Cos s=-2/5 and sin t=4/5, s and t are in quadrant II
find cos(s+t) and cos(s-t)

•cos(s+t) = cos(s)cos(t) - sin(s)sin(t) = (-⅖).(-⅗) - (√21 /5).(⅘) = +6/25 - 4√21 /25 = (6-4√21)/25

•cos(s-t) = cos(s)cos(t) + sin(s)sin(t) = (-⅖).(-⅗) + (√21 /5).(⅘) = +6/25 + 4√21 /25 = (6+4√21)/25

cos(t) = ±√(1 - sin²(t)) → -√(1 - sin²(t)) = -√(1 - (⅘)²) = -⅗

sin(s) = ±√(1 - cos²(s)) → +√(1- cos²(s)) = +√(1 - (-⅖)²) = √21 /5

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

the barter