Mylo O'Moore
2021-05-01
Answered

Determine whether the function given by the table is linear, exponential, or neither. If the function is linear, find a linear function that models the data. If it is exponential, find an exponential function that models the data.
x f(x)
-1 8/7
0 8
1 56
2 392

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lobeflepnoumni

Answered 2021-05-02
Author has **99** answers

The function is linear if there is a common difference between the y-values for a constant increase in the z-values while the function is exponential if there is a common ratio between the y-values for a constant increase in the z-values.

Since there is a common ratio of 7 between the y-values for « constant increase of 1 in the x-values, then the function is exponential.

Use the exponential model$y=a{b}^{x}$

where a is the initial value (where x= 0) and & is the common ratio. Substitute a = 8 and b= 7 so:$y=8{\left(7\right)}^{x}$

or

$f\left(x\right)=8{\left(7\right)}^{x}$

Since there is a common ratio of 7 between the y-values for « constant increase of 1 in the x-values, then the function is exponential.

Use the exponential model

where a is the initial value (where x= 0) and & is the common ratio. Substitute a = 8 and b= 7 so:

or

asked 2021-02-25

The population of California was 29.76 million in 1990 and 33.87 million in 2000. Assume that the population grows exponentially.

(a) Find a function that models the population t years after 1990.

(b) Find the time required for the population to double.

(c) Use the function from part (a) to predict the population of California in the year 2010. Look up California’s actual population in 2010, and compare.

(a) Find a function that models the population t years after 1990.

(b) Find the time required for the population to double.

(c) Use the function from part (a) to predict the population of California in the year 2010. Look up California’s actual population in 2010, and compare.

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The exponential models describe the population of the indicated country, A, in millions, t years after 2010.Which country has the greatest growth rate? By what percentage is the population of that country increasing each year?

India,

Iraq,

Japan,

Russia,

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How do you write y=3x-8 in standard form?

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logarithmic derivative of ${x}^{{e}^{({x}^{2}+\mathrm{cos}x)}}$

I'm having a hard time taking the derivative of

$f(x)={x}^{{e}^{({x}^{2}+cosx)}}.$

I'm aware that I have to take the logarithm of both sides.

$\mathrm{ln}(y)=\mathrm{ln}({x}^{{e}^{({x}^{2}+\mathrm{cos}x)}})=\mathrm{ln}(x)\cdot {e}^{({x}^{2}+\mathrm{cos}x)}$

Which I tried to untie, so lets start: First I use the product rule:

$\frac{1}{y}\cdot {y}^{\prime}=\frac{1}{x}\cdot {e}^{({x}^{2}+\mathrm{cos}x)}+\mathrm{ln}(x)\cdot {e}^{({x}^{2}+\mathrm{cos}x)}$

Next the power rule:

$\frac{1}{y}\cdot {y}^{\prime}=\frac{1}{x}\cdot {e}^{({x}^{2}+\mathrm{cos}x)}+\mathrm{ln}(x)\ast {e}^{({x}^{2}+cosx)}\cdot {x}^{2}+\mathrm{cos}x\cdot {e}^{({x}^{2}+\mathrm{cos}x)-1}$

Then I bring the y to the right:

${y}^{\prime}=y\cdot (\frac{1}{x}\cdot {e}^{({x}^{2}+\mathrm{cos}x)}+\mathrm{ln}(x)\cdot {e}^{({x}^{2}+\mathrm{cos}x)}\cdot {x}^{2}+\mathrm{cos}x\cdot {e}^{({x}^{2}+\mathrm{cos}x)-1}\cdot 2x-\mathrm{sin}x)$

And exchange y with the term:

${y}^{\prime}={x}^{{e}^{({x}^{2}+\mathrm{cos}x)}}\cdot (\frac{1}{x}\cdot {e}^{({x}^{2}+\mathrm{cos}x)}+\mathrm{ln}(x)\cdot {e}^{({x}^{2}+\mathrm{cos}x)}\cdot {x}^{2}+\mathrm{cos}x\cdot {e}^{({x}^{2}+\mathrm{cos}x)-1}\cdot 2x-\mathrm{sin}x)$

This is extremely overwhelming for me and I have absolutely no clue if this is right, I looked in to the result of wolfram and It doesn't seem to be correct. Any help would be appreciated.

