Let P be the plane that intersects
- x-axis at point (-5,0,0);
- y-axis at point (0,-2,0);
- z-axis at point (0,0,5).
Write the equation of the plane P:
Then
The coefficients at variables x, y and z are the coordinates of perpendicular vector to the plane. Thus
Answer: 
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:
x=-1
Step-by-step explanation:
Answer:
-16p³ + 48p² - 16p
Explanation:
In order to simplify the given expression, we need to multiplied 8p with (-2p²+6p-2)
8p*(-2p²+6p-2)
we need to distribute 8p over the brackets or parenthesis.
=8p*(-2p²) + 8p*(6p) + 8p*(-2)
= 8*(-2)*p*p² + 8*6p*p + 8*(-2)* p
= -16p³ + 48p² - 16p
We can't simplify it more, so that's the final answer.
I hope it's help.
Answer:
Step-by-step explanation:
