The answer to that question is 15.51
The answer is 31. Hope it help!
Step-by-step explanation:
- Since,in question 21 the trspezium is an isoseles tapizium
- therefore,9x-17=4x+28
- 5x=45
- x=9
- therefore q=9x-17
- =9*9-17
- =81-17
- q=64
- therefore 2t=360-128(since its isoseles triangle therefore adjacent base angles are equal)
- 2t=23
- t=116
- 2(K+M)=360
- 2(3x+16 + 11x-18)=360
- 2(14x-2)=360
- 28x-4=360
- 28x=364
- x=13
- J=K
- Isoleses angles are equal
- 3x+16=J
- 39+16=J
- J=55
- Bd=Ec (Congeruent sides of congurent trisnle are equal,BED~=CDE)
- 8x-27=2x+33
- 6x=60
- x=10
- Therefore,Bd=8x-27
- Bd=80-27
- Bd=53
From the intervals from (-2,0) and (4,7) I am not sure if you are required to use (,) or [,] but I will change my answer if needed.
Thank you!
![\bf 2sin(x)cos(x)=sin(x)\sqrt{2}\implies 2sin(x)cos(x)-sin(x)\sqrt{2}=0 \\\\\\ sin(x)~[2cos(x)-\sqrt{2}]=0\\\\ -------------------------------\\\\ sin(x)=0\implies \measuredangle x=0~~,~~\pi \\\\ -------------------------------\\\\ 2cos(x)-\sqrt{2}=0\implies 2cos(x)=\sqrt{2}\implies cos(x)=\cfrac{\sqrt{2}}{2} \\\\\\ \measuredangle x=\frac{\pi }{4}~~,~~\frac{7\pi }{4}](https://tex.z-dn.net/?f=%5Cbf%202sin%28x%29cos%28x%29%3Dsin%28x%29%5Csqrt%7B2%7D%5Cimplies%202sin%28x%29cos%28x%29-sin%28x%29%5Csqrt%7B2%7D%3D0%0A%5C%5C%5C%5C%5C%5C%0Asin%28x%29~%5B2cos%28x%29-%5Csqrt%7B2%7D%5D%3D0%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0Asin%28x%29%3D0%5Cimplies%20%5Cmeasuredangle%20x%3D0~~%2C~~%5Cpi%20%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A2cos%28x%29-%5Csqrt%7B2%7D%3D0%5Cimplies%202cos%28x%29%3D%5Csqrt%7B2%7D%5Cimplies%20cos%28x%29%3D%5Ccfrac%7B%5Csqrt%7B2%7D%7D%7B2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cmeasuredangle%20x%3D%5Cfrac%7B%5Cpi%20%7D%7B4%7D~~%2C~~%5Cfrac%7B7%5Cpi%20%7D%7B4%7D)
now, we're not including the III and II quadrants, where the cosine has an angle of the same value, but is negative, because the exercise seems to be excluding the negative values of √(2).
