Point slope form is y-3 = -5( x + 2 )
Slope intercept form is y= -5x - 7
Answer:
a) See the file below, b)
, c) 
Step-by-step explanation:
a) Points moves clockwise as t increases. See the curve in the file attached below. The parametric equations describe an ellipse.
b) The arc length formula is:
![s = \int\limits^{0.5\pi}_{-0.25\pi} {[\left( 3\cdot \cos t\right)^{2}+\left(-5\cdot \sin t \right)^{2}]} \, dx](https://tex.z-dn.net/?f=s%20%3D%20%5Cint%5Climits%5E%7B0.5%5Cpi%7D_%7B-0.25%5Cpi%7D%20%7B%5B%5Cleft%28%203%5Ccdot%20%5Ccos%20t%5Cright%29%5E%7B2%7D%2B%5Cleft%28-5%5Ccdot%20%5Csin%20t%20%5Cright%29%5E%7B2%7D%5D%7D%20%5C%2C%20dx)
c) The perimeter of that arc is approximately:


See the attached figure to better understand the problem
we know that
in the triangle ABC
tan 48°=AB/AC--------> AC=AB/tan48°------> 1200/tan 48°------> 1080.48 ft
in the triangle ABD
tan 33°=AB/AD---------> AD=AB/tan 33°-----> 1200/tan 33°------> 1847.88 ft
<span>the distance between the ships is AD-AC---> 1847.88-1080.48----> 767.4 ft
the answer is
</span>
the distance between the ships is 767.4 ft<span>
</span>
Third term = t3 = ar^2 = 444 eq. (1)
Seventh term = t7 = ar^6 = 7104 eq. (2)
By solving (1) and (2) we get,
ar^2 = 444
=> a = 444 / r^2 eq. (3)
And ar^6 = 7104
(444/r^2)r^6 = 7104
444 r^4 = 7104
r^4 = 7104/444
= 16
r2 = 4
r = 2
Substitute r value in (3)
a = 444 / r^2
= 444 / 2^2
= 444 / 4
= 111
Therefore a = 111 and r = 2
Therefore t6 = ar^5
= 111(2)^5
= 111(32)
= 3552.
<span>Therefore the 6th term in the geometric series is 3552.</span>