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Archana Sathappan
3/1/2011 10:48:25 am
For number 11 pg 717, I got 46 degrees for the reference angle and 134 degrees for the angle in standard position. So when I graph it, should my arrow go in the negative or positive direction...since the original given angle is -2746 degrees.
Archana Sathappan
3/1/2011 11:39:48 am
For numbers 19 and 20, which ways are the arrows supposed to go?
Randy Boyd
3/3/2011 11:52:57 am
For # 18 on page 726, when you make the coordinates for when the angle is 270, (0,-1) is it read as (cosine,sine) or (sine,cosine)?
Ishta
3/6/2011 02:00:05 am
Archana, I think the reference should be drwn in the positive direction because the numbers are positive, I think. Check with other people as well. That is what I put.
Archana Sathappan
3/6/2011 04:42:20 am
I am a bit confused on p 726 number 5...can someone plz help me out?
Brandon Jachimiec
3/8/2011 07:24:04 am
Archana, For page 726 #5,
Ishta
3/8/2011 08:32:10 am
On the 13-5 homework, does the theta sign mean the same thing as x, so like y=cos(theta) is the same as y=cos(x)?
Archana Sathappan
3/9/2011 08:58:13 am
Ishta,
Christian Carvallo
3/9/2011 09:31:30 am
I'm so lost can some one explain to me how you get the period in these trigonometry graphs.
Anna Reimers
3/9/2011 09:35:49 am
on pg. 749 #5 i understand you move the graph up 3 and the amp is 2. But what about the 1/5? Should you multiply that by (x-pi) ?
Archana Sathappan
3/9/2011 09:48:54 am
Christian,
Archana Sathappan
3/9/2011 09:50:07 am
I need help to graph number 5 pg.749. Can someone plz help me?
Kristina Chapman
3/9/2011 11:18:13 am
Anna, yeah i just had that be the "b" value when figuring out the period for the graph. and if anyone knows how to graph number five, i'm so confused on it!
Melina Kanji
3/9/2011 01:08:59 pm
So for tonight's homework, on page 749, when we are trying to figure out the period, and for example, if we're doing number 5, it says y= 3+ cos1/5 (x-π), would the coefficient of x, be 1/5? or would it just be 1? so to figure out the period, would you distribute the 1/5, then solve?
Archana Sathappan
3/9/2011 09:04:19 pm
Melina,
Noelle
3/10/2011 07:27:36 am
I was wondering, are the asymtotes only for secant, cosecant, and cotangent?
Archana Sathappan
3/10/2011 08:03:18 am
Noelle,
Tyler Yanisch
3/10/2011 02:51:05 pm
Can i get some help on #8 on page 749
Noelle Linden
3/11/2011 02:46:33 am
for writing the equations, how do you find the horizontal shift?
Archana Sathappan
3/12/2011 09:26:32 am
Can someone please tell me what the equation is for number 4 pg 752 and show me how to do a couple of the steps? Because for the answer, I got y=-10cos pi/10(x+7)-40. But, when I put it in the calculator to check it, it's not coming out right.
Archana Sathappan
3/12/2011 09:32:14 am
Sorry! I meant to say pi/15, not pi/10.
Christian Carvallo
3/13/2011 05:39:52 am
On skills practice 105 worksheet how do you graph the equations where it's phase shift is in degrees, yet it's period is in radians. Thanks
Sydney Stevens
3/13/2011 10:31:27 am
Noelle you find the horizontal shift by using the number thats inside the parenthesis (the opposite of it). and I had the same question as Christian too so if anyone can help with that thankyou!
Erin Hohman
3/13/2011 02:26:24 pm
For the worksheet skills practice 105 number 9...how do you graph a period of 1?
Archana Sathappan
3/13/2011 02:53:26 pm
For skills practice 106 number 2, I am confused on which points to choose in order to make an equation.
Emma Burck
3/14/2011 08:56:00 am
Archana: You could really start on any of the points and be able to write the equation, but what made most sense to me was starting at -30 and ending at 150 so that you have a negative sine graph.
Sam bostick
3/14/2011 09:16:28 am
to Christian: to find the period in radians you use the equation (2π/b). and 2π=360 degrees so the equation to find the period in degrees is (360/b).
