Not sure if you mean to ask for the first order partial derivatives, one wrt x and the other wrt y, or the second order partial derivative, first wrt x then wrt y. I'll assume the former.
Or, if you actually did want the second order derivative,
![\dfrac{\partial^2}{\partial y\partial x}(2x+3y)^{10}=\dfrac\partial{\partial y}\left[20(2x+3y)^9\right]=180(2x+3y)^8\times3=540(2x+3y)^8](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Cpartial%5E2%7D%7B%5Cpartial%20y%5Cpartial%20x%7D%282x%2B3y%29%5E%7B10%7D%3D%5Cdfrac%5Cpartial%7B%5Cpartial%20y%7D%5Cleft%5B20%282x%2B3y%29%5E9%5Cright%5D%3D180%282x%2B3y%29%5E8%5Ctimes3%3D540%282x%2B3y%29%5E8)
and in case you meant the other way around, no need to compute that, as
by Schwarz' theorem (the partial derivatives are guaranteed to be continuous because
is a polynomial).
Answer:
y=68
Step-by-step explanation:
54=y−14
54+14=y
68=y
We need to evaluate the value of fraction 
.
We also given j=12.
In order to evaluate the value of fraction , we need to plug j=12.
Plugging j=12, we get
We have 12 in numerator and 4 in denominator.
We always divide top number by bottom number.
So, we need to divide 12 by 4.
On dividing 12 by 4 we get 3.
<h3>Therefore,
</h3>
Answer:
Option (c) "
"
Step-by-step explanation:
The given equation is :
q = c4(h + r)
We need to solve it for r.
Dividing both sides of the given equation by 4c.
Now, subtract h both sides of the equation.
Hence, this is the required solution.
Answer: (-2y+3)(y+11)
Step-by-step explanation:
33 is 3×11 so expect something like
(y+3)(y+11) = y^2+14y+33
leading term has coefficient -2 so
one factor is like (-2y+__)
Try (-2y+3)(y+11) = -2y^2-19y+33
which happens to be the answer
