ALGEBRA C4

COORDINATE GEOMETRY C10

Find the slope of a line which passes through (3, -2) and (6,-7).

Solution

x₁ = 3,  y₁ =-2,  x₂ = 6,  y₂ = -7.

m = y2 –y1

      x2 – x1

m =   -9 –(-7)

          6 –3

m =   -9 + 7

           3

m =    - 2

           3

Hence the slope is -2/3

TRY THIS.........

Find the slope of a line which passes through (3, -9) and (6,-12).

 

PROBABILITY C10

A bag contains 10 red pencils and 7 blue pencils. Two pencils  are taken from the bag. What is the probability that they are both blue?

Solution

(This is a problem with replacement)

n(R) = 10, n(B) = 7, n(S) = 18

P(B) = n(B)

           n(S)

1st pick = 7/18

2nd pick = 7/18 as well.

P(B) =  7      x    7

           18          18

P(B) =    49    

             324         

Therefore Probability of drawing a blue pencil is 49/324     


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

 A bag contains 6 red pencils and 10 blue pencils. Two pencils  are taken from the bag. What is the probability that they are both red?
 

LOGARITHMS B32

If log_x81 – log₂64 = -2; find x

solution

log_x81 – log₂64 = -2

log_x81 – log₂2⁶ = -2

log_x81 – 6log₂2 = -2

log_x81 – 6 x 1 = -2

log_x81 – 6 = -2

log_x81 = -2 + 6

log_x81 = 4

81 = x⁴

3⁴ = x⁴      Powers cancel out

x = 3

Hence x = 3.

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

 If log_x81 – log₅625 = 0; find x

 

CIRCLES C3

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

Solution

<ABD = <DAC

∴ <ABD = 51 [angles in alternate segments are equal.]

               Consider ∆ABD

<ABD + <BAD+ <BDA = 180⁰ [total degrees of a triangle]

51 + (60 + 51) + <BDA= 180⁰

51 + 111 + <BDA= 180⁰

162 + <BDA= 180⁰

<BDA= 180⁰ -162

<BDA= 18⁰  

∴ <BDA= 18⁰  

ALGEBRA C3

COORDINATE GEOMETRY C9

Find the slope of a line which passes through (3, -2) and (6, 6).

Solution

x₁ = 3,  y₁ =-2,  x₂ = 6,  y₂ = 6

m = y2 –y1

      x2 – x1

m =   -9 – 6

         6 – 3

m =   -15

           3

m =    -5

           

Hence the slope is -5

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

Find the slope of a line which passes through (-9, -2) and (6, -5).