BODMAS F1

 

Evaluate 90 + 50 x 3 – 80 ÷ 4

 

Solution

 

We apply BODMAS.

 

= 90 + 50 x 3 – 80 ÷ 4

 

= 90 + 50 x 3 – 20    [After dividing]

 

= 90 + 150 – 20    [After multiplying]

 

= 240 – 20    [After adding]

 

= 220       [After subtracting]

 

Hence 90 + 50 x 3 – 80 ÷ 4 = 220      

 

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

 

Evaluate 200 + 10 x 30 – 60 ÷ 4

PERIMETERS F1

 Area of rectangle is 48m². Find its perimeter if width is 4m. 

 

Solution

 

A = L x w

 

48 = L x 4

 

48 = 4L

4      4

 

12 = L

 

Now; P =2(L + w)

 

= 2(12 + 4)

 

= 2 x 16

 

= 32m

 

Hence perimeter is 32m

 

TRY THIS………..

 

Area of rectangle is 80m². Find its perimeter if width is 4m.  

LOGARITHMS F2

 If log_x 2401 – log₃729 = -4; find x

 

Solution

 

log_x 2401 – log₃729 = -2

 

log_x 2401 – log₃3⁶ = -2

 

log_x 2401 – 6log₃3 = -2

 

log_x 2401 – 6 x 1 = -2

 

log_x 2401 – 6 = -2

 

log_x 2401 = -2 + 6

 

log_x 2401 = 4 

 

2401 = x⁴        2401=7x7x7x7=7⁴ by prime factors

 

7⁴ = x⁴      Powers cancel out

 

x = 7

 

 

Hence x = 7.

 

TRY THIS………….

 

If log_m2048 – log₆216 = 8; find m

 

VECTORS F1

 

If u = 16i + 2j and v = 8i + 10j find 4u - 3v

 

solution

 

= 4u - 3v

 

= 4(16i + 2j) - 3(8i + 10j)

 

= 64i + 8j - 24i - 30j

 

= (64i - 24i) + (8j - 30j)    [grouping like terms]

 

= 40i - 22j

 

Hence 4u - 3v = 40i - 22j .

 

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

 

If u = 10i + 3j and v = 11i + 4j;  find 6u - 2v.