Answer:
(D)
<u>Use the law of sines:</u>
- 13 / sin x = 12 / sin 67°
- sin x = 13 sin 67° / 12
- sin x = 0.997
- x = arcsin (0.997)
- x = 85.6°
(E)
<u>Find the third angle:</u>
<u>Use the las of sines:</u>
- x / sin 61° = 21 / sin 71°
- x = 21 sin 61° / sin 71°
- x = 19.4 cm
Answer:
48%
Step-by-step explanation:
P (not getting a speckled rock) = 12/18 = 2/3
P (not getting a green rock) = 13/18
2/3 * 13/18 = 26/54 = 13/27 ≈ 48%
First, notice that:
![2\tan (\frac{x}{2})=2\cdot(\pm\sqrt[]{\frac{1-cosx}{1+\cos x})}](https://tex.z-dn.net/?f=2%5Ctan%20%28%5Cfrac%7Bx%7D%7B2%7D%29%3D2%5Ccdot%28%5Cpm%5Csqrt%5B%5D%7B%5Cfrac%7B1-cosx%7D%7B1%2B%5Ccos%20x%7D%29%7D)
And in the denominator we have:

then, we have on the original expression:
![\begin{gathered} \frac{2\tan(\frac{x}{2})}{1+\tan^2(\frac{x}{2})}=\frac{2\cdot\pm\sqrt[]{\frac{1-\cos x}{1+cosx}}}{\frac{2}{1+\cos x}}=\frac{2\cdot(\pm\sqrt[]{1-cosx})\cdot(1+\cos x)}{2\cdot(\sqrt[]{1+cosx})} \\ =(\sqrt[]{1-\cos x})\cdot(\sqrt[]{1+\cos x})=\sqrt[]{(1-\cos x)(1+\cos x)} \\ =\sqrt[]{1-\cos^2x}=\sqrt[]{\sin^2x}=\sin x \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%5Cfrac%7B2%5Ctan%28%5Cfrac%7Bx%7D%7B2%7D%29%7D%7B1%2B%5Ctan%5E2%28%5Cfrac%7Bx%7D%7B2%7D%29%7D%3D%5Cfrac%7B2%5Ccdot%5Cpm%5Csqrt%5B%5D%7B%5Cfrac%7B1-%5Ccos%20x%7D%7B1%2Bcosx%7D%7D%7D%7B%5Cfrac%7B2%7D%7B1%2B%5Ccos%20x%7D%7D%3D%5Cfrac%7B2%5Ccdot%28%5Cpm%5Csqrt%5B%5D%7B1-cosx%7D%29%5Ccdot%281%2B%5Ccos%20x%29%7D%7B2%5Ccdot%28%5Csqrt%5B%5D%7B1%2Bcosx%7D%29%7D%20%5C%5C%20%3D%28%5Csqrt%5B%5D%7B1-%5Ccos%20x%7D%29%5Ccdot%28%5Csqrt%5B%5D%7B1%2B%5Ccos%20x%7D%29%3D%5Csqrt%5B%5D%7B%281-%5Ccos%20x%29%281%2B%5Ccos%20x%29%7D%20%5C%5C%20%3D%5Csqrt%5B%5D%7B1-%5Ccos%5E2x%7D%3D%5Csqrt%5B%5D%7B%5Csin%5E2x%7D%3D%5Csin%20x%20%5Cend%7Bgathered%7D)
therefore, the identity equals to sinx
Answer:
Option C
Step-by-step explanation:
We are given a coefficient matrix along and not the solution matrix
Since solution matrix is not given we cannot check for infinity solutions.
But we can check whether coefficient matrix is 0 or not
If coefficient matrix is zero, the system is inconsistent and hence no solution.
Option A)
|A|=![\left[\begin{array}{ccc}4&2&6\\2&1&3\\-2&3&-4\end{array}\right] =0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%262%266%5C%5C2%261%263%5C%5C-2%263%26-4%5Cend%7Barray%7D%5Cright%5D%20%3D0)
since II row is a multiple of I row
Hence no solution or infinite
OPtion B
|B|=![\left[\begin{array}{ccc}2&0&-2\\-7&1&5\\4&-2&0\end{array}\right] \\=2(10)-2(10)=0](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%260%26-2%5C%5C-7%261%265%5C%5C4%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2810%29-2%2810%29%3D0)
Hence no solution or infinite
Option C
![\left[\begin{array}{ccc}6&0&-2\\-2&0&6\\1&-2&0\end{array}\right] \\=2(36-2)=68](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D6%260%26-2%5C%5C-2%260%266%5C%5C1%26-2%260%5Cend%7Barray%7D%5Cright%5D%20%5C%5C%3D2%2836-2%29%3D68)
Hence there will be a unique solution
Option D
=0
(since I row is -5 times III row)
Hence there will be no or infinite solution
Option C is the correct answer