Saturday, 25 March 2017

ARITHMETIC 1


Evaluate 3962 – 3862

Solution

We apply difference of two squares: a2 - b2 = (a+b)(a-b)

3962 – 3862  = (396 + 386)( 396 – 386)

                   = (782)( 10)

                   = 7820   [only add a zero at 782]

3962 – 3862 = 13820

TRY THIS…………….
  

Evaluate 5082 – 4082

LOGARITHMS 2


If Logax = 0.9, find Loga(1/x)

Solution

= Loga(1/x)

= Logax-1

= -1 x Logax

= -1 x 0.9

= -0.9


TRY THIS………..


If Logaw= 3.6, find Loga(1/w)

SLOPE OR GRADIENT 1


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

Solution

x1 = -1,  y1 =-8,  x2 = 6,  y2 = -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-8t2.

Solution

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

Maximum = 4ac-b2
                        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-5t2.

POLYGONS 2

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

Solution

n = 62

Total degrees = (n - 2)1800

                       = (62 – 2)1800

                       = 60 x 1800

                       = 108000

Total degrees = 108000


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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

= Log104 +  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 = (23 + 512 + 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 3600.
= 28 x 3600.
   60

= 28 x 36006
   601

= 28 x 6

= 1680.

Hence the angle is 168

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 740 greater than an exterior angle. Find the interior angle.

Solution

Let i = interior angle, e = exterior angle.

Now i  + e=1800…………………(1)

But i = e+740 …………………(2)

Substitute (2) in (1) above.

e+740   + e=1800

e+ e+740   =1800

2e+ 740   =1800

2e=1800 - 740   

2e=1060

2e=1060          dividing by 2 both sides.
2      2

e = 530

But i  + e=1800…………………(1)

i  + 530=1800.

i  =1800 - 530

i = 1270

Hence i = 127


TRY THIS………………………   


An interior angle of a regular polygon is 780 greater than an exterior angle. Find the interior angle.