A) Find csc ∅, given that cot ∅ = -7/2 and cos ∅ < 0 b) Find cot ∅, given that cos ∅ = 15/17 and ∅ is in Quadrant IV

a) csc∅ = √53/2

b) cot ∅ = 15/√514

Step-by-step explanation:

a) The value of cot ∅ = -7/2

cos ∅ < 0

We note that cos(∅) is less than 0 in Quadrant II and III, and cot(∅) is less than 0 in the Quadrant II, therefore, we have;

∅ is in the Quadrant II

csc∅ = (√((-7)² + 2²))/2 =√53/2 ≈ 3.64

csc∅ = √53/2

b) cos∅ 15/17

∅ is in Quadrant IV

∴ cot ∅ = 15/(√(15² + 17²)) = 15/√514 ≈ 0.662

cot ∅ = 15/√514

answer: ouch. thats difficult

step-by-step explanation:

hello :

70y^8 + 30y^6 = 2 ( 35 y^8 + 15y^6)

the lenght is : 35 y^8

the width is : 15y^6