All categories

2 years ago

Pls help me plsssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssssss

Mathematics

2 answers:

Lelu [443]2 years ago

8 0

The answer is y < 2

explanation:

Svet_ta [14]2 years ago

3 0

Answer:

x ≤ 2

Step-by-step explanation:

You might be interested in

Can some one Plz do this for me ( 28 Points )

Harrizon [31]

Answer:

Step-by-step explanation:

Moon- 109 / 4

Saturn- 59 / 1 , 177 / 3

Venus- 243 / 1 , 729 / 3

7 0

2 years ago

Read 2 more answers

Given solve triangle ABC. Round the answer to the nearest hundredth. A. A- 9.590, B8841°, c-11.89 B. A 9.590, B-8841, c 12.17 C.

nikitadnepr [17]

I think the answer is b

3 0

3 years ago

Evaluate −b2−2bx2−x when x=−2.

Y_Kistochka [10]

The answer...... is 6x+2

6 0

3 years ago

porfavor ayudenme dare corona a quien responda bien con procedimiento una sustancia tiene un volumen de 20 ml y una densidad de

valentina_108 [34]

Answer:

Step-by-step explanation:

6 0

3 years ago

Use the Chain Rule (Calculus 2)

atroni [7]

1. By the chain rule,

$\dfrac{\mathrm dz}{\mathrm dt}=\dfrac{\partial z}{\partial x}\dfrac{\mathrm dx}{\mathrm dt}+\dfrac{\partial z}{\partial y}\dfrac{\mathrm dy}{\mathrm dt}$

I'm going to switch up the notation to save space, so for example, $z_x$ is shorthand for $\frac{\partial z}{\partial x}$ .

$z_t=z_xx_t+z_yy_t$

We have

$x=e^{-t}\implies x_t=-e^{-t}$

$y=e^t\implies y_t=e^t$

$z=\tan(xy)\implies\begin{cases}z_x=y\sec^2(xy)=e^t\sec^2(1)\\z_y=x\sec^2(xy)=e^{-t}\sec^2(1)\end{cases}$

$\implies z_t=e^t\sec^2(1)(-e^{-t})+e^{-t}\sec^2(1)e^t=0$

Similarly,

$w_t=w_xx_t+w_yy_t+w_zz_t$

where

$x=\cosh^2t\implies x_t=2\cosh t\sinh t$

$y=\sinh^2t\implies y_t=2\cosh t\sinh t$

$z=t\implies z_t=1$

To capture all the partial derivatives of $w$ , compute its gradient:

$\nabla w=\langle w_x,w_y,w_z\rangle=\dfrac{\langle1,-1,1\rangle}{\sqrt{1-(x-y+z)^2}}}=\dfrac{\langle1,-1,1\rangle}{\sqrt{-2t-t^2}}$

$\implies w_t=\dfrac1{\sqrt{-2t-t^2}}$

2. The problem is asking for $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ . But $z$ is already a function of $x,y$ , so the chain rule isn't needed here. I suspect it's supposed to say "find $\frac{\partial z}{\partial s}$ and $\frac{\partial z}{\partial t}$ " instead.