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fenix001 [56]
2 years ago
12

Which of the following exponential functions represents the graph below?

Mathematics
1 answer:
mario62 [17]2 years ago
3 0

Answer:

the graph corresponds to function "D"  f(x)=3\,(\frac{1}{5} )^x

Step-by-step explanation:

Since the graph shown corresponds to an exponential "decay" (the function decreases as we move from left to right), the base of the exponent has to be a number smaller than 1 (one). So we examine the only two options that give such (options C and D which have fractions as the base - 1/3 and 1/5 respectively)

From there, we analyze which of the two functions satisfies the crossing of the y-axis at (0,3) which is clearly shown in the graph:

We study both:

function C at x = 0 gives:

f(x)= 5\,(\frac{1}{3})^x\\f(0)=5\,(\frac{1}{3})^0 \\f(0)=5\,(1)\\f(0)=5

while function D at x = 0 gives:

f(x)= 3\,(\frac{1}{5})^x\\f(0)=3\,(\frac{1}{5})^0 \\f(0)=3\,(1)\\f(0)=3

Therefore, the graph corresponds to function "D"

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8 0
1 year ago
A worker is paid rs. 2130 for 6 days .if his total wage during a month is rs. 9230 find the number of days he worked in the mont
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Answer:

26 days

Step-by-step explanation:

If 6 days = rs. 2130

then 1 day = (2130 × 1) ÷ 6

Therefore 1 day = rs 355

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= 9230 ÷ 355

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2 years ago
The table contains the proof of the theorem of the relationship between slopes of parallel lines. What is the missing statement
Kipish [7]

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4 0
2 years ago
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Eli invested $ 330 $330 in an account in the year 1999, and the value has been growing exponentially at a constant rate. The val
Cloud [144]

Answer:

The value of the account in the year 2009 will be $682.

Step-by-step explanation:

The acount's balance, in t years after 1999, can be modeled by the following equation.

A(t) = Pe^{rt}

In which A(t) is the amount after t years, P is the initial money deposited, and r is the rate of interest.

$330 in an account in the year 1999

This means that P = 330

$590 in the year 2007

2007 is 8 years after 1999, so P(8) = 590.

We use this to find r.

A(t) = Pe^{rt}

590 = 330e^{8r}

e^{8r} = \frac{590}{330}

e^{8r} = 1.79

Applying ln to both sides:

\ln{e^{8r}} = \ln{1.79}

8r = \ln{1.79}

r = \frac{\ln{1.79}}{8}

r = 0.0726

Determine the value of the account, to the nearest dollar, in the year 2009.

2009 is 10 years after 1999, so this is A(10).

A(t) = 330e^{0.0726t}

A(10) = 330e^{0.0726*10} = 682

The value of the account in the year 2009 will be $682.

4 0
2 years ago
What is 6.37 × 104 written in standard form?   A. 63,700   B. 637,000   C. 6370   D. 637
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In short, Your Answer would be Option A

Hope this helps!
8 0
2 years ago
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