I'm having a hard time taking the derivative of

$f(x)={x}^{{e}^{({x}^{2}+cosx)}}.$

I'm aware that I have to take the logarithm of both sides.

$\mathrm{ln}(y)=\mathrm{ln}({x}^{{e}^{({x}^{2}+\mathrm{cos}x)}})=\mathrm{ln}(x)\cdot {e}^{({x}^{2}+\mathrm{cos}x)}$

Which I tried to untie, so lets start: First I use the product rule:

$\frac{1}{y}\cdot {y}^{\prime}=\frac{1}{x}\cdot {e}^{({x}^{2}+\mathrm{cos}x)}+\mathrm{ln}(x)\cdot {e}^{({x}^{2}+\mathrm{cos}x)}$

Next the power rule:

$\frac{1}{y}\cdot {y}^{\prime}=\frac{1}{x}\cdot {e}^{({x}^{2}+\mathrm{cos}x)}+\mathrm{ln}(x)\ast {e}^{({x}^{2}+cosx)}\cdot {x}^{2}+\mathrm{cos}x\cdot {e}^{({x}^{2}+\mathrm{cos}x)-1}$

Then I bring the y to the right:

${y}^{\prime}=y\cdot (\frac{1}{x}\cdot {e}^{({x}^{2}+\mathrm{cos}x)}+\mathrm{ln}(x)\cdot {e}^{({x}^{2}+\mathrm{cos}x)}\cdot {x}^{2}+\mathrm{cos}x\cdot {e}^{({x}^{2}+\mathrm{cos}x)-1}\cdot 2x-\mathrm{sin}x)$

And exchange y with the term:

${y}^{\prime}={x}^{{e}^{({x}^{2}+\mathrm{cos}x)}}\cdot (\frac{1}{x}\cdot {e}^{({x}^{2}+\mathrm{cos}x)}+\mathrm{ln}(x)\cdot {e}^{({x}^{2}+\mathrm{cos}x)}\cdot {x}^{2}+\mathrm{cos}x\cdot {e}^{({x}^{2}+\mathrm{cos}x)-1}\cdot 2x-\mathrm{sin}x)$

This is extremely overwhelming for me and I have absolutely no clue if this is right, I looked in to the result of wolfram and It doesn't seem to be correct. Any help would be appreciated.

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How do you multiply 4(3x+1)(x-2)?

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Determine whether the sequence is arithmetic. If so, identify the common difference 11, 10.2, 9.4, 8.6, 7.8, …

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logarithms equations,

Hello I have following problem: solve equation $\mathrm{log}(x-5{)}^{2}+\mathrm{log}(x+6{)}^{2}=2$and I rewrited this equation as

$2\mathrm{log}(x-5)+2\mathrm{log}(x+6)=\mathrm{log}100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2(\mathrm{log}(x-5)(x+6))=\mathrm{log}100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{log}{x}^{2}+x-30=\mathrm{log}10\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{2}+x-40=0$

and I solved this equation, but I obtained only two solutions and there should be four, so I wonder if it is necessary to create $\mathrm{log}(x-5{)}^{2}(x+6{)}^{2}=\mathrm{log}100$ or is a simplier way.

Hello I have following problem: solve equation $\mathrm{log}(x-5{)}^{2}+\mathrm{log}(x+6{)}^{2}=2$and I rewrited this equation as

$2\mathrm{log}(x-5)+2\mathrm{log}(x+6)=\mathrm{log}100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}2(\mathrm{log}(x-5)(x+6))=\mathrm{log}100\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}\mathrm{log}{x}^{2}+x-30=\mathrm{log}10\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}{x}^{2}+x-40=0$

and I solved this equation, but I obtained only two solutions and there should be four, so I wonder if it is necessary to create $\mathrm{log}(x-5{)}^{2}(x+6{)}^{2}=\mathrm{log}100$ or is a simplier way.