Plug in the values of x given in the domain:-
x + 5 = -4 + 5 = 1
-2 + 5 = 3 and so on
Are you sure you have the question right/ Because none of the choices are correct.

8 0

3 years ago

What’s 13.1 written as a fraction

MAXImum [283]

Answer:1 is the numerator 13 is the denominator

Step-by-step explanation:

8 0

4 years ago

Read 2 more answers

Let f be a function of two variables that has continuous partial derivatives and consider the points

Sergeu [11.5K]

Answer:

The directional derivative of f at A in the direction of $\vec{u}$ AD is 7.

Step-by-step explanation:

Step 1:

Directional of a function f in direction of the unit vector $\vec{u}=(a,b)$ is denoted by $D\vec{u}f(x,y)$ ,

$D\vec{u}f(x,y)=f_{x}\left ( x ,y\right ).a+f_{y}(x,y).b$ .

Now the given points are

$A(8,9),B(10,9),C(8,10) and D(11,13)$ ,

Step 2:

The vectors are given as

AB = (10-8, 9-9),the direction is

$\vec{u}_{AB} = \frac{AB}{\left \| AB \right \|}=(1,0)$

AC=(8-8,10-9), the direction is

$\vec{u}_{AC} = \frac{AC}{\left \| AC \right \|}=(0,1)$

AC=(11-8,13-9), the direction is

$\vec{u}_{AD} = \frac{AD}{\left \| AD \right \|}=\left (\frac{3}{5},\frac{4}{5} \right )$

Step 3:

The given directional derivative of f at A $\vec{u}_{AB}$ is 9,

$D\vec{u}_{AB}f=f_{x} \cdot 1 + f_{y}\cdot 0\\f_{x} =9$

The given directional derivative of f at A $\vec{u}_{AC}$ is 2,

$D\vec{u}_{AB}f=f_{x} \cdot 0 + f_{y}\cdot 1\\f_{y} =2$

The given directional derivative of f at A $\vec{u}_{AD}$ is

$D\vec{u}_{AD}f=f_{x} \cdot \frac{3}{5} + f_{y}\cdot \frac{4}{5}$

$D\vec{u}_{AD}f=9 \cdot \frac{3}{5} + 2\cdot \frac{4}{5}$

$D\vec{u}_{AD}f= \frac{27+8}{5} =7$

The directional derivative of f at A in the direction of $\vec{u}_{AD}$ is 7.

3 0

3 years ago

2b) A farmer is building a pen inside a barn. The pen will be in the shape of a right triangle.

kenny6666 [7]

Answer:

Step-by-step explanation:

the sum of any two sides of a right triangle more than any other side

14+15>29, x is less than 29

14+x>15

x>1

15+x>14

x>-1

1<x<29

lets use Pythagorean Theorem

14^(2)+15^(2)=c^2

196+225=c^2

421=c^2

plusminus sqrt(421)=c

distance can't be negative, so:

c=sqrt(421)

c=20.5182845287

sqrt(421) is the largest possible right triangle side

1<x $\leq$ sqrt(421)

B) as we can see from above, the largest possible length of the third side is 14/sqrt(421) or about 20.5182845287ft

C)sqrt(421) by 14 by 15

tan(x1)=15/14

x1=tan^-1(15/14)

x1=47 degrees

tan(x2)=14/15

x2=tan^-1(14/15)

x2=43 degrees

7 0

3 years ago