I need help with no.7
1 answer:
Answer:
(a) 315°
(b) 3°
(c) 238°
Step-by-step explanation:
Bearings are measured clockwise from north. The triangle described is illustrated in the attachment.
<h3>(a)</h3>The bearing of P from R is 180° different from the bearing of R from P it will be ...
135° +180° = 315° . . . . bearing of P from R
__
<h3>(b)</h3>The bearing of Q from R is 48° more than the bearing of P from R, so is ...
315° +48° = 363°, or 3° . . . . bearing of Q from R
__
<h3>(c)</h3>The angle QPR has a value that makes the sum of angles in the triangle equal to 180°. It is ...
180° -48° -55° = 77°
The bearing of Q from P is 77° less than the bearing of R from P, so is ...
135° -77° = 58°
As above, the reverse bearing from Q to P is ...
58° +180° = 238° . . . . bearing of P from Q
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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).
Or, if you actually did want the second order derivative,
and in case you meant the other way around, no need to compute that, as