I need HELP ON MY TEST 6+1 IS HELP
2 answers:
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Just to make sure we're using the same language, I'm going to use the function form of:
[] I would agree that k = 2, since the period is only half as long as a normal sine function. So, we so far, have y = A sin(2x) + h. We still need to find A and h.
[] The y-intercept is 3. Remember that the y-intercept happens when x = 0. So, plugging in x = 0 into our formula, we have: 3 = A sin(2*0) + h. In other words, 3 = A sin(0) + h = 0 + h = h. So, we now know that h = 3. The formula is now y = A sin(2x) + 3.
[] Finally, there is a single x-intercept. Picture what the sine function looks like right now, it is floating in the air around y = 3. We need to stretch it vertically until it just grazes the x-axis.
If A = 1, then our sine function bounces between 2 and 4 (+/- 1 around h = 3). But that doesn't touch 0, so no good.
If A = 2, then our sine function bounces between 1 and 5 (+/- 2 around h = 3). Again, not quite touching 0 yet.
The answer should be A = 3, then our sine function bounces between 0 and 6 (+/- 3 around h = 3).
The final formula is y = 3 sin(2x) + 3.
[] I would agree that k = 2, since the period is only half as long as a normal sine function. So, we so far, have y = A sin(2x) + h. We still need to find A and h.
[] The y-intercept is 3. Remember that the y-intercept happens when x = 0. So, plugging in x = 0 into our formula, we have: 3 = A sin(2*0) + h. In other words, 3 = A sin(0) + h = 0 + h = h. So, we now know that h = 3. The formula is now y = A sin(2x) + 3.
[] Finally, there is a single x-intercept. Picture what the sine function looks like right now, it is floating in the air around y = 3. We need to stretch it vertically until it just grazes the x-axis.
If A = 1, then our sine function bounces between 2 and 4 (+/- 1 around h = 3). But that doesn't touch 0, so no good.
If A = 2, then our sine function bounces between 1 and 5 (+/- 2 around h = 3). Again, not quite touching 0 yet.
The answer should be A = 3, then our sine function bounces between 0 and 6 (+/- 3 around h = 3).
The final formula is y = 3 sin(2x) + 3.
Answer:
Option D. The student did not use the correct formula to calculate the area of the segment
Step-by-step explanation:
step 1
Find the area of the isosceles triangle
Applying the law of sines
step 2
Find the area of the sector
The area of the sector is 1/6 of the area of the circle
so
substitute the value
step 3
Find the area of the segment
The area of the segment is equal to the area of sector minus the area of triangle
therefore
The student did not use the correct formula to calculate the area of the segment