Answer:
Hans:6
Eric:18
Kaitlin:74
Step-by-step explanation:
first create an equation using the information in the question
5h+68=98
first subtract 68
5h+68=98
-68 -68
5h=30
divide by 5
Hans:6
multiply by 3
Eric:18
68+6=74
Kaitlin:74
Hope this helps! Plz award Brainliest : )
Answer:
answer C i got it right 100%
Step-by-step explanation:
Answer:
The segment is 6 units long.
Step-by-step explanation:
The points (–2, 4) and (–2, –2) are vertices of a heptagon. We have to explain how to find the length of the segment formed by these endpoints.
If two points at the ends of a straight line PQ are P(
) and Q(
), then the length of the segment PQ will be given by the formula
Now, in our case the two points are (-2,4) and (-2,-2) and the length of the segment will be
units. (Answer)
(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.
