Cos s=-2/5 and sin t=4/5, s and t are in quadrant II
find cos(s+t) and cos(s-t)
1 answer:
Answer:
•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
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Answer:
31
Step-by-step explanation:
x, x+1, x+2
x + x + 1 +x + 2 = 90
3x + 3 = 90
-3 -3
3x = 87
/3 /3
x = 29
29 + 2 = 31
Answer:
because x is the middle integer of three consecutive integers
=> The remaining 2 numbers respectively are x - 1 and x + 1
=> the sum of these three integers is
x - 1 + x + x + 1 = 3x
15 i am pretty sure i’m probably gonna be wrong
Answer:
First equation is the right answer.
X÷3=0.6
1.8÷3=0.6
0.6=0.6
Hence proved
