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 Hydrogen777 Administrator5990 Posts11 Cresco 14:2 - 15.82.62818.536 days ago A Tree said:So you want a line going between those two points?I got it. A function from (1, 2) to (a, b) where a and b are rational numbers is f(x) = (x - 1)(b - a) + a. [Quote] [Link]
 Fwip said:Basically, integers are numbers that you can write without a decimal point. -5, 0, and 200182391 are all integers. 4/3, 0.5, and 3.14159... are not. Oh. That helped a lot. Thanks fwip-whip. [Quote] [Link]
 Ghost ride the Fwip. *remeniscs, remembering the fwipocolypse* Hey, can somebody provide a concise description of each of the conic sections, and how to identify them just by looking at the function? Specifically hyperbolas, and the wibby wobby, asymptoty...stuff. [Quote] [Link]
 Circles are of the form: Ax2 + Ay2 = C, where A and C are positive. Ellipses: Ax2 + By2 = C, with A != B (as well as positive). Parabolas look like: y = Ax2 + Bx + C or x = Ay2 + Bx + C where A is nonzero (positive or negative) and B and C are any (real) number. And I believe hyperbolas are: Ax2 - By2 = C or Ay2 - Bx2 = C where A and B are positive, and C is any number (might have to be nonzero?) Edit: Ah, not completely right. Or rather, technically correct but misleading. Redux: Circles: (x - A)2 + (y - B)2 = R2 (A and B are the X&Y coordinates of the center, R is the radius) Ellipses: ((x - A)2 / F2) + ((y - B)2 / G2) = 1 (A and B are again coordinates, F & G influence the size & shape of the ellipse). Hyperbolas: ((x - A)2 / F2) - ((y - B)2 / G2) = 1 (F and G are positive). [Quote] [Link]
 What abou having to draw both parts of the hyperbola and the asymptotes. My Geometry teacher and my Algerbra 2/Trig teacher failed to teach this thoroughly. [Quote] [Link]
 Hmm, I don't remember the right way to do it. I usually just plug in a couple values for x or y, plot them, and eyeball the asymptotes. Easier if you don't mind using your calculator to get a table of values. :b [Quote] [Link]
 eofpi Administrator656 Posts11 Cresco 14:2 - 19.45.23818.354 days ago
 Would some nice person be able to explain to me how to get an equation (let's say 3sinx + 4cosx) in the form Asin(x+a)? [Quote] [Link]
 A Tree said:Would some nice person be able to explain to me how to get an equation (let's say 3sinx + 4cosx) in the form Asin(x+a)? asinx+bcosx=sqrt(a2+b2)*sin((x+[The angle formed from the x-axis to point (a,b)])) 3sinx+4cosx=5sin(x+sin(4/5)) #You can use cos, tan, or sin for the angle [Quote] [Link]
 Umm.. don't you use auxiliary method? So let's say we have 3sinx + 4cosx. We need to put it in the form Asin(x+a)? Well, 3sinx + 4cosx = Asin(x+a) Asin(x+a) = A[sinxcosa + sinacosx] =Asinxcosa + Asinacosx You match the coefficients. So 3sinx matches with Asinxcosa, and 4cosx matches with Asinacosx. 3 is the coefficient of sinx, and so is Acosa. So Acosa=3 4 is the coefficient of cosx, and so is Asina. So Asina=4 You cancel out the A's and divide sina/cosa to get tana. So tana=4/3. Therefore a=53 degrees. So you have Asin(x+53) To find A you just root the sum of the square coefficients of the first equation (3sinx + 4cosx) so root(3^2+4^2) = 5. So A=5. Therefore 3sinx + 4cosx = 5sin(x+53). Sorry if it didn't make sense, best way i could explain. [Quote] [Link]
 I have a problem that's in a random textbook that my mom gave me: Find the area inside the weird nut-shaped thing. By the way, it's supposed to be perfectly curved, instead of wiggly like mine. EDIT: I forgot something. The curved thing is in a square. [Quote] [Link]
 I assumed they were quarter circles. page stretching solution image [Quote] [Link]
 Can somebody explain simply how to use the unit circle? [Quote] [Link]
 Hydrogen777 Administrator5990 Posts11 Vigeo 1:3 - 6.59.75805.997 days ago The unit circle is useful for seeing the values of sin(θ) and cos(θ) for a given angle θ. If you draw a line from the origin through the unit circle such that the line forms an angle θ with the positive x-axis, the line will intersect the unit circle at a point P. The coordinates of P are (cos(θ), sin(θ)). Usually, common angles and their corresponding points on the unit circle are graphed so that you can see the pattern of the values that the sine and cosine take as you move around the circle. [Quote] [Link]
 I somewhat remember that if you have a quadratic polynomial and one root, there is a simple relationship to the other root. Anybody recall what that is? I only ask because I am using my BAII plus to solve a quadratic and it quickly gives one root via the IRR button (after putting in the proper Cash Flow worksheet). [Quote] [Link]
 Use the sum or product of the roots? [Quote] [Link]
 Never mind. I found it: c/a=x1*x2 [Quote] [Link]
 Eitakxner said:Can someone please help me with this stupid algebra problem? I assume that x, y, and z need to work for all three of the presented problems. If that's the case, I think I give up now. I don't like working around fractions. [Quote] [Link]

Step 1 - Destroy the denominators

2x + 4 - 3y - 12 + z + 1 = 0
4x - 16 + 3y + 3 - 6z + 12 = -12
2x + 2 + 2y + z - 1 = 3

Step 2 - Move constant terms off to the right

2x - 3y + z = 7
4x + 3y - 6z = -11
2x + 2y + z = 2

Step 3 - Subtract the 3rd equation from the 1st

2x - 2x - 3y - 2y + z - z = 7 - 2
-5y = 5
y = -1

2x + z = 4
4x - 6z = -8

Step 5 - Subtract the 2nd equation from 2 times the 1st

4x - 4x + 2x + 6z = 8 + 8
8z = 16
z = 2

It's 1

x = 1
y = -1
z = 2

Yay.