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Zolol [24]
2 years ago
15

What are some similarities and differences of the rates of change of linear, quadratic, and exponential functions?

Mathematics
1 answer:
gregori [183]2 years ago
5 0

Answer:

linear functions change a constant rate. i.e. the slope of the graph is constant.

d/dx (mx+b) = m


quadratic functions change at a linear rate, slower than the function itself.

The slope of the tangent to the graph is proportional to the x coordinate.

d/dx (ax^2) = 2ax

Slope of tangent to ax^2 at point (x,ax^2) is 2ax.


Exponential functions change at the same rate as the function itself. The slope of the tangent to the graph is proportional to the y coordinate.


The slope of tangent at point (x,exp(x)) is exp(x).


Definition of exp(x) is just exp(0) = 1 and d/dx exp(x) = exp(x).


d/dx exp(ax) = a exp(ax) is not part of the definition, but follows from from the chain rule


Step-by-step explanation:


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valina [46]

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Answer:

  x-intercept: 7/6

Step-by-step explanation:

The x-intercept is found by solving for x when y=0.

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  7/6 = x . . . . divide both sides by 6

The x-intercept is 7/6 = 1 1/6 ≈ 1.167.

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The y-intercept is the constant in the given equation: -7. The slope of 6 tells you the line rises 6 units for 1 unit to the right. So, the point (1, -1) is another point on the graph. The line through those two points, (0, -7) and (1, -1), will intersect the x-axis at x=7/6.

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3 years ago
Write an equation in standard form of the line that contains the point (-1,2) amd is perpendicular to y = 3x - 1
Strike441 [17]
Since it is perpendicular to y=3x-1, the slope must be a negative reciprocal.
there the slope of the perpendicular line is -\frac{1}{3} x

So now we have, 
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To find b, we use the point that the line passes through and substitute it to the y and x values.

2=- \frac{1}{3} (-1)+b

which would give us b=5/3

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Gina puts $ 4500 into an account earning 7.5% interest compounded continuously. How long will it take for the amount in the acco
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5150=4500e^{0.075\cdot t} \implies \cfrac{5150}{4500}=e^{0.075t}\implies \cfrac{103}{90}=e^{0.075t} \\\\\\ \log_e\left( \cfrac{103}{90} \right)=\log_e(e^{0.075t})\implies \log_e\left( \cfrac{103}{90} \right)=0.075t \\\\\\ \ln\left( \cfrac{103}{90} \right)=0.075t\implies \cfrac{\ln\left( \frac{103}{90} \right)}{0.075}=t\implies\stackrel{\textit{about 1 year and 291 days}}{ 1.8\approx t}

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1 year ago
Perform the operation 8-1/2+5-1/8
IrinaVladis [17]

Answer:

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Answer: 9/2π

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