Answer:
Sorry dude, I don't know, but maybe someone else will come and answer!!!
Step-by-step explanation:
Answer:
![L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20%5Cdfrac%7B6%7D%7BS%5E2%2B1%7D%20%5B%5Csqrt%7B3%7D%20%5C%20S%20%2B1%20%5D)
Step-by-step explanation:
Given that:

recall that:
cos (A-B) = cos AcosB + sin A sin B
∴
![f(t) = 12 [cos\ t \ cos \dfrac{\pi}{6}+ sin \ t \ sin \dfrac{\pi}{6}]](https://tex.z-dn.net/?f=f%28t%29%20%3D%2012%20%5Bcos%5C%20%20t%20%5C%20%20cos%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%2B%20sin%20%5C%20t%20%20%5C%20sin%20%5Cdfrac%7B%5Cpi%7D%7B6%7D%5D)
![f(t) = 12 [cos \ t \ \dfrac{3}{2}+ sin \ t \ sin \dfrac{1}{2}]](https://tex.z-dn.net/?f=f%28t%29%20%3D%2012%20%5Bcos%20%5C%20%20t%20%5C%20%5Cdfrac%7B3%7D%7B2%7D%2B%20sin%20%20%5C%20t%20%20%5C%20sin%20%5Cdfrac%7B1%7D%7B2%7D%5D)

![L(f(t)) = L ( 6 \sqrt{3} \ cos \ (t) + 6 \ sin \ (t) ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20L%20%28%206%20%5Csqrt%7B3%7D%20%5C%20cos%20%5C%20%28t%29%20%2B%206%20%5C%20sin%20%5C%20%28t%29%20%5D)
![L(f(t)) = 6 \sqrt{3} \ L [cos \ (t) ] + 6\ L [ sin \ (t) ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%206%20%5Csqrt%7B3%7D%20%5C%20L%20%5Bcos%20%5C%20%28t%29%20%5D%20%2B%206%5C%20L%20%5B%20sin%20%5C%20%28t%29%20%5D)



![L(f(t)) = \dfrac{6}{S^2+1} [\sqrt{3} \ S +1 ]](https://tex.z-dn.net/?f=L%28f%28t%29%29%20%3D%20%5Cdfrac%7B6%7D%7BS%5E2%2B1%7D%20%5B%5Csqrt%7B3%7D%20%5C%20S%20%2B1%20%5D)
Yes, if FJ || GH, and also if FG || HJ, since a parallelogram has two pairs of parallel sides.
now, we know FG = HJ = 10, but if it's a parallelogram, then the other two sides must also have the same length, so chances are FJ = GH = 15.
Answer:
x=16/13
y= -9/13
Step-by-step explanation:
A: -4y=-x+4
B: y+3x=3
4B: 4y+12x=12
A+4B: 12x= -x+16
-----> 13x=16
x=16/13
-4y= -16/13+4
4y=16/13-4
y=4/13-1= -9/13
Range = 6.9
mode = 4.2
median= 3