9514 1404 393
Answer:
y = (x -1)² -16
Step-by-step explanation:
Note the location of the vertex (point A) on the graph. Its x-coordinate is readily identifiable as 1. Its y-coordinate is some value between -15 and -20, closer to -15. (If you go to the trouble of finding the vertex coordinates, you discover they are (1, -16).)
Once you have determined what the vertex is, you can compare the offered answer choices to the vertex form ...
y = (x -h)² +k
where (h, k) are the vertex coordinates. That is, you are looking for an answer choice that is something like ...
y = (x -1)² -16
(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.
Answer: Hmm Ok i do this-
Step-by-step explanation:
Wait are we trying to find x?
Ok so for the first one x = 5 and for the second one x = 2
Answer:A
Step-by-step explanation: In exponents,
a
1
m
=
m
√
a
Hence
x
1
2
=
2
√
x
The answer is 136 because 7 x 18 is 126 then + 10 is 136
