(a) Yes all six trig functions exist for this point in quadrant III. The only time you'll run into problems is when either x = 0 or y = 0, due to division by zero errors. For instance, if x = 0, then tan(t) = sin(t)/cos(t) will have cos(t) = 0, as x = cos(t). you cannot have zero in the denominator. Since neither coordinate is zero, we don't have such problems.
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(b) The following functions are positive in quadrant III:
tangent, cotangent
The following functions are negative in quadrant III
cosine, sine, secant, cosecant
A short explanation is that x = cos(t) and y = sin(t). The x and y coordinates are negative in quadrant III, so both sine and cosine are negative. Their reciprocal functions secant and cosecant are negative here as well. Combining sine and cosine to get tan = sin/cos, we see that the negatives cancel which is why tangent is positive here. Cotangent is also positive for similar reasons.
Given:
descends at a rate of 320 feet per minute for 3 minutes.
320 feet per minute * 3 minutes = 960 feet
Total change in altitude is 960 feet. Since it is descending, I believe it would be a negative 960 feet.
Answer:
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Step-by-step explanation:
about your question I'll just put it in the comment section:)
Answer:
34 is 85% of 40
Step-by-step explanation:
Answer:
V2=-20.0155m/s
Step-by-step explanation:
To answer this question we will use the principle of conservation of momentum which is
MtVt=M1V1+M2V2
1) let's find the initial velocity.
We will use,
v=a*t
v=10m/s2*2s = 20 m/s
d=vot+1/2*a*t2
d=0+1/2*10*4
d=20m
2) For the speed of the lighter piece we use,
v22=v12+2ad
0=v12+2*-9.81*(530-20)
v12=10006.2
v1=100.031m/s
3) Computing this in our conservation of momentum equation it will be-
MtVt=M1V1+M2V2
1500kg*20m/s=(1500*1/3)kg*100.031m/s + (1500*2/3)+V2
30000=50015.5+1000V2
Therefore V2=-20.0155m/s.