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]

From the given set of conditions, it's likely that you are asked to find the values of

and

at the point

.
By the chain rule, the partial derivative with respect to

is

and so at the point

, we have


Similarly, the partial derivative with respect to

would be found via

Answer:
12
Step-by-step explanation:
x^2 - a
If a is a perfect square we could factor
x^2 - b^2 = (x-b)(x+b)
So a cannot be 36,49,81 since they are perfect squares
a must be 12
Answer:
x = 7
Step-by-step explanation:
6 - 4x = -22
-6 - 6
-4x = -28
x = -28/-4
x = 7
Answer:
Rounding to nearest hundredths gives us r=0.06.
So r is about 6%.
Step-by-step explanation:
So we are given:

where


.


Divide both sides by 1600:

Simplify:

Take the 6th root of both sides:
![\sqrt[6]{\frac{23}{16}}=1+r](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B%5Cfrac%7B23%7D%7B16%7D%7D%3D1%2Br)
Subtract 1 on both sides:
![\sqrt[6]{\frac{23}{16}}-1=r](https://tex.z-dn.net/?f=%5Csqrt%5B6%5D%7B%5Cfrac%7B23%7D%7B16%7D%7D-1%3Dr)
So the exact solution is ![r=\sqrt[6]{\frac{23}{16}}-1](https://tex.z-dn.net/?f=r%3D%5Csqrt%5B6%5D%7B%5Cfrac%7B23%7D%7B16%7D%7D-1)
Most likely we are asked to round to a certain place value.
I'm going to put my value for r into my calculator.
r=0.062350864
Rounding to nearest hundredths gives us r=0.06.