Draw a triangle ,label them ,sorry it kinda hard to see
Answer:
D. (-x+5)^2/9+(y-4)^2/4=1 (I assume there was a mistype)
Step-by-step explanation:
The first step is isolating the sine and cosine functions.
x=5-3cos(t)
x + 3cos(t) = 5
3cos(t) = 5 - x
cos(t) = (5 - x)/3
y=4+2sin(t)
y - 4 = 2sin(t)
(y - 4)/2 = sin(t)
Then, square at both sides of the equal sign
cos²(t) = (5 - x)²/3² = (5 - x)²/9
sin²(t) = (y - 4)²/2² = (y - 4)²/4
Recall the trigonometric identity and replace.
cos²(t) + sin²(t) = 1
(5 - x)²/9 + (y - 4)²/4 = 1
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).
I say the answer is 9 because he wrapped 8 games and 1 console and 8+1=9.
Answer:
299 miles per hour
Step-by-step explanation:
![v=\frac{234}{\sqrt[3]{\frac{p}{w}}}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B234%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7Bp%7D%7Bw%7D%7D%7D)
![v=\frac{234}{\sqrt[3]{\frac{1311}{2744}}}](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B234%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B1311%7D%7B2744%7D%7D%7D)

Therefore, the velocity of the car at the end of a drag race is 299 miles per hour