Answer:
a)
.
b) 
Step-by-step explanation:
Given a function
, this function has the following gradient:
.
(a) find the gradient of f
We have that
. So


.
.

(b) find the directional derivative of f at (2, 4, 0) in the direction of v = i + 3j − k.
The directional derivate is the scalar product between the gradient at (2,4,0) and the unit vector of v.
We have that:

.

The vector is 
To use v as an unitary vector, we divide each component of v by the norm of v.

So

Now, we can calculate the scalar product that is the directional derivative.

97.8ft^2 but if you round it to the nearest 10 it will be 100ft^2
Answer:
I only know the first one....
4+2+1=7
49p/7= 7p
4units:2units:1units
4×7p= 28p
2×7p= 14p
1×7p= 7p
divided it is...
28p:14p:7p
Answer:
C. 27
Step-by-step explanation:
<span> first, write the equation of the parabola in the required form: </span>
<span>(y - k) = a·(x - h)² </span>
<span>Here, (h, k) is given as (-1, -16). </span>
<span>So you have: </span>
<span>(y + 16) = a · (x + 1)² </span>
<span>Unfortunately, a is not given. However, you do know one additional point on the parabola: (0, -15): </span>
<span>-15 + 16 = a· (0 + 1)² </span>
<span>.·. a = 1 </span>
<span>.·. the equation of the parabola in vertex form is </span>
<span>y + 16 = (x + 1)² </span>
<span>The x-intercepts are the values of x that make y = 0. So, let y = 0: </span>
<span>0 + 16 = (x + 1)² </span>
<span>16 = (x + 1)² </span>
<span>We are trying to solve for x, so take the square root of both sides - but be CAREFUL! </span>
<span>± 4 = x + 1 ...... remember both the positive and negative roots of 16...... </span>
<span>Solving for x: </span>
<span>x = -1 + 4, x = -1 - 4 </span>
<span>x = 3, x = -5. </span>
<span>Or, if you prefer, (3, 0), (-5, 0). </span>