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expeople1 [14]
3 years ago
8

What is the decay factor of the exponential function represented by the table?

Mathematics
1 answer:
bulgar [2K]3 years ago
8 0

Given:

The table of values of an exponential function.

To find:

The decay factor of the exponential function.

Solution:

The general form of an exponential function is:

y=ab^x              ...(i)

Where, a is the initial value and 0 is the decay factor and b>1 is the growth factor.

The exponential function passes through the point (0,6). Substituting x=0,y=6 in (i), we get

6=ab^0

6=a(1)

6=a

The exponential function passes through the point (1,2). Substituting x=1,y=2,a=6 in (i), we get

2=6(b)^1

2=6b

\dfrac{2}{6}=b

\dfrac{1}{3}=b

Here, b=\dfrac{1}{3} lies between 0 and 1. Therefore, the decay factor of the given exponential function is \dfrac{1}{3}.

Hence, the correct option is A.

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PLSS HELP
Dmitry [639]

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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3 years ago
Solve k−7≤0. Graph the solution
leonid [27]

Answer:

k ≤ 7

Step-by-step explanation:

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3 years ago
Zoe compares the price of a sweatshirt at two stores. At Store A, the sweatshirt costs $25. A friend told her that the sweatshir
Rudiy27

Answer:

<u>A) (1+0.3)25</u>

<u />

Explanation:

Zoe's expression <u>25+0.3(25)</u> equals to 32.5. So the only expression that equals to 32.5 is <u>A) (1+0.3)25</u>.

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iogann1982 [59]

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