In which quadrant is θ located if sin θ is negative and sec θ is positive? i, ii, iii, iv?

Quadrant II.

Step-by-step explanation:

1. The sine is negative in quadrant III and quadrant IV. Then:

sinθ<0 (Quadrant III and quadrant IV).

2. Secant is equal to and the cosine is positive in quadrant I and quadrant IV:

cosθ>0 (Quadrant I and Quadrant IV)  

Secθ>0 (Quadrant I and Quadrant IV)

3. We look for a quadrant where the sine is negative and the cosine is positive. We know that the sine is negative in the IV quadrant and that the cosine and secant are positive in the IV quadrant.

Therefore the correct answer is the last one: Quadrant IV

true

step-by-step explanation:

answer: 204

step-by-step explanation:

we are told that the revenue is

x⋅(4080−10x)=−10x^2+4080x

this is a quadratic function, so to maximize it we should find the x-value of its vertex. recall that the x-value of the vertex of the parabola ax^2+bx+c is given by the formula −b/2a.

so in this case, we have a=−10 and b=4080, so the x-value of the vertex is

−b/2a= −(4080)/2(−10)=204