A. yes, the data represents a function because u have no repeating x values. A function cannot have repeating x values...they can have repeating y values, just not the x ones

Step-by-step explanation:

B. table : (8,8)(12,12)(14,16)(16,16)

look at ur points...when x = 8, y = 8...so the table, when x = 8 has a

value of 8

relation : f(x) = 8x - 5....when x = 8

f(8) = 8(8) - 5

f(8) = 64 - 5

f(8) = 59....and the relation has a value of 59

Therefore, the relation has a greater value when x = 8 <==

C. f(x) = 8x - 5...when f(x) = 19

19 = 8x - 5

19 + 5 = 8x

24 = 8x

24/8 = x

3 = x <==

7 0

3 years ago

Use the Chain Rule to find the indicated partial derivatives. z = x^4 + xy^3, x = uv^4 + w^3, y = u + ve^w Find : ∂z/∂u , ∂z/∂v

k0ka [10]

I'll use subscript notation for brevity, i.e. $\frac{\partial f}{\partial x}=f_x$ .

By the chain rule,

$z_u=z_xx_u+z_yy_u$

$z_v=z_xx_v+z_yy_v$

$z_w=z_xx_w+z_yy_w$

We have

$z=x^4+xy^3\implies\begin{cases}z_x=4x^3+y^3\\z_y=3xy^2\end{cases}$

and

$\begin{cases}x=uv^4+w^3\\y=u+ve^w\end{cases}\implies\begin{cases}x_u=v^4\\x_v=4uv^3\\x_w=3w^2\\y_u=1\\y_v=e^w\\y_w=ve^w\end{cases}$

When $u=1,v=1,w=0$ , we have

$\begin{cases}x(1,1,0)=1\\y(1,1,0)=2\end{cases}\implies\begin{cases}z_x(1,2)=12\\z_y(1,2)=12\end{cases}$

and the partial derivatives take on values of

$\begin{cases}x_u(1,1,0)=1\\x_v(1,1,0)=4\\x_w(1,1,0)=0\\y_u(1,1,0)=1\\y_v(1,1,0)=1\\y_w(1,1,0)=1\end{cases}$

So we end up with

$\boxed{\begin{cases}z_u(1,1,0)=24\\z_v(1,1,0)=60\\z_w(1,1,0)=12\end{cases}}$

3 0

3 years ago