Monday 31 October 2016

RELATIONS 1


Given that x є A where A= {-3≤x≤3} produce a pictorial representation of the relation, R:x onto (2x-11)
Solution
x
-3
-2
-1
0
1
2
3
4
2x-11
-17
-15
-13
-11
-9
-7
-5
-3

The diagram is shown below:




TRY THIS……….

Given that x
є A, draw a pictorial representation  of
 x onto (x-7)  [2x10]


FUNCTIONS 1


Find a linear function f(x) with gradient -10 which is such that f(5)=12.

        Solution.

m=-10, points = [5,12] and [x, f(x)]

m = y2-y1/x2-x1

-10 = f(x) – 12/x-5

f(x)-12=-10(x-5) [after cross multiply]
f(x)-12=-10x-50
f(x)=-10x+50+12
f(x)=-10x+62

Hence a  linear function is f(x) = -10x+62.

TRY THIS……….

Give out a linear function f(x) with

Gradient 5 and f(2)=14

LOGARITHMS 1


Evaluate Log2(2048 x 32).

Solution

= Log2(2048 x 32)

= Log22048 +  Log232          (applying the product rule)

= Log2211 +  Log225             (2048= 211 and 32=25 )

= 11Log22 +  5Log22       ( remember  Logaac = cLogaa )

= (11 x 1) +  (5 x 1)          ( remember  Logaa = 1 )

= 11 + 5

= 16

hence Log2(2048 x 32) = 16

TRY THIS………………



Evaluate Log2(8 x 256). 

FACTORIZING 1


Factorize 625x- 81y2

Solution

we use difference of two squares a2 – b2 = (a - b)(a + b)

625x2- 81y2  = 252x2 - 92y2

                   = (25x)2 - (3y)2

                   = (25x - 3y)( 25x + 3y)

Hence 625x2 - 81y2 = (25x - 3y)(25x + 3y)

TRY THIS………………………….



factorize 49m2- 36n2

MIDPOINT 1



Find the midpoint of a line from (19, 6) to (5, 10)

Solution

x1=19, x2=5, y1=6, y2=10.

Mid point = (x1 + x2, y1 + y2)
                         2             2

Mid point = (19 + 5,  6 + 10)
                        2          2

Mid point = (24, 16)
                     2    2

Hence midpoint = (12, 6)

TRY THIS……………………..



Find the midpoint of a line from (14, -8) to (-6, 22)