(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.
---------------------------------------------------------------------------------------
(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.
Answer:
Step-by-step explanation:
The range, in a layman language, is a set of values for which f(x) exist. Hence suppose our pre-image is 0-6,and since f(x)=20x+4 ,then
When x=0, f(x)= 4
When x=1, f(x)=24
When x=2, f(x)=44
When x=3, f(x)= 64
When x=4, f(x)=84
When x=5,f(x)=104
When x=6, f(x)=124
Hence the answer is B.
Answer:
Step-by-step explanation:
11 * 1/4
11/4
or 2 3/4
The Answer to your question ∑ (2k -10) from 3 to k=6 is -4
volume of a cone = (1/3)*pi*r^2*h
180 = pi(r^2)h
180/pi(r*2) = h
