Answer:
sec 0 = 5/(2√6)
Step-by-step explanation:
sin 0 = -1/5
Fourth quadrant Cos positive Rest negative
cos²0 + sin²0 = 1
cos²0 = 1-sin²0
cos²0 = 1 - (1/5)²
cos²0 = 1- 1/25
cos²0 = 24/25
cos0 = ±(2√6)/5
quadrant IV cos is positive.
Therefore cos0 = (2√6)/5
Now, sec0 = 1/cos0
sec0 = 1/{(2√6)/5)}
sec0 = 5/(2√6)
.
Hope it helps
\left[a _{3}\right] = \left[ \frac{ - b^{2}}{6}+\frac{\frac{ - b^{4}}{3}+\left( \frac{-1}{3}\,i \right) \,\sqrt{3}\,b^{4}}{2^{\frac{2}{3}}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}+\frac{\frac{ - \sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{24}+\left( \frac{1}{24}\,i \right) \,\sqrt{3}\,\sqrt[3]{\left( -1296 - 432\,b^{2} - 16\,b^{6}+\sqrt{\left( 1679616+1119744\,b^{2}+186624\,b^{4}+41472\,b^{6}+13824\,b^{8}\right) }\right) }}{\sqrt[3]{2}}\right][a3]=⎣⎢⎢⎢⎢⎡6−b2+2323√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))3−b4+(3−1i)√3b4+3√224−3√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))+(241i)√33√(−1296−432b2−16b6+√(1679616+1119744b2+186624b4+41472b6+13824b8))⎦⎥⎥⎥⎥⎤
Unfortunately, you are wrong. 2x + 3y
x = 4 y = -6
2(4) + 3(-6)
8 + -18
Answer is -10.
If you found this helpful, please brainliest me!
The tangent ratio itself is OppositeLeg/AdjacentLeg, if that's what you're asking for.