Tuesday 9 October 2018

MID POINT 1


Find a and b if the midpoint of a line from (a, 6) to (8, b) is
(30, 7)

Solution

x1=a, x2=8, y1=6, y2=b.

(30, 7)= (x1 + x2, y1 + y2)
                   2             2

(30, 7)= (a + 8,  6 + b)
                  2          2


Equating equal values of x;

30= a + 8
         2      

2 x 30= (a + 8)  x 2 1
                2 1     

60 = a + 8

60-8 = a

a = 52.

Equating equal values of y;

7 = 6 + b
        2

2 x 7= (6 + b)  x 2 multiplying by 2 both sides.
                2 1     

14 = 6 + b

14 – 6 = b

b = 8

Hence a = 52 and b=8

TRY THIS……………………..



Find m and n if the midpoint of a line from (m, 19) to (13, n) is (9, 11)

POLYGONS 1


Find the exterior angle of a regular polygon with 18 sides.

Solution

n = 6

e =   360
         n

e =   360
        18

e =   360 20
        181

e = 200

Hence the exterior angle is 200

TRY THIS………………………………………..



Find the exterior angle of a regular polygon with 12 sides.

PRIME FACTORS 1




Write 560 as a product of prime factors.

Solution
2
560
2
280
2
140
2
70
5
35
7
7
1

 Then, 480 = 2 x 2 x 2 x 2 x 5 X 7

Hence   480 = 24 x 5 x 7.

TRY THIS........

Write 1000 as a product of prime factors.

SETS 1



If n(B)= 60 , n(AuB) = 170 and n(AnB)=10, find n(A)

Solution

n(AuB) = n(A) + n(B) - n(AnB)

  170    = n(A)  + 60– 10

  170     = 60 -10 + n(A) 

  170     = 50 + n(A) 

  170 - 50        = n(A)

  120        = n(A)

Hence n(A) = 120 answer


TRY THIS…………………………….




If n(B)= 55 , n(AuB) = 169 and n(AnB)=16, find n(A)

ANGLES 1


If A and B are supplementary angles such that A = 27° and B = x + 55° , find the value of x .

Solution

A + B = 1800. [Since supplementary angles add up to 1800]

270 + x + 550 = 1800.

x + 820 = 1800.

x = 1800 - 820

x = 980  answer.

TRY THIS.............................. 



If A and B are supplementary angles such that A = 53° and B = x + 42° , find the value of x .


Monday 8 October 2018

EXPONENTIALS 1


If a2 = 3; find the value of a10 + a8.

solution

=a10 + a8.

=(a2)5 + (a2)4.

=(3)5  +  (3)4

=243 + 81

=324 answer  
  

TRY THIS.........................


If a2 = 5; find the value of a10 - a6.



ALGEBRA 2


The sum of five consecutive numbers is 525. Find the 4th number.

Solution

Let the numbers be as shown in the table below

1st number
2nd number
3rd number
4th number
5th number
TOTAL
n
n+1
n+2
n+3
n+4
525

Then, n + (n+1) + (n+2) + (n+3) + (n+4) = 525

5n + 1+2+3+4= 525

5n + 10= 525

5n = 525- 10

5n = 515 

5n = 515
5       5

n = 103

4th number = n + 3

                   = 103 + 3

                   = 106

Hence the 4th number is 106

TRY THIS………………….. 



The sum of five consecutive numbers is 1815. Find the largest number.  

FUNCTIONS 1




FACTORIZE 1


Factorize 1-25m2

Solution

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

1-25m2 = 12-52m2       

            = 12 - (5m)2       

            = (1 - 5m)(1 + 5m)

Hence 1-25m2 = (1 - 5m)(1 + 5m)  answer


TRY THIS…………….



Factorize 1 – 49y2

LOGARITHMS 1


If Logax = -20, find Loga(1/x6)

Solution

= Loga(1/x6)                        

= Logax-6                   [since 1/cn =c-n]

= -6 x Logax

= -6 x -20

= 120


TRY THIS………..


If LogaW= -70, find Loga(1/W4)

ALGEBRA 1


Express as a single fraction m - 3 + m+5
                                               6         2
solution

= m - 3 + m+5
      6          2

= m - 3 + 3(m+5)
            6         

= m - 3 + 3m+15
            6         

= m + 3m + 15- 3
            6         

= 4m +  12
         6         

= 4(m +  3)
         6         

= 2(m +  3)
         3         

= 2m +  6       answer
         3         


TRY THIS...................

Express as a single fraction 4n - 1 + n+1
                                                3          2