5 for the first one and 7 for the second hope this helps
The triangle pay $32 more for that day than it paid per day during the first period of time.
Step-by-step explanation:
The given is,
Triangle Construction pays Square Insurance $5,980
To insure a construction site for 92 days
To extend the insurance beyond the 92 days costs $97 per day
Triangle extends the insurance by 1 day
Step:1
Insurance per day from the 92 days period,

Where, Total insurance for 92 days = $ 5,980
Period = 92 days
From the values, equation becomes,

= $ 65 per day
Step:2
Insurance per day after the 92 days,
= $ 97
Amount Pay for that day than it paid per day during the first period of time,

= $32
Result:
The triangle pay $32 more for that day than it paid per day during the first period of time, if the Triangle Construction pays Square Insurance $5,980
to insure a construction site for 92 days and to extend the insurance beyond the 92 days costs $97 per day.
Answer: sin u = -5/13 and cos v = -15/17
Step-by-step explanation:
The nice thing about trig, a little information goes a long way. That’s because there is a lot of geometry and structure in the subject. If I have sin u = opp/hyp, then I know opp is the opposite side from u, and the hypotenuse is hyp, and the adjacent side must fit the Pythagorean equation opp^2 + adj^2 = hyp^2.
So for u: (-5)^2 + adj^2 = 13^2, so with what you gave us (Quad 3),
==> adj of u = -12 therefore cos u = -12/13
Same argument for v: adj = -15,
opp^2 + (-15)^2 = 17^2 ==> opp = -8 therefore sin v = -8/17
The cosine rule for cos (u + v) = (cos u)(cos v) - (sin u)(sin v) and now we substitute: cos (u + v) = (-12/13)(-15/17) - (-5/13)(-8/17)
I am too lazy to do the remaining arithmetic, but I think we have created a way to approach all of the similar problems.