Since "per cent" means "of 100", any fraction of the pattern n/100 is equal to n percent. Therefore, 9/100 is equal to nine percent.
It is any fraction of the form (30*n)/(100*n) where n is a non-zero integer or the reciprocal of a common factor of 30 and 100.
That would be equal to the sum of all natural numbers less than or equal to 100 minus the sum of all natural numbers less than or equal to 89. The sum of all natural numbers less than or equal to x can be calculated with this equation: x ∑n = n(n + 1) / 2 n=1 We can take the numbers given above then, and plug them into that sum, giving us the equation: x = 100(100 + 1)/2 - 89(89 + 1) / 2 x = 5050 - 4005 x = 1045
90 is the 100%.100% :N = 30% : 2730%N = 27 (100%)N = 27(100%)/30%N = 2700/30N = 90
100*n/64100*n/64100*n/64100*n/64
100% is equal to 625000. 80% : 500000 = 100% : N 80%N/80%= (500000 * 100%)/80% N= 5000000/8 N= 625000
Since "per cent" means "of 100", any fraction of the pattern n/100 is equal to n percent. Therefore, 9/100 is equal to nine percent.
114
If 10^n = 100, then n would equal 2 because 10^2 = 100. To find n^2, you would square the value of n, which in this case is 2. Therefore, n^2 would equal 2^2, which is 4.
300
50n = 5000 n = 5000/50 n = 100
It is any fraction of the form (30*n)/(100*n) where n is a non-zero integer or the reciprocal of a common factor of 30 and 100.
$n$ is equal to 140 since 60% of $n$ is equal to 84. To find $n$, you need to divide 84 by 0.6 (or multiply by 100 and then divide by 60). This gives you $n$ = 140.
100 N due to newton's third law that every action has an opposite and equal reaction.
The wall pushes back on you with an equal force of 100 N, as described by Newton's Third Law of Motion which states that for every action, there is an equal and opposite reaction.
At terminal velocity, the force of air resistance is equal in magnitude to the weight of the object. Since the weight of the bag of nails is 100 N, the air resistance on the bag of nails falling at terminal velocity is also 100 N.
Let the five consecutive numbers be represented as n-2, n-1, n, n+1, and n+2. The sum of these numbers can be expressed as (n-2) + (n-1) + n + (n+1) + (n+2) = 5n. Setting this sum equal to 100 gives us 5n = 100. Solving for n, we find n = 20. Therefore, the five consecutive numbers that add up to 100 are 18, 19, 20, 21, and 22.