.Use the graph below to answer the question that follows: What are the amplitude, period, and midline of the function? Amplitude: 8; period: π; midline: y = 1 Amplitude: 8; period: 2π; midline: y = 5 Amplitude: 4; period: 2π; midline: y = 5 Amplitude: 2; period: π; midline: y = 1
Pick any point on the graph, say (0,-1)
now trace along the graph (going from left to right) and tell me when the point you trace gets to be at the same horizontal level as (0,-1) what is the x coordinate of this next point?
Pi.
so when we get to x = pi, the whole process repeats itself again that means the period is pi
So, its either A or D ;o
what is the distance from the highest peak to the lowest valley (only focus on the vertical y component)
5?
the lowest it goes is y = -1 the highest it goes is y = 3
subtract the two 3 - (-1) = 3 + 1 = 4
So the total distance is 4 units cut that in half to get the amplitude basically the distance from center midline to peak (or center midline to valley)
So. 2?
yes
SO ITS D :D
and the midline is basically the midpoint between the peak & valley
Can you help me with more, PLEASEEE?
sure
Same question? or new one?
either way works
At the farmer's market, there is a scale that measures the weight of fruit and vegetables. The needle is 16 cm long and it rotated 68° when Danielle placed six oranges on it. How far did the needle travel as it rotated? 272 pi over 45 45 pi over 272 1088π 544π
what symbol comes after "and it rotated 68"
degrees?
its the degree sign. yeah.
oh ok
what is the circumference of this whole circle?
Is that 2(pi)(r)> Is the radius 4?
?*
"The needle is 16 cm long" so r = 16
So 2(3.14)(16)?
correct, which is what approximately?
100.48
that is the total distance it would travel if it traveled 360 degrees
however, it only does 68 degrees so it does 68/360 of the entire circle
that means you multiply that fraction by the circumference to get the distance you travel just along that little arc
Which is like 18.9.
correct
Which answer choice would that be?
hmm let me think
it looks like they used radians
Yeah, i got confused.
so we convert 68 degrees to radians by multiplying it by pi/180 68*(pi/180) = 68pi/180 = 17pi/45 ---------------- Now that we have the central angle in radians, we can just multiply it by the radius to get the arc length (radius)*(angle in radians) = (16)*(17pi/45) = (16*17pi/45) = 272pi/45
OO. So A c:
the nice thing about radians is that you can use them to find arc length with that formula however, radians are a bit harder to think about compared to degrees in my mind
yes A
You dont know how much you're helping me. xD
Which of the following functions has the greatest y-intercept? f(x) x y −3 9 −2 4 −1 1 0 0 1 1 2 4 g(x) = 5 cos(3x) − 5 f(x) g(x) f(x) and g(x) have the same y-intercept Not enough information
|dw:1409274948632:dw| Dont mind the boxes in the first.
what's the y intercept of f(x)?
f(x) x y −3 9 −2 4 −1 1 0 0 1 1 2 4
Yeah, i dont know. :c
the y-intercept always occurs when x = 0
this is because this is when the graph crosses that vertical y axis
so what is the y value when x = 0?
0.
therefore, the y intercept of f(x) is 0
for g(x), plug in x = 0 to get ???
0
so they both have the same y intercept
and this happens to be at the origin
What are the x-coordinates for the maximum points in the function f(x) = 4 cos(2x − π) from x = 0 to x = 2π? x = 3 pi over 2, x = π x = pi over 2, x = 3 pi over 2 x = 0, x = 2π x = 0, x = π
Where the question marks are, are pi signs.
is this a calculus class?
Nu, Trig xD
ok, we'll have to rely on the calculator then
what kind of calculator do you have?
My mom has mine >.<
But it was a Ti-84.
so you currently don't have one?
Nope, i have my ipod.
and can you draw out the function? I see a lot of boxes where symbols are supposed to go
its like my browser can't display them
Mine cant either. and okay. \[f(x)=4\cos(2x-\Pi) from x=0 \to x=2\Pi \]
thanks
The question marks are just pi signs xD
If we were to graph that function, we would get this http://www.wolframalpha.com/input/?i=4*cos%282x-pi%29
hmm that got cut off, let me find another graph program
Ive been doing this test for 3 hours xD
oh I can't help with tests
that's something you need to do on your own
Its a pre-test, not the actual thang.
then getting it wrong doesn't hurt your grade as it helps the teacher see where you are in terms of the content if you get it right, then the teacher doesn't need to go over it if you get it wrong, then the teacher will know where to focus his/her energy so either way, you'll be fine
Naw, this goes against my grade. Because i just learned it. so we have to take this, before we take our ACTUAL test. which really sucks. because im a very unfortunate when it comes to math, and if i dont get help, i fail.
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