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Elodia [21]
3 years ago
5

7. How does the recognition of revenue on account (accounts receivable) affect the income statement compared to its effect on th

e statement of cash flow?
Mathematics
1 answer:
snow_tiger [21]3 years ago
8 0

Answer:

Whenever there is a credit sale, revenue is recognized at the time of sale, and accounts receivables is created, and then revenue increases with the amount of sale, in income statement.

As accounts receivables is increased in indirect method such net increase in accounts receivables from previous year end is deducted under cash flow from operating activities,

Further, recognizing revenue by creating accounts receivable individually does not impact cash flow statement as no cash is affected until amount is collected from accounts receivables.

Thus on a net result, the effect shall be as follows:

Effect on income statement = Increase in income.

Effect on cash flow statement = No effect as does not involve cash

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$6x + $15 ≤ $70<br><br><br> I have no idea- xd
Delvig [45]

Answer:

x = 9

Step-by-step explanation:

6 × 9 = 54

54 + 15 = 69, which is less than 70.

3 0
3 years ago
a company manufactures and sells novelty Mugs. the manufacturing cost consist of a fixed cost of R8000 and a variable cost of R1
kvasek [131]
Ok so

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then
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How to determine the amount of zeroes of a function?
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4 years ago
What is the period and midline?
OverLord2011 [107]
\bf \qquad \qquad \qquad \qquad \textit{function transformations}&#10;\\ \quad \\&#10;% function transformations for trigonometric functions&#10;\begin{array}{rllll}&#10;% left side templates&#10;f(x)=&{{  A}}cos({{  B}}x+{{  C}})+{{  D}}&#10;\\ \quad \\&#10;&#10;\end{array}\qquad

\bf \begin{array}{llll}&#10;% right side info&#10;\bullet \textit{ stretches or shrinks}\\&#10;\quad \textit{horizontally by amplitude } |{{  A}}|\\&#10;\bullet \textit{ horizontal shift by }\frac{{{  C}}}{{{  B}}}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is negative, to the right}\\&#10;\qquad  if\ \frac{{{  C}}}{{{  B}}}\textit{ is positive, to the left}\\&#10;\bullet \textit{vertical shift by }{{  D}}\\&#10;\qquad if\ {{  D}}\textit{ is negative, downwards}\\&#10;\qquad if\ {{  D}}\textit{ is positive, upwards}\\&#10;&#10;&#10;\end{array}

\bf \begin{array}{llll}&#10;\bullet \textit{function period}\\&#10;\qquad \frac{2\pi }{{{  B}}}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\&#10;\qquad \frac{\pi }{{{  B}}}\ for\ tan(\theta),\ cot(\theta)&#10;\end{array}&#10;

so if you notice yours \bf \begin{array}{llll}&#10;3.2cos&\left( \frac{5}{3}\theta \right)+&6.1\\&#10;&\ \uparrow&\uparrow \\&#10;&B&D &#10;\end{array}

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now, with D = 6.1, that moves the midline  up vertically that much

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notice the picture below
the vertical shift by the D component, or 6.1, moved the midline to y = 6.1 :)

6 0
3 years ago
What was two. thirds and fifth twelve
Harrizon [31]
The answer is 1 and 1/12 because you have to have a common denomonator.
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3 years ago
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