I need these tomorrow morning (test) so if anyone is on break now and can get these done I'd be forever indebted.
1. Find two solutions for the triangle with angle A=61, side b = 18, side a = 17.
Note from me: There are two solutions, the one I got was angle B = 68, angle C = 51, and side c = 15
2. Let AB be a vector with initial point (-4, -1) and terminal point (-2, -5)
- Write vector in component form
- Write vector as a linera combination
- Find Magnitude
3. Find the angle between/ state the relationship of the vectors
a) v=i+2j w=2i-j
b) v=3i-7j w=-2i+14/3j
c) v=<5.4, -2.3> w=<13.5, -6.9>
4. Find a unit vector in the direction of v=3i-2j
5. Given vector a = <-5, 12>
a) Find the vector parallel to vector a but with a magnitude of 39
b) Find the vector opposite of vector a but with a magnitude of 26
c) Find the vector orthogonal to vector a but with magnitude of 52
algebra 2?
i just took my final yesterday so im brain dead right now
I know how to do them, but I don't care enough to do them.
I did like, half of those on my math final today, though. :)
1) The other triangle, using the ambiguous case, is A=61, a=17, B=112, b=18, C=7, c =2.
I assume you wanted everything rounded to the nearest whole, but just in case...
A=61, a=17, B=112.17005612591556, b=18, C=6.829943874084435, c=2.3115065868979543
Edit: We know that A=61, a=17, and b = 18. a/sinA=b/sinB. All I did was used the supplement of the first value of B I found (68) in place of the original. Sin 68 = Sin 112. I just went from there.
2)a) (2, -4)
b) 2i-4j
c)sqrt(20)
Edit: a) If I did this correctly, all I had to do was find out how far it moved. ((-2)-(-4),(-5)-(-1))
b)Once I had part a, I just changed it to linear form or whatever it's called.
c) Pythagorean theorem.
3)a)90 The vectors have the same magnitude, but are at right angles to each other.
b)0 The 2nd vector is 2/3s the first and is opposite.
c)176 I didn't really see a relationship. Taking a shot in the dark, but is w supposed to be w=16.2, -6.9?
Edit: I can't really explain these. I drew rough sketches of them and filled in the angles and side lengths.
If you need me to draw the pictures to show, I will.
4)(3/sqrt(13), -2/sqrt(13))
Because the magnitude is sqrt(13), we have to divide everything by sqrt(13) to get the vector to be a unit vector.
5)a)(-15, 36)
b)(10,-24)
c)Not sure if there's supposed to be 2, but here goes: (48, 20) and (-48, -20)
And yes, I did all of this to raise my post count.
Edit: Magnitude = 13
Parallel means same direction, but it had to be 3x larger, so I multiplied by 3.
Opposite means, well, opposite direction, so I changed the signs. It had to be 2x the magnitude, so I multiplied by 2.
Orthogonal means at a right angle to, or perpidicular to. So, I found the slope of the first vector was -5/12. Therefore, the slope of the vector I was looking for was 12/5. However, it had to be 4x larger, so I multiplied by 4. I put the second one on there, (-48, -20), because it is opposite to the first one I found.
Quote from: BigAznDaddy on June 07, 2006, 10:17:26 PM
algebra 2?
i just took my final yesterday so im brain dead right now
analysis.
Thanks dark_drake, you're awesome. Do you have your work saved? If not, no big. If you do could you scan/copy and paste/whatever it?
Quote from: OG Trust on June 08, 2006, 12:05:58 AM
Quote from: BigAznDaddy on June 07, 2006, 10:17:26 PM
algebra 2?
i just took my final yesterday so im brain dead right now
analysis.
Thanks dark_drake, you're awesome. Do you have your work saved? If not, no big. If you do could you scan/copy and paste/whatever it?
I didn't really write much down, and on top of that, I don't have a scanner, but I'll try to say how I found it in each one.
Edit: There you go. Not sure if my explanations helped, but I tried.