The tables represent the functions f(x) and g(x).

PLEASE HELP WITH THESE 4 QUESTIONS IF CORRECT ILL MARK BRAINLIEST

Serjik [45]

Answer:

the first box maybe

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Use lagrange multipliers to find the shortest distance, d, from the point (4, 0, −5 to the plane x y z = 1

Varvara68 [4.7K]

I assume there are some plus signs that aren't rendering for some reason, so that the plane should be $x+y+z=1$ .

You're minimizing $d(x,y,z)=\sqrt{(x-4)^2+y^2+(z+5)^2}$ subject to the constraint $f(x,y,z)=x+y+z=1$ . Note that $d(x,y,z)$ and $d(x,y,z)^2$ attain their extrema at the same values of $x,y,z$ , so we'll be working with the squared distance to avoid working out some slightly more complicated partial derivatives later.

The Lagrangian is

$L(x,y,z,\lambda)=(x-4)^2+y^2+(z+5)^2+\lambda(x+y+z-1)$

Take your partial derivatives and set them equal to 0:

$\begin{cases}\dfrac{\partial L}{\partial x}=2(x-4)+\lambda=0\\\\\dfrac{\partial L}{\partial y}=2y+\lambda=0\\\\\dfrac{\partial L}{\partial z}=2(z+5)+\lambda=0\\\\\dfrac{\partial L}{\partial\lambda}=x+y+z-1=0\end{cases}\implies\begin{cases}2x+\lambda=8\\2y+\lambda=0\\2z+\lambda=-10\\x+y+z=1\end{cases}$

Adding the first three equations together yields

$2x+2y+2z+3\lambda=2(x+y+z)+3\lambda=2+3\lambda=-2\implies \lambda=-\dfrac43$

and plugging this into the first three equations, you find a critical point at $(x,y,z)=\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)$ .

The squared distance is then $d\left(\dfrac{14}3,\dfrac23,-\dfrac{13}3\right)^2=\dfrac43$ , which means the shortest distance must be $\sqrt{\dfrac43}=\dfrac2{\sqrt3}$ .

7 0

3 years ago

The speed of the current in a creek is 5 mph. A person clan kayak 3 mi upstream in the same time that it takes him to kayak 13 m

Brut [27]

Answer:

8

Step-by-step explanation:

0) the basic formula is: L=v*t, where L - distance, v - speed/velocity; t - time;

1) if the person's speed in still water is 'v' and the speed of water is 5 (according to the condition), then the upstream speed is 'v-5' and the downstream speed is 'v+5';

2) according to the condition the upstream time and the downstream time are the same, it means t₁=t₂=t, where t₁=upstream time and t₂=downstream time;

3) according to the items above it is possible to make up the equation of the upstream travel: t(v-5)=3; ⇒ t=3/(v-5);

4) according to the items above it is possible to make up the equation of the downstream travel: t(v+5)=13; ⇒ t=13/(v+5);

5) if t=3/(v-5) and t=13/(v+5), then

$\frac{3}{v-5} =\frac{13}{v+5}; \ => \ v=8.$