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arsen [322]
3 years ago
15

Which of the following must be present for a check to be legitimate? I. The recipient’s bank II. The date III. A memo or note a.

I and II b. II only c. III only d. I, II, and III
Mathematics
2 answers:
katrin [286]3 years ago
6 0
Its b .........................................
atroni [7]3 years ago
3 0

Answer:

Option b. II only is the right answer.

Step-by-step explanation:

The DATE should be present for a check to be legitimate.

A check's date is very important because any expiration time period is based on the date written on the check. When a date is written and its deposited, a countdown starts to cash the check. This does not depend on the date when the check was actually sent or received, it all depends on the date. If the person writes a wrong year  dated check , it cannot be cashed. Maximum six months prior checks are considered to be cashed.

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A bank account earns a fixed rate of interest compounded quarterly (every 3 month)
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Answer:

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Step-by-step explanation:

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Read 2 more answers
June has 30 days. A family of 3 used 3,960 gallons of water during the month of June. How many gallons of water did each person
lys-0071 [83]

Answer:

44 gallons

Step-by-step explanation:

3960/3=1320

Each person used 1320 in one month

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6 0
3 years ago
I NEED THIS DONE ASAPPPP<br> expand 4(m+2)<br><br> 4(m+2) =
77julia77 [94]

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3 0
3 years ago
If A+B+C=<img src="https://tex.z-dn.net/?f=%5Cpi" id="TexFormula1" title="\pi" alt="\pi" align="absmiddle" class="latex-formula"
seraphim [82]

Answer:

a + b + c = \pi \\  =  > c=  \pi - a - b \\  <  =  >  \tan(c)  =  \tan(\pi - a - b)  =  -\tan(a + b)

Step-by-step explanation:

we have:

\tan(a)  +  \tan(b)  +  \tan(c)  \\  =  \tan(a)  +  \tan(b)  -  \tan(a + b)  \\  =  \tan( a)  +  \tan(b)  -  \frac{ \tan(a) +  \tan(b)  }{1 -  \tan(a)  \tan(b) }  \\  =  \frac{ ( \tan(a) +  \tan(b)  ) \tan(a) \tan(b)  }{ \tan(a) \tan(b)  - 1 } (1)

we also have:

\tan(a)  \tan(b)  \tan(c)  \\  =  -  \tan(a)  \tan(b)  \tan(a + b)  \\  =  \frac{ -(\tan( a  )   + \tan(b) ) \tan(a)  \tan(b) }{1 -  \tan(a)  \tan(b) }  \\  =  \frac{( \tan(a)  +  \tan(b)) \tan(a)   \tan(b) }{ \tan(a) \tan(b)  - 1 } (2)

from (1)(2) => proven

5 0
3 years ago
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