(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.
We let x be the number of tickets that Sara and Jake sold individually. With this representation, the number of tickets sold by Cole is equal to 2x+6.
(a) x + x + (2x + 6) = 538
The value of x from the generated equation is 133.
(b) Sara sold a total of 133 tickets.
Answer:
(-1, 2) and (-1, 3.5)
Step-by-step explanation:
The triangle ΔDEF spans 4 squares horizontally.
So, the midsegment of ΔDEF will coincide with the line 4/2 = 2 squares from the vertex F.
Note that the <em>x </em>coordinate of the vertex F is -3 and 2 units to the right of F is -1.
Therefore, the midsegment of ΔDEF coincides with the line <em>x</em> = -1.
So, the <em>x</em> coordinates of the end points of the midsegment are -1.
Let's find the <em>y</em> coordinates of the end points.
From the given figure, it is clear that the mid point of FD is half way between 3 and 4 and hence it is 3.5.
Mid point of FE is 2.
So, the co-ordinates of the end points of the midsegment are (-1, 2) and (-1, 3.5).
A regular polygon has congruent sides.
An octagon has 8 sides of equal length.
P = 8s = 8 * 8 cm = 64 cm
Answer: 64 cm
