 }

__TRY THIS………………………__
In a survey of 30 people, 23 people said
their favourite sport was football. What angle in a pie chart would this represent?

### POLYGONS 1

An
interior angle of a regular polygon is 74

^{0}greater than an exterior angle. Find the interior angle.
Solution

Let
i = interior angle, e = exterior angle.

Now
i + e=180

^{0}…………………(1)
But
i = e+74

^{0 }…………………(2)
Substitute
(2) in (1) above.

e+74

^{0 }+ e=180^{0}
e+
e+74

^{0 }=180^{0 }
2e+
74

^{0 }=180^{0 }
2e=180

^{0 }- 74^{0 }
2e=106

^{0 }__2__e=

__106__

^{0 }dividing by 2 both sides.

2 2

e
= 53

^{0}
But
i + e=180

^{0}…………………(1)
i + 53

^{0}=180^{0}.
i =180

^{0}- 53^{0}
i = 127

^{0}
Hence
i = 127

^{0 }

__TRY THIS………………………__
An
interior angle of a regular polygon is 78

^{0}greater than an exterior angle. Find the interior angle.
Subscribe to:
Posts (Atom)