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Maru [420]
3 years ago
13

Which of these debts could possibly be forgiven under Chapter 7 bankruptcy? A.Child support. B.A student loan. C.Alimony. D.A mo

rtgage
Mathematics
2 answers:
Leto [7]3 years ago
8 0
The Debt that could be forgiven under bankruptcy is: D. mortgage A mortgage is a type of debt that is used to purchase a property. In case of bankruptcy, the debt could easily be forgiven because the owner of the property could easily take it over again and they basically loss nothing

Mkey [24]3 years ago
4 0

For APEX it's credit card :D

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A desk is purchased for $277. Calculate its worth after depreciating for 5 years if it’s depreciation rate is 20% per year.
Over [174]
20% x 5 = 100%
Depreciation means decreased of a value
8 0
4 years ago
Given the sequence 5,1,3 which term of sequence is -75<br>​
dimaraw [331]
Probably maybeeeeeeee
3 0
4 years ago
DESCRIBE HOW TO WRITE 3 DIGIT NUMBERS IN THREE DIFFERENT WAYS
victus00 [196]
Just put each number in different positions.
425
542
254
All of these numbers are different and use the same 3 digits
3 0
3 years ago
- Compare. Write &lt;, &gt;, or=.<br> -1/2 ___1.25<br> 0___-2.25<br> -1.5___-0.25
Phoenix [80]

Answer:

1.25 is greatrer

0 is greater

/025 is greater

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
Given AG bisects CD, IJ bisects CE, and BH bisects ED. Prove KE = FD.
OverLord2011 [107]

When a line is bisected, the line is divided into equal halves.

See below for the proof of \mathbf{KE \cong FD}

The given parameters are:

  • <em>AC bisects CD</em>
  • <em>IJ bisects CE</em>
  • <em>BH bisects ED</em>

<em />

By definition of segment bisection, we have:

  • \mathbf{CK \cong KE}
  • \mathbf{EF \cong FD}
  • \mathbf{CE \cong ED}

By definition of congruent segments, the above congruence equations become:

  • \mathbf{CK = KE}
  • \mathbf{EF = FD}
  • \mathbf{CE = ED}

By segment addition postulate, we have:

  • \mathbf{CE = CK + KE}
  • \mathbf{ED = EF + FD}

Substitute \mathbf{ED = EF + FD} in \mathbf{CE = ED}

\mathbf{CK + KE = EF + FD}

Substitute \mathbf{CK = KE} and \mathbf{EF = FD}

\mathbf{KE + KE = FD + FD}

Simplify

\mathbf{2KE = 2FD}

Apply division property of equality

\mathbf{KE = FD}

By definition of congruent segments

\mathbf{KE \cong FD}

Read more about proofs of congruent segments at:

brainly.com/question/11494126

6 0
3 years ago
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