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Ray Of Light [21]
3 years ago
14

Colin maxed out his credit card at $4,000

Mathematics
2 answers:
Svet_ta [14]3 years ago
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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64/9 to a mixed number
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Elizabeth lives in San Francisco and works in Mountain View. In the morning, she has 3 transportation options (take a bus, a cab
Lesechka [4]

Answer:

4/9

Step-by-step explanation:

The possibilities of transportation: (The first will be for morning, second will be for afternoon)

B, B

B, C

B, T

C, B

C, C

C, T

T, B

T, C

T, T

It is clearly seen that there are 9 transportation options.

(Using cab 1 time we have BC, CB, CT, TC. So four of the transportation methods use cab one time.)

Therefore, the probability that she will use a cab only once is 4/9.  

6 0
3 years ago
Pls help with 29..tank you
trapecia [35]

Answer:

See below

Step-by-step explanation:

Let the  2 roots be  A and A^2.

Then A^3 = c/a and A + A^2 = -b/a.

Using the identity a^3 b^3 = (a + b)^3 - 3ab^2 - 3a^2b:-

A^3 + (A^2)^3 =  ( A + A^2)^3 - 3 A.A^4 - 3 A^2. A^2

= (A + A^2)^3 - 3A*3( A + A^2)

Substituting:-

c/a + c^2/a^2 = (-b/a)^3 - 3 (c/a)(-b/a)

Multiply through by a^3:-

a^2c + ac^2 = -b^3 + 3abc

Factoring:-

ac(a + c) = 3abc - b^3

This is not the formula required in the question but let's see if the formula in the question reduces to this. If it does we have completed the proof.

a(c - b)^3 = a(c^3 - 3c^2b + 3cb^2 - b^3) = ac^3 - 3ac^2b + 3acb^2 - ab^3

c(a - b)^3 = c(a^3 - 3a^2b + 3ab^2 - b^3) = ca^3 - 3a^2bc + 3acb^2 - cb^3.

These are equal so we have

ca^3 - 3a^2bc + 3acb^2 - cb^3 = ac^3 - 3abc^2 + 3acb^2 - ab^3

The 3acb^2 cancel out so we have:-

a^3c - ac^3 =  3a^2bc - 3abc^2 + b^3c - ab^3

ac(a^2 - c^2) = 3abc( a - c) + b^3(c - a)

ac(a + c)(a -c) = 3abc(a - c) - b^3 (a - c)

Divide through by (a - c):-

ac(a + c) = 3abc - b^3 , which is the result we got earlier.

This completes the proof.

 

3 0
3 years ago
Read 2 more answers
1. The ratio of diamonds to stars was 3 to 11, and three times the number of diamonds was 2 less than the number of stars. How m
Dvinal [7]

Answer:

There were 3 diamonds.

Step-by-step explanation:

\frac{Diamonds(D)}{Stars(S)}=\frac{3}{11}

Therefore, 11D=3S

S=3D+2

Therefore,

\frac{11D}{3}=3D+2

11D=9D+6

2D=6

D=3

4 0
3 years ago
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