Colin maxed out his credit card at $4,000

2 answers:

8 0

It wants to know the interest payed over 15 months, he payed $300 each month and of that 300, there was 12.35% of it taken out. if you do the math...
300 x (12.35/100) it comes out to 37.05 that means that each month he payed $37.05 to interest when he made his payments, and at this point we can ignore the $300 he pays each month and focus on the interest alone, we know he is going to pay the same amount for 15 months which means he pays $37.05 towards interest 15 times over, with some simple math...
$37.05 x 15 which comes out to $555.75
meaning over the total 15 months he payed $555.75 total in interest
Thus your answer is $555.75

The other person who answered wasnt exactly wrong it's just the fact that if your paying $300 a month you dont pay it a little each day you pay the full $300 at one time each month, so the balance will never actually hit $0 it will blow past it, meaning you can't do the math how he answered it, you have to reverse it and figure it out from the interest first.

Hope This Helps.

FinnZ [79.3K]3 years ago

7 0

So Colin is required to pay $300 for 15 total months in order to get his credit card balance down to $0.

In order to find this total, we need to take the amount he is going to pay per month ($300) and multiply is by the total number of months that he is going to be paying that amount (15):

$300 * 15 months = $4,500

The total amount of money that he is going to pay after his card balance reaches $0 is $4,500.

However, Colin only maxed out his credit card to $4,000 - the rest of the money is due to interest. So to find this amount, we take the total amount that he paid ($4,500) and subtract it from the total amount he initially maxed out his card with ($4,000):

$4,500 - $4,000 = $500

The amount of interest that Colin has paid when his credit card balance is $0 is $500.

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Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x' (t) is its velocity, and x'

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Answer:

a) $v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

b) $0$

c) $a(t) = x''(t) = v'(t) =6t-12$

When the acceleration is 0 we have:

$6t-12=0, t =2$

And if we replace t=2 in the velocity function we got:

$v(t) = 3(2)^2 -12(2) +9=-3$

Step-by-step explanation:

For this case we have defined the following function for the position of the particle:

$x(t) = t^3 -6t^2 +9t -5 , 0\leq t\leq 10$

Part a

From definition we know that the velocity is the first derivate of the position respect to time and the accelerations is the second derivate of the position respect the time so we have this:

$v(t) =x'(t) = \frac{dx}{dt} = 3t^2 -12t +9$

$a(t) = x''(t) = v'(t) =6t-12$

Part b

For this case we need to analyze the velocity function and where is increasing. The velocity function is given by:

$v(t) = 3t^2 -12t +9$

We can factorize this function as $v(t)= 3 (t^2- 4t +3)=3(t-3)(t-1)$

So from this we can see that we have two values where the function is equal to 0, t=3 and t=1, since our original interval is $0\leq t\leq 10$ we need to analyze the following intervals: