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3 years ago

What are the domain and range of the logarithmic function f(x) = log7x? Use the inverse function to justify your answers.

Mathematics

2 answers:

Pepsi [2]3 years ago

8 0

Answer:

The domain of f is x > 0.

The range of f is all real numbers.

The domain of the inverse function is all real numbers.

The range of the inverse function is y > 0.

The domain of f is the same as the

range of the inverse function.

The range of f is the same as the

domain of the inverse function.

Fofino [41]3 years ago

5 0

Answer:

The domain of f is x > 0.

The range of f is all real numbers.

and

The domain of f is the same as the range of the inverse function.

The range of f is the same as the domain of the inverse function.

Step-by-step explanation:

All log functions without any additional terms added to it outside the operation have domain x>0 and all real number range.

When the inverse is found, (x,y) are switched to (y,x). This measn the range becomes the domain the domain becomes the range.

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1161.90 to the nearest 100

chubhunter [2.5K]

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4 years ago

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Kisachek [45]

Answer:

15h + 8

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Step-by-step explanation:

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3 years ago

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find the volume of the solid formed by revolving the region bounded by the graphs of y = 4x - x^2 and f(x) = x^2 from [0,2] abou

Neko [114]

Answer:

$v = \frac{32\pi}{3}$

$v=33.52$

Step-by-step explanation:

Given

$f(x) = 4x - x^2$

$g(x) = x^2$

$[a,b] = [0,2]$

Required

The volume of the solid formed

Rotating about the x-axis.

Using the washer method to calculate the volume, we have:

$\int dv = \int\limit^b_a \pi(f(x)^2 - g(x)^2) dx$

Integrate

$v = \int\limit^b_a \pi(f(x)^2 - g(x)^2)\ dx$

$v = \pi \int\limit^b_a (f(x)^2 - g(x)^2)\ dx$

Substitute values for a, b, f(x) and g(x)

$v = \pi \int\limit^2_0 ((4x - x^2)^2 - (x^2)^2)\ dx$

Evaluate the exponents

$v = \pi \int\limit^2_0 (16x^2 - 4x^3 - 4x^3 + x^4 - x^4)\ dx$

Simplify like terms

$v = \pi \int\limit^2_0 (16x^2 - 8x^3 )\ dx$

Factor out 8

$v = 8\pi \int\limit^2_0 (2x^2 - x^3 )\ dx$

Integrate

$v = 8\pi [ \frac{2x^{2+1}}{2+1} - \frac{x^{3+1}}{3+1} ]|\limit^2_0$

$v = 8\pi [ \frac{2x^{3}}{3} - \frac{x^{4}}{4} ]|\limit^2_0$

Substitute 2 and 0 for x, respectively

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ \frac{2*0^{3}}{3} - \frac{0^{4}}{4} ])$

$v = 8\pi ([ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ] - [ 0 - 0])$

$v = 8\pi [ \frac{2*2^{3}}{3} - \frac{2^{4}}{4} ]$

$v = 8\pi [ \frac{16}{3} - \frac{16}{4} ]$

Take LCM

$v = 8\pi [ \frac{16*4- 16 * 3}{12}]$

$v = 8\pi [ \frac{64- 48}{12}]$

$v = 8\pi * \frac{16}{12}$

Simplify

$v = 8\pi * \frac{4}{3}$

$v = \frac{32\pi}{3}$

$v=\frac{32}{3} * \frac{22}{7}$

$v=\frac{32*22}{3*7}$

$v=\frac{704}{21}$

$v=33.52$

8 0

3 years ago

on a piece of paper,graph c(x)=3X+2.00.Then determine which answer matches the graph you drew,including the correct axis labels.

Setler [38]

Answer:

Attached

Step-by-step explanation:

The equation for the graph is

c(x)=3x+2

rewrite as

y=3x+2.............................................(1)

Then graph equation to visualize as attached.

Let the x-axis to represent minutes and the y-axis to represent cost

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3 years ago

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LuckyWell [14K]

The Answer Is
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3 years ago