Answer: it’s (3,2)
Step-by-step explanation:
Cos60 degrees=1/2, so AK/AB=1/2. Since AK=KD, AK=1/2AD=1/2AB. Therefore, AB=AD. This is a rhombus, with four equal sides. Triangle ABK is congruent to triangle DBK (SAS), since AK=KD, angle AKB=angle BKD=90, and BK=BK. Therefore, BD=AB. The sum of four side lengths is 24. Each side length is equal to 24/4=6. BD=6.
Answer:
70-35=35
35/70*100
=50%
Step-by-step explanation:
Trig was never my strong point but ok
remember the quotient rule
![\frac{dy}{dx} \frac{f(x)}{g(x)}=\frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cfrac%7Bf%28x%29%7D%7Bg%28x%29%7D%3D%5Cfrac%7Bf%27%28x%29g%28x%29-g%27%28x%29f%28x%29%7D%7B%28g%28x%29%29%5E2%7D)
so
remember the pythagorean identity sin²(x)+cos²(x)=1
so
![\frac{dy}{dx} \frac{1+sin(x)}{1-cos(x)}=\frac{cos(x)(1-cos(x))-sin(x)(1+sin(x))}{(1+cos(x))^2}=](https://tex.z-dn.net/?f=%5Cfrac%7Bdy%7D%7Bdx%7D%20%5Cfrac%7B1%2Bsin%28x%29%7D%7B1-cos%28x%29%7D%3D%5Cfrac%7Bcos%28x%29%281-cos%28x%29%29-sin%28x%29%281%2Bsin%28x%29%29%7D%7B%281%2Bcos%28x%29%29%5E2%7D%3D)
![\frac{cos(x)-cos^2(x)-sin(x)-sin^2(x)}{(1+cos(x))^2}=](https://tex.z-dn.net/?f=%5Cfrac%7Bcos%28x%29-cos%5E2%28x%29-sin%28x%29-sin%5E2%28x%29%7D%7B%281%2Bcos%28x%29%29%5E2%7D%3D)
![\frac{cos(x)-sin(x)-sin^2(x)-cos^2(x)}{(1+cos(x))^2}=](https://tex.z-dn.net/?f=%5Cfrac%7Bcos%28x%29-sin%28x%29-sin%5E2%28x%29-cos%5E2%28x%29%7D%7B%281%2Bcos%28x%29%29%5E2%7D%3D)
![\frac{cos(x)-sin(x)-(sin^2(x)+cos^2(x))}{(1+cos(x))^2}=](https://tex.z-dn.net/?f=%5Cfrac%7Bcos%28x%29-sin%28x%29-%28sin%5E2%28x%29%2Bcos%5E2%28x%29%29%7D%7B%281%2Bcos%28x%29%29%5E2%7D%3D)
![\frac{cos(x)-sin(x)-(1)}{(1+cos(x))^2}=](https://tex.z-dn.net/?f=%5Cfrac%7Bcos%28x%29-sin%28x%29-%281%29%7D%7B%281%2Bcos%28x%29%29%5E2%7D%3D)
![\frac{cos(x)-sin(x)-1}{(1+cos(x))^2}=](https://tex.z-dn.net/?f=%5Cfrac%7Bcos%28x%29-sin%28x%29-1%7D%7B%281%2Bcos%28x%29%29%5E2%7D%3D)
![\frac{-sin(x)+cos(x)-1}{(1+cos(x))^2}=](https://tex.z-dn.net/?f=%5Cfrac%7B-sin%28x%29%2Bcos%28x%29-1%7D%7B%281%2Bcos%28x%29%29%5E2%7D%3D)
taht is the last option
thanks to jdoe0001 for showing me which identity to use