Answer:
He must invest R297 521 today.
Step-by-step explanation:
The compound interest formula is given by:

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per year and t is the time in years for which the money is invested or borrowed.
Banabas must pay his ex-wife an amount of R350 000 in two years’ time.
This means that 
Interest rate of 8.15% per annum compounded monthly:
This means that
.
Amount he must invest today:
This is P. So




He must invest R297 521 today.
Step-by-step explanation:
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Answer:
The relation for the distance cover by Tina and Lia is 20 = 4 × m
Option B
Step-by-step explanation
Given as :
The distance cover by Tina =
= 20 miles
The distance cover by Lia =
= m miles
According to question
Tina walked 4 times as far as Lia
So, The distance cover by Tina = 4 × The distance cover by Lia
Or,
= 4 ×
I.e 20 miles = 4 × m miles
or. 20 = 4 × m
So, relation is 20 = 4 × m
Hence, The relation for the distance cover by Tina and Lia is 20 = 4 × m
I.e option B Answer
There are 500 even numbers from 1 to 1000 starting with 2 and ending with 998. The sequence of even numbers is an arithmetic sequence with a common difference of 2. Using the equation for the arithmetic series,
S = (a1 + an)(n/2)
and substituting,
S = (2 + 998)(500/2)
we get,
S = 250,000
Therefore, the sum of all even numbers between 1 and 1000 is 250,000.