RELATIONS 2

Let R={(10,3), (11,1), (-8,-6), (10,-20)}. Find the domain and range of R.

solution

For domain we check on the values of x in each point.

Domain={10, 11, -8}

For range we check on the values of y in each point.

Range={3, 1, -6, -20}

Hence Domain={10, 11, -8} and Range = {3, 1, -6, -20}

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

Let R={(7,3), (-41,-8), (-6,-6), (17, 0)}. Find the domain and range of R.

ALGEBRA 1

Multiply 2x - 7y by -9a

solution

= -9a(2x - 7y)

= (-9a x 2x) + (-9a x -7y)

= -12ax + 63ay     [since -9a x -7y = +63ay ]

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

Multiply 12p - 7q by  -4a

LOGARITHMS 3

Simplify Log₂128 - Log₃27

solution

= Log₂128 - Log₃27

= Log₂2⁷ - Log₃3³

= 7Log₂2 - 3Log₃3

= (7 x 1) - (3 x 1)

= 7 - 3

= 4

Hence Log₂128 - Log₃27 = 4

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

Simplify Log₂256 - Log₃27

GRADIENT 1

Find the slope of a line which passes through (-9, -4) and (8,-11)

Solution

x₁ = -9,  y₁ =-4,  x₂ = 8,  y₂ = -11

m = y2 –y1

       x2 – x1

m =   -11 –(-4)

          8 –(-9)

m =   -11 + 4

          8 + 9

m =    - 7

          17

Hence the slope is -7/17

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

Find the slope of a line which passes through (-11, -4) and (4,-11)

ARC LENGTH 1

Find radius if the length of an arc of the circle is ^2∏/₁₅ cm.

Solution

L= ^2∏/₁₅  , r = ?

L = ∏r

    180

^2∏/15  = ∏r

             180

180  x ^2∏/15  = ∏r  x   180    [multiplying by 180 both sides]

                      180

360 ∏  = ∏r 

     15      

                    

360 ∏  = ∏r 

    15∏  ∏

360   =  r 

    15      

r = 24cm

Hence radius is 24 cm

TRY THIS………………..

Find radius if the length of an arc of the circle is ^7∏/₃₀ cm.

ANGLES 1

If x-15⁰ and 4x + 45⁰ are complementary angles, find the value of x.

solution

Complementary angles add up to 90⁰.

so,  x-15⁰ + 4x + 45⁰ = 90⁰.

so,  x + 4x + 45⁰ - 15⁰ = 90⁰.

so,  5x + 30⁰ = 90⁰.

so,  5x = 90⁰ -  30⁰

so,  5x = 60⁰

so,  5x = 60¹²

       5       5

Hence x = 12

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

If 3x - 12⁰ and 2x + 77⁰ are complementary angles, find the value of x.

POLYNOMIALS 2

If f(x) = x⁴ + kx² + 3x + 7 has a remainder of 22 when divided by x+2; find k.

solution

f(x) = x⁴ + kx² + 3x + 7

x + 2 = 0

x = -2

f(x) = (-2)⁴ + k(-2)² + 3(-2) + 7 = 22

16 + 4k + (-6) + 7 = 22

16 + 4k + 1 = 22

4k + 17 = 22

4k = 22 - 17

4k = 5

4k = 5

4      4

k = 5/4

hence k=5/4

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

If f(x) = x⁴ - kx² + 3x - 8 has a remainder of 20 when divided by x-3; find k.

LOGARITHMS 2

Evaluate Log100000 +  log 0.001 + log₃243

Solution

= Log100000 +  log 0.001 + log₃243

= Log10⁵ +  log 10^-3 + log₃3^-5

= 5Log10 +  (-3log 10) + (-5log₃3)    [since log_aaⁿ  = nlog_aa]

= (5x1) + (-3x1) + (-5x1)             [since log_aa = 1]

= 5 + (-3) + (-5)

= 5 - 8

= -3

Hence Log100000 +  log 0.001 + log₃243 = -3

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

Evaluate Log100,000 +  log 0.01 + log₃729

CIRCLES 1

Given the figure below, prove that an angle inscribed in a semi –circle is a right angle.

solution

Given:

        diameter AOB

        O as centre of circle

        C is any point in the circumference

Required to prove:   

<ACB =90⁰       

Proof:      

< at centre = 2(< at circumference)

<AOB=2(<ACB) ------------    (i)

But <AOB is a straight angle

Straight angle=180⁰

Hence, <AOB = 180⁰

Now, from (i)

<ACB = 2(<ACB)

180⁰=2(<ACB)

90⁰= <ACB     [after dividing by 2 both sides]

Hence ACB = 90⁰ proved!!!

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

Given the figure below, prove that an angle inscribed in a semi –circle is a right angle.

RELATIONS 1

Let R={(10,3), (11,1), (-8,-6), (10,-20)}. Find the domain and range of R.

solution

For domain we check on the values of x in each point.

Domain={10, 11, -8}

For range we check on the values of y in each point.

Range={3, 1, -6, -20}

Hence Domain={10, 11, -8} and Range = {3, 1, -6, -20}

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

Let R={(7,3), (-41,-8), (-6,-6), (17, 0)}. Find the domain and range of R.