To find each intercept, you must let the other variable be equal to zero to find the point of the other axis.
For example, if you were to find the x intercept on this function you allow:
y = 0
3x + 4(0) = -12
3x + 0= -12
x = -4
and to find the y intercept you let x =0
3(0) + 4y = -12
0 + 4y =-12
y = -3
so you’re points of intercept are (-4,0) for the x intercept. and (0,-3) for the y intercept
Hi! Sorry for the delay in answering your question! Here’s a solution to the problem, I even did a little extra to show you how you can develop an idea of how the function would look when plotted.
If you need a higher quality picture try this one here
Inspired by this twocubes’ post and asked to make a animation of it, I made a gif.
:3 this is pretty
(you could have @mentioned me or tagged me so that I could have noticed this earlier tho :V)
Hello, I have a couple of specific questions and a general question. Currently, I’m in the process of learning ‘bearings’ in trigonometry and in general I don’t think I understand the concept despite several rereads of my textbook. I must get something, as I get my problems half right [as shown above in blue and purple]. However, I don’t understand how, in example either problem above, there is no South measurement. Due to not understanding this I’ll usually add a south measurement and get the problem wrong. The exact prompt that goes the with the blue and purple problems is: ‘An observer for a radar station is located at the origin of a coordinate system.For each of the points in these exercises, find the bearing of an airplane located at that point. Express bearing using both methods.’
Aside from that, there’s a question shortly after that that I am completely lost on. I’ve never been good at word problems, so I don’t even know where to begin.
The ray y=x, x is greater than or equal to 0, contains the origin and all points in the coordinate system whose bearing is 45 degrees. Determine the equation of a ray consisting of the origin and all points whose bearing is 240 degrees.
The given solution to this last problem is in red.
Thank you so much for your help!
The current generation of detectors at the SuperCDMS (Cryogenic Dark Matter Search) experiment, known as iZIP (interleaved Z-sensitive Ionization Phonon). More info (PDF).
Source: SuperCDMS Gallery
(Proud of that my former university is part of the team.)