Question

Please help? Will make you the Brainliest!

Answer 1

Answer:

D. 19/2   D.-4/$\sqrt{15}$   D.cos^2x  look at explanation for last one

Step-by-step explanation:

1. Find the value of sec if cos delta  = 2/19

because sec is basically 1/cos, and since it's Quad I there's nothing weird, you can just flip the fraction

therefore it's  D. 19/2

2. Find the value of csc given cos delta = -1/4 and triangle is in third Quad

so cosine is adj over hypo: since the hypo can't ever be negative, the triangle must have a = -1 and c =4. to find bc use the c^2 = a^2 + b^2 to solve for b^2, and since it's third quad, b is negative, so b = -$\sqrt{15}$2. NOw that you know all three sides, finding csc is ez. So csc is hypo over opposite bc its opposite of sine, so therefore csc delta = $-4/\sqrt{15}$

3. simplify sin^2qcot^2

ok, so if cot is opposite of tan, and tan is sine over cos, cot must be cos over sine :D! SO cot^2 = $\frac{cos^2x }{sin^{2}x}$. multiply that by sin^2 and the sines will cancel, leaving you with just cos^2x! sorry, I used x insead of q out of habit... oops!

4. determine the quad if sine delta = -1/4

sine is opposite over hypo, and as mentioned before, hypo can't equal to negative, so the opposite side must be negative. therefore, the triangle must lie in quad III or quad IV

ahem, these are your answers, idk why is it all D lol, I hope this helped. sorry this was so long!

Answer 2

D. 19/2 is ur answer.