Answer:
3
Step-by-step explanation:no
Answer:
D.
Step-by-step explanation:
Remember that the limit definition of a derivative at a point is:
![\displaystyle{\frac{d}{dx}[f(a)]= \lim_{x \to a}\frac{f(x)-f(a)}{x-a}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%28a%29%5D%3D%20%5Clim_%7Bx%20%5Cto%20a%7D%5Cfrac%7Bf%28x%29-f%28a%29%7D%7Bx-a%7D%7D)
Hence, if we let f(x) be ln(x+1) and a be 1, this will yield:
![\displaystyle{\frac{d}{dx}[f(1)]= \lim_{x \to 1}\frac{\ln(x+1)-\ln(2)}{x-1}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%7B%5Cfrac%7Bd%7D%7Bdx%7D%5Bf%281%29%5D%3D%20%5Clim_%7Bx%20%5Cto%201%7D%5Cfrac%7B%5Cln%28x%2B1%29-%5Cln%282%29%7D%7Bx-1%7D%7D)
Hence, the limit is equivalent to the derivative of f(x) at x=1, or f’(1).
The answer will thus be D.
Answer:
(4,5)
Step-by-step explanation:
The "feasible region" has vertices (0,0), (7,0), (5,4), and (4,5)
P = 5x + 6y
Plug in each vertices in P and find out which give maximum value
(0,0) => P= 5(0) + 6(0) = 0
(7,0) => P= 5(7) + 6(0) = 35
(5,4) => P= 5(5) + 6(4) = 49
(4,5) => P= 5(4) + 6(5) = 50
We got maximum P=50 for vertex (4,5)
So the coordinates of the point that has the maximum value is (4,5)
2.3, 2.6, 2 4/5
2 4/5= 2.8
A quadratic equation has the general form
of: <span>
y=ax² + bx + c
It can be converted to the vertex form in order
to determine the vertex of the parabola. It has the standard form of:
y = a(x+h)² - k
This can be done by completing a square. The steps are as follows:
</span><span>y = 3x2 + 9x – 18
</span>y = 3(x2 <span>+ 3x) – 18
</span>y + 27/4= 3(x2 <span>+ 3x+ 9/4) – 18
</span>y = 3(x2 + 3/2)^2 – 99<span>/4
</span>
Therefore, the first step is to group terms with the variable x and factoring out the coefficient of x^2.
