Answer:
i would say C
Step-by-step explanation:
9514 1404 393
Answer:
see attached
Step-by-step explanation:
Most of this exercise is looking at different ways to identify the slope of the line. The first attachment shows the corresponding "run" (horizontal change) and "rise" (vertical change) between the marked points.
In your diagram, these values (run=1, rise=-3) are filled in 3 places. At the top, the changes are described in words. On the left, they are described as "rise" and "run" with numbers. At the bottom left, these same numbers are described by ∆y and ∆x.
The calculation at the right shows the differences between y (numerator) and x (denominator) coordinates. This is how you compute the slope from the coordinates of two points.
If you draw a line through the two points, you find it intersects the y-axis at y=4. This is the y-intercept that gets filled in at the bottom. (The y-intercept here is 1 left and 3 up from the point (1, 1).)
2.8.1

By definition of the derivative,

We have

and

Combine these fractions into one with a common denominator:

Rationalize the numerator by multiplying uniformly by the conjugate of the numerator, and simplify the result:

Now divide this by <em>h</em> and take the limit as <em>h</em> approaches 0 :

3.1.1.
![f(x) = 4x^5 - \dfrac1{4x^2} + \sqrt[3]{x} - \pi^2 + 10e^3](https://tex.z-dn.net/?f=f%28x%29%20%3D%204x%5E5%20-%20%5Cdfrac1%7B4x%5E2%7D%20%2B%20%5Csqrt%5B3%5D%7Bx%7D%20-%20%5Cpi%5E2%20%2B%2010e%5E3)
Differentiate one term at a time:
• power rule


![\left(\sqrt[3]{x}\right)' = \left(x^{1/3}\right)' = \dfrac13 x^{-2/3} = \dfrac1{3x^{2/3}}](https://tex.z-dn.net/?f=%5Cleft%28%5Csqrt%5B3%5D%7Bx%7D%5Cright%29%27%20%3D%20%5Cleft%28x%5E%7B1%2F3%7D%5Cright%29%27%20%3D%20%5Cdfrac13%20x%5E%7B-2%2F3%7D%20%3D%20%5Cdfrac1%7B3x%5E%7B2%2F3%7D%7D)
The last two terms are constant, so their derivatives are both zero.
So you end up with

Answer:
Multiply the base by the exponent
Step-by-step explanation:
This ratio in fraction form is 10/9