Step-by-step explanation:
Before we answer, the area of a Circle given it radius is
The area of a circle given it diameter is
So let answers question 1-4.
1. The area is
2. The area is
3.
4.
For the 5th problem, circumference is
where d is the diameter and
where r is the radius.
A circumference of 100 m means the radius is 50 because
So this means the area is
(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.
The easiest way, I think, is to convert the mixed number into an improper fraction, then multiply by 3.
3 1/2 = 7/2
7/2 · 3 = 21/2
now just change the improper fraction back to a mixed number by dividing and putting the remainder into fraction form
21/2 = 10 1/2
You could also multiply the whole number by 3 and the fraction by 3, ending up with 9 3/2, but then have to convert the improper fraction into a mixed number
3/2 = 1 1/2
then add the numbers together
9 + 1 1/2 = 10 1/2
either way works, whatever is easiest for you.
Answer:
A. because -4 is 4 units away from 0.