Zain Rahman
3/14/2011 09:38:52 am
christian: i think you can change the period to degrees also and just graph it that way. Like in the skills practice 105 for number 1 the period was pi, (which equals 180 degrees). its going to be the same thing either way then you just move the numbers around to do the horizontal shift.
Ishta
3/14/2011 11:04:27 am
For the math models, I still can't get how to plug an equation in? Can someone help me?
Ishta
3/14/2011 11:05:00 am
In the calculator that is.
Amir Khalid Richardson
3/14/2011 11:37:23 am
@ Ishta
Nikita Hariharan
3/14/2011 04:54:32 pm
I am confused on #2 on Pg. 760 especially b and d. How do u solve this problem?
Ishta
3/15/2011 10:22:50 am
For the inverse circular functions homework, I do not understand how the arc sin, arc cos, arc tan and so forth. Can someone explain it to me?
Archana Sathappan
3/15/2011 01:16:51 pm
Ishta,
Nidhi Chintalapani
3/15/2011 02:38:23 pm
I don't understand the last 3 problems in the worksheet for 13-7. I got the equations for the rest of them though but i just can't get the last ones.
Rohan Roy
3/15/2011 02:41:44 pm
To Ishta,
Anna Reimers
3/15/2011 03:19:29 pm
Nikita - to find the equation break it down into the different parts. To find the period you find the distance between 0.3 seconds and 1.8 seconds because that is one cycle. Then you can find the other parts from the graph since you know the highest point is at 60 and the lowest is at 40. Then for part d. you move the graph over so it starts at 0 instead of 0.3 in order to find the distance.
Brandon Jachimiec
3/16/2011 08:54:47 am
On page 779 for the questions like #70, or 71, how do you choose between degrees or radians? Is it radians because their is no degree sign, or is there something else to it? I know, I can just do both then see which one is write in the back of the book, but I was just wondering if I was forgetting something.
Nidhi Chintalapani
3/16/2011 12:29:00 pm
Brandon- I'm pretty sure you use degrees because in class that is what we've been using for this section. And Mrs.Johnson has been using theda, so i'm pretty sure it's degrees. But I'm not entirely sure.
Craig Cook
3/16/2011 12:59:17 pm
To Nidhi,
Kristina Chapman
3/16/2011 01:14:34 pm
Brandon and Nidhi-
Archana Sathappan
3/16/2011 01:19:39 pm
I skipped problem 2 on pg. 761 b/c I didn't get it, and I still don't get it...like, how to graph it...I'm kinda confused on the graphing part, but I think that I got the rest with setting up the equation and all. The graph is the part that I need help with.
Rohan Roy
3/16/2011 02:45:57 pm
For problem number 81 on page 779, I dont understand how they are getting -1 as an answer. Doesnt the cotangent mean it is cos/sin?
Anna Reimers
3/16/2011 03:26:14 pm
Rohan - The cotangant also means 1/tangent correct? The angle is 315 degrees, so the tangent is -1, so the cotangent would be 1/-1 which equals -1.
ryan v
3/17/2011 01:00:17 pm
anyone know how to do number 69 on pg 779, or any problem similar to that?
Randy Boyd
3/17/2011 01:26:09 pm
Ryan, you just plug Sine(-0.9692) into the calculator. i believe the back of the book gives the answer in radians though.
Randy Boyd
3/17/2011 01:58:31 pm
How do you put the equations of the math models in the calculator. my graph never shows up
Andy
3/17/2011 02:01:08 pm
Randy you have to distribute the period so your calculator knows through what 'time' per say to graph a full cycle through, if it's still not showing up zoom in or out
Randy Boyd
3/17/2011 02:03:20 pm
ooo, gotcha. Thanks Andy
sahana s
3/17/2011 03:13:36 pm
so when trying to find the equation of a graph the amplitude is from the axis to the high node but to get the real axis (vertical shift) its just the middle of the top and bottom most coordinate points?
Nikita Hariharan
3/17/2011 03:16:29 pm
You subtract or add to the amplitude so the placement would be different on the graph.
Nikita Hariharan
3/17/2011 03:17:02 pm
Sahana: Yes.
Nikita Hariharan
3/17/2011 03:18:52 pm
Is the answer to R3) d. 1/pi?
Sahana S
3/17/2011 03:21:55 pm
no nikita because the cosine of 180 degrees is -1 and secant is 1 over cosine so 1 over -1 = -1 Comments are closed.
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