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gogolik [260]
2 years ago
15

Which is a short-term consequence of making a late payment on your bill? it will be harder to buy a house because it affects you

r credit history. there will be a late fee added to the bill. it will be harder to secure a new loan at a low rate. the bank will charge a fee for having a negative balance.
Mathematics
2 answers:
iris [78.8K]2 years ago
3 0

There will be a late fee added to the bill as a short-term consequence of making a late payment on your bill. Option B is correct.

<h3>What is the late payment?</h3>

A late payment is a payment sent to a lender or service provider after the due date or after the grace period for the payment has expired.

The consequence of making a late payment on your bill is;

A-it will be harder to buy a house because it affects your credit history.

B-here will be a late fee added to the bill

C-it will be harder to secure a new loan at a low rate

D- The bank will charge a fee for having a negative balance

All the given options are the consequence of making a late payment on your bill. But the options A, C, and D are the long-term consequences. While option B is a long-term consequence.

There will be a late fee added to the bill as a short-term consequence of making a late payment on your bill.

Hence option B is correct.

To learn more about the late payment refer to the link;

brainly.com/question/23509549

guapka [62]2 years ago
3 0

B - there will be a late fee added to the bill

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Vinil7 [7]

Answer:

The factors of  2(x+y)^2-9(x+y)-5 is ((x+y)-5)(2x+2y+1)

Step-by-step explanation:

Given polynomial

=>2(x+y)^2-9(x+y)-5

To Find:

The factors of the polynomial =?

Solution:

Lets assume  k = (x+y)

Then 2(x+y)^2-9(x+y)-5 can be written as 2k^2-9k-5

Now by using quadratic formula

k =\frac{-b\pm\sqrt{(b^2-4ac}}{2a}

where

a= 2

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c= -5

Substituting the values, we get

k =\frac{-b\pm\sqrt{(b^2-4ac)}}{2a}

k =\frac{-(-9) \pm \sqrt{((-9)^2-4(2)(-5)}}{2(2))}

k =\frac{-(-9) \pm \sqrt{(81+40)}}{4}

k =\frac{-(-9) \pm \sqrt{(121)}}{4}

k =\frac{-(-9) \pm 11}}{4}

k= \frac{ 9 \pm 11}{4}

k =  \frac{20}{4}                         k =  \frac{-2}{4}    

k_1 =5                                      k_2 = -\frac{1}{2}

2k^2-9k-5= 2(k-5)(k+\frac{1}{2})

Solving the RHS we get

\frac{2}{2}(k-5)(2k+1)

(k-5)(2k+1)

Substituting k = x+y

((x+y)-5)(2(x+y+1)

((x+y)-5)(2x+2y+1)

5 0
3 years ago
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