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harina [27]
3 years ago
10

The number has 6 digits.

Mathematics
1 answer:
lawyer [7]3 years ago
3 0
<span>It has 6 digits, and smallest number and biggest number is given.
Smallest digit= 0.01 or the tenths =1
Largest digit is 8000 = thousands
tens value = hundreds – 2 = X
hundreds = tens + Hundreds = 8 (thousands value)
ones = largest even digit = 8
tenths= the same value with every place
hundredths = 1

Thus,
Thousands = 8
hundreds = 3 + 5 = 8 hundreds = 5
tens = 5 – 2  = 3
ones = 8
.
tenths = 3
Hundredths = 1

Therefore the mystery number is 8538.31</span>



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Step-by-step explanation:

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Formula for 4,9,14 19​
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Answer:

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a_{n} = a₁ + (n - 1)d

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a_{n} = 4 + 5(n - 1) = 4 + 5n - 5 = 5n - 1

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The graph of an exponential function is given. Which of the following is the correct equation of the function?
katen-ka-za [31]

Answer:

If one of the data points has the form  

(

0

,

a

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, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

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)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

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EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).If one of the data points has the form  

(

0

,

a

)

, then a is the initial value. Using a, substitute the second point into the equation  

f

(

x

)

=

a

(

b

)

x

, and solve for b.

If neither of the data points have the form  

(

0

,

a

)

, substitute both points into two equations with the form  

f

(

x

)

=

a

(

b

)

x

. Solve the resulting system of two equations in two unknowns to find a and b.

Using the a and b found in the steps above, write the exponential function in the form  

f

(

x

)

=

a

(

b

)

x

.

EXAMPLE 3: WRITING AN EXPONENTIAL MODEL WHEN THE INITIAL VALUE IS KNOWN

In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.

SOLUTION

We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation  

N

(

t

)

=

80

b

t

to find b:

⎧

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎨

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎪

⎩

N

(

t

)

=

80

b

t

180

=

80

b

6

Substitute using point  

(

6

,

180

)

.

9

4

=

b

6

Divide and write in lowest terms

.

b

=

(

9

4

)

1

6

Isolate  

b

using properties of exponents

.

b

≈

1.1447

Round to 4 decimal places

.

NOTE: Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.

The exponential model for the population of deer is  

N

(

t

)

=

80

(

1.1447

)

t

. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)

We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem,  

(

0

,

8

0

)

and  

(

6

,

18

0

)

. We can also see that the domain for the function is  

[

0

,

∞

)

, and the range for the function is  

[

80

,

∞

)

.

Graph of the exponential function, N(t) = 80(1.1447)^t, with labeled points at (0, 80) and (6, 180).

Step-by-step explanation:

4 0
2 years ago
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noname [10]
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6 0
2 years ago
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Answer:

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