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pishuonlain [190]
2 years ago
13

PLSS HELP

Mathematics
1 answer:
Dmitry [639]2 years ago
3 0

Answer:

$8,000

Step-by-step explanation:

Let the store earned $x in December.

Therefore,

Money spent to buy new inventory =\frac{1}{4} x

Remaining money = x - \frac{1}{4} x =\frac{3}{4} x

Money used to pay bills =\frac{1}{2} \times \frac{3}{4} x=\frac{3}{8} x

Money still left over = $3,000

Total money earned in December = \frac{1}{4} x+ \frac{3}{8} x+3,000

\therefore x= \frac{1}{4} x+ \frac{3}{8} x+3,000

\therefore x= \frac{2}{8} x+ \frac{3}{8} x+3,000

\therefore x= \frac{5}{8} x+ 3,000

\therefore x- \frac{5}{8} x=3,000

\therefore \frac{8x-5x}{8} =3,000

\therefore \frac{3x}{8} =3,000

\therefore x =3,000\times \frac {8}{3}

\therefore x =1,000\times 8

\therefore x =\$8,000

Thus, total money earned in December is $8,000.

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The numbers of teams remaining in each round of a single-elimination tennis tournament represent a geometric sequence where an i
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Answer:

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

Step-by-step explanation:

We are given the following in the question:

The numbers of teams remaining in each round follows a geometric sequence.

Let a be the first the of the geometric sequence and r be the common ration.

The n^{th} term of geometric sequence is given by:

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the first term can be calculated as:

16=a(\dfrac{1}{2})^3\\\\a = 16\times 6\\a = 128

Thus, the required geometric sequence is

a_n = 128\bigg(\dfrac{1}{2}\bigg)^{n-1}

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