Friday 28 February 2014

FACTORIZATION C2

Factorize 2x2 - 7x - 30

Solution

= 2x2 - 7x – 30.  We split the middle term (-7x) to be (-12x + 5x).

= 2x2 - 12x + 5x – 30     (-7x =-12x + 5x).

= (2x2 -12x) + (5x - 30)

= 2x(x - 6) + 5(x - 6)

= (2x + 5) (x -6)

Hence2x2 - 7x – 30 = (2x + 5) (x -6) answer.



TRY THIS…..


Factorize 4x2 - 3x – 10.

DIFFERENCE OF 2 SQUARES C1

Evaluate 7802 – 6802

Solution

7802 – 6802 = (780 + 680)( 780 – 680)

                   = (1460)( 100)

                   = 146000


Hence 7802 – 6802 = 146000

LOGARITHMS B38

Evaluate the following giving your answer in standard form.

38.3 x 10-4
   4 x 10-20

Solution

38.3 x 10-4
         4 x 10-20

38.3  x   10-4
     4         10-20

=  9.575 x 10-4 –(-20)

=  9.575 x 10-4 + 20    

=  9.575 x 1016     


Hence   38.3 x 10-4   = 9.575 x 1016     

                 4 x 10-20

RADICALS C10


INEQUALITIES C1

2x – 3 x + 15 6x

Solution

2x – 3 x + 15 and  x + 15 6x

2x – x 3+ 15 and  15 6x - x

x 18 and  15 5x (divide by5 both sides)

x 18 and  3 x


x 18 and  x 3  answer

Wednesday 26 February 2014

RADICALS C9


FACTORIZATION C11


Factorize 2x2 - 15x + 7.

Solution

= 2x2 - 15x + 7. We split the middle term (-15x) to be (-x - 14x).

= 2x2 -x - 14x + 7

= (2x2 – x) – (14x + 7)

= x(2x – 1) – 7(2x - 1)

= (x - 7) (2x - 1)

Hence 2x2 - 15x + 7= (x - 7) (2x - 1)answer.



TRY THIS…..


9x2 - 8x - 1.

MATRIX B1



PROBABILITY C14

Given the interval of 1-10 inclusive, Find the probability of having an even number or a prime number.

Solution

This is a non-mutually exclusive event.

n(S) = number of sample space = 10
Let P(E)= probability of even numbers = 5/10
Let P(R)= probability of prime  numbers = 4/10
P(EnR) = probability of having both even and prime numbers= 1/10

P(EuR) = P(E) + P(R)  - P(EnR).

 P(EuR) =  5  +  41
                10    10   10

P(EuR) =  9  -   1 
                10    10  

P(EuR) =  8 
                10       


Hence probability of an even number or a prime number is 8/10.


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

Given the interval of 11-20 inclusive, Find the probability of having an odd number or a prime number.


CIRCLES C8

In the figure below, find the value of <FGE.



Solution

<FEG = <HFG

<FEG = 540 [angles in alternate segments are equal.]

Consider GEF

<EFG + <FGE+ <GEF = 1800 [total degrees of a triangle]

 (390 + 540) + <FGE + 54= 1800

930 + <FGE + 540 = 1800

1470 + <FGE= 1800

<FGE= 1800 - 1470

<FGE= 330  


<FGE = 330  


LOGARITHMS B37

If logx625 – log264 = -4; find x

solution

logx625 – log264 = -2

logx625 – log226 = -2

logx625 – 6log22 = -2

logx625 – 6 x 1 = -2

logx625 – 6 = -2

logx625 = -2 + 6

logx625 = 4

625 = x4

54 = x4      Powers cancel out

x = 5


Hence x = 5.

Tuesday 25 February 2014

FACTORIZATION C10

Factorize 3x2 - 16x + 5.

Solution

= 3x2 - 16x + 5. We split the middle term (-16x) to be (-x - 15x).

= 3x2 - x - 15x + 5.

= (3x2 – x) – (15x + 5)

= x(3x – 1) – 5(3x - 1)

= (x - 5) (3x - 1)

Hence 3x2 - 16x + 5= (x - 5) (3x - 1)answer.


TRY THIS…..


6x2 - 17x - 14.


RADICALS C8




EXPONENTS A11


simplify 5
              8-2/3

Solution

5
    8-2/3

=  5 x 1
           8-2/3

=  5 x 82/3        [Since 1/a-n = an]

=  5 x (81/3)2        [Since 81/3 = cube root of 8 = 2]

=  5 x (2)2        

=  5 x 4

= 20        
          
Hence   5        = 20

              8-2/3


EXPONENTS A10

Simplify (D11)10

solution

= (D11)10

=  D11x10

=  D110

TRY THIS


Simplify  (G2)12

FACTORIZATION C9

Factorize 2x2 - 3x - 35.

Solution

= 2x2 - 3x - 35. We split the middle term (-3x) to be (+7x - 10x).

= 2x2 + 7x - 10x - 35.

= (2x2 + 7x) – (10x – 35).

= x(2x + 7) – 5(2x + 7).

= (x - 5) (2x + 7)

Hence 2x2 - 3x – 35 = (x - 5) (2x + 7) answer.


TRY THIS…..


8x2 - 22x + 5.

