Answer:
True
Step-by-step explanation:
Any two sides of a triangle is greater than the third side.
This equation only has one answer. Think of it this way, wherever the graph intersects, that's your solution. And in this case, the graph is only intersecting in one place, which is (1,4)
Answer:
2.
Step-by-step explanation:
The answer is the seocnd one.
Your system of equations is
![\left[ \begin{array}{ccc}1&1&0\\2&0&-1\\0&1&-3\end{array}\right]\cdot \left[ \begin{array}{c}x\\y\\z\end{array} \right] = \left[ \begin{array}{c}10\\9\\-5\end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C2%260%26-1%5C%5C0%261%26-3%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%20%5Cright%5D%20%3D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D10%5C%5C9%5C%5C-5%5Cend%7Barray%7D%20%5Cright%5D)
Then the solution is
![\left[ \begin{array}{c}x\\y\\z\end{array} \right] = \left[ \begin{array}{ccc}1&1&0\\2&0&-1\\0&1&-3\end{array}\right]^{-1}\cdot \left[ \begin{array}{c}10\\9\\-5\end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%20%5Cright%5D%20%3D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D1%261%260%5C%5C2%260%26-1%5C%5C0%261%26-3%5Cend%7Barray%7D%5Cright%5D%5E%7B-1%7D%5Ccdot%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D10%5C%5C9%5C%5C-5%5Cend%7Barray%7D%20%5Cright%5D)
Your graphing calculator or any of several web sites can compute the inverse matrix for you.
![\left[ \begin{array}{c}x\\y\\z\end{array} \right] = \dfrac{1}{7}\cdot \left[ \begin{array}{ccc}1&3&-1\\6&-3&1\\2&-1&-2\end{array}\right]\cdot \left[ \begin{array}{c}10\\9\\-5\end{array} \right] = \left[ \begin{array}{c}6\\4\\3\end{array} \right]](https://tex.z-dn.net/?f=%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7Dx%5C%5Cy%5C%5Cz%5Cend%7Barray%7D%20%5Cright%5D%20%3D%20%5Cdfrac%7B1%7D%7B7%7D%5Ccdot%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bccc%7D1%263%26-1%5C%5C6%26-3%261%5C%5C2%26-1%26-2%5Cend%7Barray%7D%5Cright%5D%5Ccdot%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D10%5C%5C9%5C%5C-5%5Cend%7Barray%7D%20%5Cright%5D%20%3D%20%5Cleft%5B%20%5Cbegin%7Barray%7D%7Bc%7D6%5C%5C4%5C%5C3%5Cend%7Barray%7D%20%5Cright%5D)
The solution is (x, y, z) = (6, 4, 3).
The tan is negative in the 2nd and 4th quadrants, knowing this, the angle that equals the value that tan = 1 in the first quadrant is the tan 45° = 1, but the question restrict for 3pi/2 < theta < 2pi, so it's the same as 270° < theta < 360° (4° Quadrant), then moving the tan 45° = to this quadrant the value will be negative, tan in 4°Q = tan(360 - 45=315°) = -1.
sec theta = 1/cos theta
cos 45° =
![\sqrt{2} \div 2](https://tex.z-dn.net/?f=%20%5Csqrt%7B2%7D%20%5Cdiv%202)
cos values in 4°Q will be positives, so it doesn't change the signal.
sec 315° = 1/cos 315°
sec 315° = 1/square root (2)/2
sec 315 = 1/1 x 2/sqroot(2)
![sec \: 315 = 2 \div \sqrt{2}](https://tex.z-dn.net/?f=sec%20%5C%3A%20315%20%3D%202%20%5Cdiv%20%5Csqrt%7B2%7D%20)
![sec \: 315 = (2 \div \sqrt{2}) \: \times ( \sqrt{2} \div \sqrt{2})](https://tex.z-dn.net/?f=sec%20%5C%3A%20315%20%3D%20%282%20%5Cdiv%20%5Csqrt%7B2%7D%29%20%5C%3A%20%5Ctimes%20%28%20%5Csqrt%7B2%7D%20%5Cdiv%20%5Csqrt%7B2%7D%29)
![sec \: 315 = 2 \sqrt{2} \div 2](https://tex.z-dn.net/?f=sec%20%5C%3A%20315%20%3D%202%20%5Csqrt%7B2%7D%20%5Cdiv%202)