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Mathematics

2 answers:

ycow [4]3 years ago

4 0

$p ^{-m-n}$

Ronch [10]3 years ago

3 0

P-m × n is the answer

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Let w(s,t)=f(u(s,t),v(s,t)) where u(1,0)=−6,∂u∂s(1,0)=5,∂u∂1(1,0)=7 v(1,0)=−8,∂v∂s(1,0)=−8,∂v∂t(1,0)=6 ∂f∂u(−6,−8)=−1,∂f∂v(−6,−8

Blababa [14]

$w(s,t)=f(u(s,t),v(s,t))$

From the given set of conditions, it's likely that you are asked to find the values of $\dfrac{\partial w}{\partial s}$ and $\dfrac{\partial w}{\partial t}$ at the point $(s,t)=(1,0)$ .

By the chain rule, the partial derivative with respect to $s$ is

$\dfrac{\partial w}{\partial s}=\dfrac{\partial f}{\partial u}\dfrac{\partial u}{\partial s}+\dfrac{\partial f}{\partial v}\dfrac{\partial v}{\partial s}$

and so at the point $(1,0)$ , we have

$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial s}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial s}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial s}\bigg|_{(s,t)=(1,0)}=(-1)(5)+(2)(-8)=-21$

Similarly, the partial derivative with respect to $t$ would be found via

$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=\dfrac{\partial f}{\partial 
u}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial u}{\partial t}\bigg|_{(s,t)=(1,0)}+\dfrac{\partial f}{\partial 
v}\bigg|_{(u,v)=(-6,-8)}\dfrac{\partial v}{\partial t}\bigg|_{(s,t)=(1,0)}$
$\dfrac{\partial w}{\partial t}\bigg|_{(s,t)=(1,0)}=(-1)(7)+(2)(6)=5$

6 0

3 years ago

The product of two posotive consecutive intergers is 132. Find the intergers.

Sergio [31]

Answer:

11 and 12

Step-by-step explanation:

because 11 and 12 are two positive consecutive integers. and when you multiply 11 and 12 you get 132.

$11*12=132\\$