The number of times Ken visits, given the £3.6 pool charge, and 1/3 savings per visit is 5 times.
<h3>Which method can be used to find the number of times Ken visits the pool to get back his £5?</h3>
The entry fee = £3.60
Amount saved on entry fee by having a membership card = 1/3 of the entry fee
Amount Ken spends on the membership card and the reduced entry fee = £5
Therefore;
Reduced entry fee = £3.60×(1 - 1/3) = £2.4
Amount saved per visit = £3.6 - £2.4 = £1.2
The number of visits, <em>n</em>, before he gets back his £5 is therefore;
- n = £5/(£1.2/visit) ≈ 4.17 visits
Ken has to visit the swimming pool more than 4 times to get his £5 back.
Rounding to the next larger whole, therefore;
- The number of visits, n, before he gets back his £5 is 5 visits
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For this case we have the following function:
Let's rewrite the function:
We set the denominator to zero to see the values of x for which it is not defined:
From here, we get:
There is a removable discontinuity at x = -2, since by rewriting the function we have:
Answer:
the removable discontinuity of f (x) is located at:
x = -2
So the original price is 40 dollars. We can see how much she saved by multiplying the original amount with the discount. 40 x .60=24. So she saved 24 dollars. We can then subtract 24 from 40 to get a price of 16 dollars.
% means out of 100 so 1%=1/100
so 3%=3/100
$30.00=3000 cents
so you multiply 3/100 by 3000 =(3*3000)/100
you can cross out two zeros at the bottom and on the top and get
(3*30)/1 or 90/1 or 90 cents
I am in the middle is class.
