RADICALS B1

 

LOGARITHMS B5

 

Evaluate Log₂(1024 x 64 x 256).

 

Solution

 

= Log₂(1024 x 64 x 256)

 

= Log₂1024 +  Log₂64 +  Log₂256         (applying the product rule)

 

= Log₂2¹⁰+  Log₂2⁶    +  Log₂2⁸           (Since1024= 2¹⁰, 64=2⁶ ,32=2⁵)

 

= 10Log₂2 +  6Log₂2    +  8Log₂2         ( remember  Log_aa^c = cLog_aa )

 

= (10 x 1) +  (6 x 1) +  (8 x 1)                  ( remember  Log_aa = 1 )

 

= 10 + 6 + 8

 

= 24 answer

 

 

hence Log₂(1024 x 64 x 256) = 22

 

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

 

Evaluate Log₂(512 x 128 x 1024). 

ALGEBRA B4

Expand -10(y – 11)

 

Solution

 

= -10(y – 11)

 

= (-10 x y) – (-10 x 11)

 

= 10y – ^-110

 

= 10y + ⁺50

 

= 10y + 50 answer

 

TRY THIS………..

 

Expand ^-10(f– 7)

BODMAS B1

 

Evaluate 17 x 237 + 463 x 17.

 

Solution

 

= 17 x 237 + 463 x 17

 

= 17 x (237 + 463)

 

= 17 x 700

 

= 11900 answer

 

 

TRY THIS………….

 

 

Evaluate 125 x 516 + 484 x 125.

BANKING B1

Ann deposited the amount of 140,000/= in a bank which gives an interest rate of 5% for 5 years. Find the simple interest she got?

 

Solution

 

I = ?,  P = 140,000/=, T = 5 years, R = 5%

 

I  = PRT

      100

 

I  = 140000 x 5 x 5

            100

 

I  = 140000 x 5 x 5

            100

 

I  = 1400 x 5 x 5

          

I  = 350,000/=

 

Hence the simple interest was Tsh 350,000/=

 

TRY THIS………………..

 

Veda deposited the amount of 48000/= in a bank which gives an interest rate of 3% for 6years. Find the simple interest she got?

 

 

EXPONENTIALS B3

 

If (x⁵)(y²) = 7500; find x and y.

 

solution

 

(x⁵)(y²) = 7500  [7500=2x2x5x5x5x5x5 by prime factorization].

 

(x⁵)(y²) = (5⁵ )x (2²)     [ since 7500=5⁵ x 2²].

 

equating equal powers,

 

(x⁵) = (5⁵)     

 

x   =   5    [since equal powers cancel out]

 

also,

 

(y²) = (2²) [since equal powers cancel out]

 

 y    =   2

 

∴ x   =   5 and y    =   2

 

TRY THIS…………

 

If (x⁵)(y³) = 15,000; find x and y.

 

PERIMETERS B1

Find the perimeter of a regular polygon with 7 sides inscribed in a circle of radius 6cm.

Solution

P = 2nrSin(180⁰/n)

n= 7 sides and r=6cm

P = 2 x 7 x 6 x Sin(180⁰/7)

P = 84 x Sin25.71⁰

P = 84 x 0.4338

Hence P = 36.44cm

TRY THIS…………….. 

Find the perimeter of a regular polygon with 12 sides inscribed in a circle of radius 7cm.

MID POINT B1

Find the midpoint of a line from (13, 6) to (9, 10)

 

Solution

 

x1=13, x2=9, y1=6, y2=10.

 

Midpoint = (x1 + x2, y1 + y2)

                          2             2

 

Midpoint = (13 + 9, 6 + 10)

                          2          2

 

Midpoint = (22, 16)

                      2    2

 

Hence midpoint = (11, 6)

 

TRY THIS……………………..

 

 

Find the midpoint of a line from (22, -8) to (-10, 16) 

LOGARITHMS B4

 

If log 2= 0.3010; find the value of log 50,000 without using tables.

 

solution

 

=log50,000

 

=log(100,000÷ 2)

 

=log100,000 – log2

 

=log10⁵ – log2

 

=5log10 – log2

 

=(5x1) – (0.3010)

 

=5 – 0.3010)

  

=4.699

 

Hence log50,000=4.699

 

 

TRY THIS……………

 

If log 2= 0.3010; find the value of log 5,000 without using tables.

 

FUNCTIONS B3

 If F(x) = 5x + 10; Find F^-1(40).

 

Solution

 

HINT: F^-1(x) means inverse.

 

PROCEDURE:

Make x the subject and then interchange x and y variables.

 

Let y=F(x)

 

So,  y= 5x + 10

 

y – 10 = 5x

 

y – 10  = ¹5x

   5          ₁5

 

y – 10   = x

    5

 

x  =   y – 10     after rearranging

            5

 

F^-1(x) = x – 10     after interchanging x and y variables.

                5

Now we calculate F^-1(40) as hereunder;

 

F^-1(40) = 40 – 10     

                  5

F^-1(40) =  30      

                 5

F^-1(40) =  6      [after dividing 30 by 5]

               

Hence, F^-1(40) =  6     

                             

TRY THIS……………………………

 

If F(x) = 7x - 20; Find F^-1(6).