RADICALS C7





LOGARITHMS B36


If logx49 – log2128 = -5; find x

solution

logx49 – log2128 = -5

logx49 – log227 = -5

logx49 – 7log22 = -5

logx49 – 7 x 1 = -5

logx49 – 7 = -5

logx49 = -5 + 7

logx49 = 2

49 = x2

72 = x2      Powers cancel out

x = 7


Hence x = 7.

CIRCLES C7


In the figure below, find the value of <RUS.



Solution

<RSU = <TRU

<RSU = 540 [angles in alternate segments are equal.]

Consider SRU

<SRU + <RUS+ <RSU = 1800 [total degrees of a triangle]

 (460 + 540) + <RUS + 54= 1800

1000 + <RUS + 540 = 1800

1540 + <RUS= 1800

<RUS= 1800 - 1540

<RUS= 260  


<RUS = 260  


PROBABILITY C13

A bag contains 6 red stones and 8 blue stones. Two balls are taken from the bag. What is the probability that they are both red?

Solution

(This is a problem with replacement)

n(R) = 6, n(B) = 8, n(S) = 14

P(R) = n(R)
            n(S)

1st pick = 6/14
2nd pick = 6/14 as well.


P(R) =      x    6
          14          14

P(R) =    36    =      4 
             196           49  


Therefore Probability of drawing a red stone is 4/49     


LOGARITHMS B35

If logx27 + log2256 = 11; find x

solution

logx27 + log2256 = 11

logx27 + log228 = 11

logx27 + 8log22 = 11

logx27 + 8 x 1 = 11

logx27 + 8 = 11

logx27 = 11- 8

logx27 = 3

27 = x3

33 = x3      Powers cancel out

x = 3


Hence x = 3.

CIRCLES C6

In the following figure, prove that the angles in the same segment of a circle are equal.



Solution

Given : A circle with centre O and the angles ∠USV and ∠UTV in the same segment formed by the chord UV (or arc UAV)

Required to prove: ∠USV = ∠UTV

Construction : Join OU and OV.

Proof :

Try to remember that angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle, hence we have,

∠UOV = 2 ∠USV ...............(i)

and,

∠UOV = 2∠UTV …………...(ii)


Since (i) = (ii), we get,

2∠USV = 2∠UTV

2∠USV = 2∠UTV                 (dividing by 2 both sides)
2             2

∠USV = ∠UTV

∴ ∠USV =∠UTV   proved!

Monday 24 February 2014

PROBABILITY C12

A certain man takes 2 cards from a standard deck of 52 cards.
What will be the probability of getting a pair of hearts?

Solution

(This is a problem with replacement)

n(E) = 13, n(S) = 52

P(E) = n(E)
           n(S)

1st pick = 13/52
2nd pick = 13/52 as well.


P(2 hearts) =   13      x    13
                       52            52

P(2 hearts) =    169    =      1 
                        2704         16  


Therefore Probability of drawing 2 heart cards is 1/16     



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

A certain man takes 2 cards from a standard deck of 52 cards.
What will be the probability of getting a pair of spades?


LOGARITHMS B34


If logx625 + log381 = 8; find x

solution

logx625 + log381 = 8

logx625 + log334 = 8

logx625 + 4log33 = 8

logx625 + 4 x 1 = 8

logx625 + 4 = 8

logx625 = 8- 4

logx625 = 4

625 = x4

54 = x4      Powers cancel out

x = 5


Hence x = 5.

CIRCLES C5

In the figure below, find the value of <HLJ.



Solution

<HJL = <LHK

<HJL = 540 [angles in alternate segments are equal.]

Consider HJL

<JHL + <HLJ+ <HJL = 1800 [total degrees of a triangle]

 (440 + 540) + <HLJ + 54= 1800

980 + <HLJ + 540 = 1800

1520 + <HLJ= 1800

<HLJ= 1800 - 1520

<HLJ= 280  


<HLJ= 280  


PROBABILITY C11


A bag contains 6 red balls and 8 blue balls. Two balls are taken from the bag. What is the probability that they are both blue?

Solution

(This is a problem with replacement)

n(R) = 6, n(B) = 8, n(S) = 14

P(B) = n(B)
            n(S)

1st pick = 8/14
2nd pick = 8/14 as well.


P(B) =      x    6
          14          14

P(B) =    36    =      4 
             196           49  


Therefore Probability of drawing a blue ball is 4/49     



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

A bag contains 10 purple stones and 12 blue stones. Two stones are taken from the bag. What is the probability that they are both blue?

LOGARITHMS B33

If logx625 + log327 = 7; find x

solution

logx625 + log327 = 7

logx625 + log333 = 7

logx625 + 3log33 = 7

logx625 + 3 x 1 = 7

logx625 + 3 = 7

logx625 = 7- 3

logx625 = 4

625 = x4

54 = x4      Powers cancel out

x = 5


Hence x = 5.

Friday 14 February 2014

CIRCLES C4



In the figure below, find the value of <MNQ.



Solution

<MQN = <PMN.

<MQN = 520 [angles in alternate segments are equal.]

Consider MQN

<QMN + <MNQ+ <MQN = 1800 [total degrees of a triangle]

 (380 + 520) + <MNQ + 52= 1800

900 + <MNQ + 520 = 1800

1420 + <MNQ= 1800

<MNQ= 1800 -1420

<MNQ= 380  

<MNQ= 380