BINARIES 1

If p*k = 4pk - p – 2k; find 3*2.

Solution

p*k = (4 x p x k) - p – (2xk)   [rewriting the given expression more clearly]

3*2 = (4 x 3 x 2) - 7 – (2x2)   [substituting 7 for p and 2 for k]

3*2 = 24 - 7 – 4

3*2 = 24 – 11   [since( -7-4) = -11]

3*2 = 13.

Hence   (3*2) = 13.

 TRY THIS........

If p*k = 5p + k – 5pk; find (-2*7).

SUM OF AP 2

The first term of an AP is 21 and the last term is 97. If the arithmetic progression consists of 20 terms, calculate the sum of all the terms. 

Solution

A₁ = 21, n=20, A_n = 97

S_n = n(A₁ + A_n)

       2

S₂₀ = 20(21 + 97)

         2

S₂₀ = 10 x 118

S₂₀ = 1180

Hence the sum of all 20 terms is 1180.

TRYTHIS………………….

The sum of first 20 terms of an AP is 1380, and the last term is 97. Find the 1^st term. 

EXPONENTIALS 2

Simplify the following by writing in power form:

w⁶³

w⁴⁰

Solution

= w^{63 - 40} 

=w²³

TRY THIS……………….

Simplify the following by writing in power form:

m⁵⁵

m²¹ 

VARIATION 1

x is inversely proportional to y. x=30 while y=5. Find y when x is 15.

Solution

x ⍺ k

      y

x = k

      y

30 = k

        5

k = 30 x 5

k = 150

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

k = 150, y= ?, x = 15.

x = k

      y

15 = 150

         y

y x 15 = 150  x  y   (multiply by y on both sides)

                y

15y = 150 ¹⁰        dividing by 15 both sides

15      15

Hence y = 10  .

TRY THIS…………….

x is inversely proportional to y. x=80 while y=4. Find y when x is 40.

SUM OF AP 1

The sum of first 20 terms of an AP is 1380, and the last term is 97. Find the 1^st term. 

Solution

A₁ = ?, n=20, A_n = 97, S_n =1380

S_n = n(A₁ + A_n)

       2

1380 = 20(A₁ + 97)

             2

1380 = 10(A₁ + 97)

1380 = 10A₁ + 970

1380 - 970= 10A₁

410= 10A₁

410= 10A₁

10     10

41= A₁

Hence 1^st term is 41.

TRYTHIS……………

The sum of first 30 terms of an AP is 2475, and the last term is 105. Find the 1^st term. 

EXPONENTIALS 1

Simplify the following by writing in power form:

(m⁵⁵ x m⁵)/ m⁴⁶ 

   

Solution

= (m⁵⁵ x m⁵)/ m⁴⁶ 

= m⁵⁵⁺⁵

    m²¹ 

= m⁶⁰

    m²¹ 

= m^60-21

= m³⁹

TRY THIS……………….

Simplify the following by writing in power form:

 (y⁶⁵ x y⁵¹)/ y⁹⁴ 

LOGARITHMS 2

If Log_ax = 0.9, find Log_a(x⁴)

Solution

= Log_a(x⁴)

= 4Log_ax

= 4 x Log_ax

= 4 x 0.9

= 3.6

TRY THIS………..

If Log_aw= 3.6, find Log_a(w³)

LOGARITHMS 1

Evaluate Log1000000 +  log 0.001 -  log 0.00000001

Solution

= Log1000000 +  log 0.001 -  log 0.00000001

= Log10⁶ +  log 10^-3 – log 10^-8

= 6Log10 + (-3 log 10) – (-8log 10)

= (6x1) + (-3x1) - (-8x1)

=6 + (-3) – (-8)

= 6 – 3 + 8

=11

Hence Log1000000 +  log 0.001-  log 0.00000001 = 11

POLYGONS 1

A regular polygon has 66 sides. Find the total degrees of that polygon.

Solution

n = 66

Total degrees = (n - 2)180⁰

                       = (66 – 2)180⁰

                       = 64 x 180⁰

                       = 11520⁰

Total degrees = 11520⁰

TRY THIS……………………….

A regular polygon has 33 sides. Find the total angles of that polygon.

FUNCTIONS 1

Find the maximum value of the quadratic equation:5-3t-8t².

Solution

a=-8, b=-3, c=5

Maximum = 4ac-b²

                        4a

Maximum = (4 x -8 x 5) - (-3)²

                          4(-8)

Maximum = (-160)- 9

                        -32

Maximum= -169                         since (-160)- 9= -169

                    -32

Maximum = 169

                     32       

Hence maximum value is 169/32

TRY THIS………………………………..

Find the maximum value of the quadratic equation:6+10t-4t².