Mega_Cabbage
Meh
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Usually to find the x-intercept, you set y to 0 and solve for x.
Yes, that is what I have been doing and I still cannot solve for it.
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Mm, I'm trying to solve the problem right now, I'll see if I can get a solution to it.
The first step is to set y = 0, then use the natural logarithm to remove the exponents. The function will probably look more familiar to you once you've done this.
You could also try to graph the function with a graphing calculator or other program.
The first step is to set y = 0, then use the natural logarithm to remove the exponents. The function will probably look more familiar to you once you've done this.
You could also try to graph the function with a graphing calculator or other program.
Yeah, I graphed the equation and know the x-intercept, but I am trying to do so without a graphing calculator. I am trying to do your method, but it requires me taking ln of both sides, so I'd have to take the ln of 0, which is illegal
Yeah, I graphed the equation and know the x-intercept, but I am trying to do so without a graphing calculator. I am trying to do your method, but it requires me taking ln of both sides, so I'd have to take the ln of 0, which is illegal
Bring - 3^-x - 4x to the other side so you can take natural log of that.
Sorry~! I made a mistake when copying the problem down. The real function is
y = e^x - 3e^-x - 4x
Anyways, I set the equation to equal 0.
0 = e^x - 3x^-x - 4x.
I then added -3x^-x - 4x to the other side.
3x^-x + 4x = e^x
ln(3x^-x + 4x) = x
ln((3+4x*e^x)/(e^x)) = x
ln(3+4x*e^x) - ln(e^x) = x
ln(3+4x*e^x) = 2x
Not sure what to do from here...
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Can anyone help me on this? Thanks!
EpicGamer12
Senior Member
Alright folks, I need help with minimizing functions. This is boolean algebra I believe and I think we are using the consensus theorem but if you know how to solve it regardless please let me know.
A + A'B + BC + BC'
The apostrophe obviously indicates the opposite of the value of that variable. If A was 1, A' would be 0.
edit: I figured it out because I'm in Computer Science boooois
A + A'B + BC + BC'
The apostrophe obviously indicates the opposite of the value of that variable. If A was 1, A' would be 0.
edit: I figured it out because I'm in Computer Science boooois
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Okay, so, let's talk Conics for a moment.
General overall form:
Ax^2 + Bxy + Cy^2 + Dx + Ey + F
Okay, now, with that out of the way, I've been looking all over for the "Standard Equations" of the following with the Bxy term included to see if it, well, exist:
NOTE: Brackets ( {} ) here are used like parentheses in conventional English.
Circles: Ax^2 + {Bxy +} Ay^2 + Dx + Ey + F
Ellipses: Ax^2 + {Bxy +} + Cy^2 + Dx ^ Ey + F
Hyperbolas: Ax^2 {+ Bxy} - Cy^2 + Dx + Ey + F
Parabolas: Ax^2 {+ Bxy} + Dx + Ey + F
First, I'd just like to know if these are all correct withing the parameters of the general form without Bxy. When including the part in the brackets, would those equations still be correct?
I can't seem to discover if they exist or not. Also, the Bxy term, if included, would have something to do with the rotation of the cone before intersected by the plane, right?
Please don't just say, "look in your textbook lol;" I'm teaching this to myself because I want to test out of a few grades of math at the end of the year and therefore I have to teach myself when given the state standards or not be taught at all.
EDIT: Well, found this with the oblique ellipse so I'm going to assume yes to both of my questions. If someone would like to confirm anyways, go for it.
General overall form:
Ax^2 + Bxy + Cy^2 + Dx + Ey + F
Okay, now, with that out of the way, I've been looking all over for the "Standard Equations" of the following with the Bxy term included to see if it, well, exist:
NOTE: Brackets ( {} ) here are used like parentheses in conventional English.
Circles: Ax^2 + {Bxy +} Ay^2 + Dx + Ey + F
Ellipses: Ax^2 + {Bxy +} + Cy^2 + Dx ^ Ey + F
Hyperbolas: Ax^2 {+ Bxy} - Cy^2 + Dx + Ey + F
Parabolas: Ax^2 {+ Bxy} + Dx + Ey + F
First, I'd just like to know if these are all correct withing the parameters of the general form without Bxy. When including the part in the brackets, would those equations still be correct?
I can't seem to discover if they exist or not. Also, the Bxy term, if included, would have something to do with the rotation of the cone before intersected by the plane, right?
Please don't just say, "look in your textbook lol;" I'm teaching this to myself because I want to test out of a few grades of math at the end of the year and therefore I have to teach myself when given the state standards or not be taught at all.
EDIT: Well, found this with the oblique ellipse so I'm going to assume yes to both of my questions. If someone would like to confirm anyways, go for it.
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@ Blue Rose
Any circle in the plane can be factored in the form (x - h)^2 + (y - k)^2 = r^2 where (h, k) is the center of the circle and r is the radius. If you expand this formula out you get
x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2
x^2 + y^2 + (-2h)x + (-2k)y + (h^2 + k^2 - r^2) = 0
Of course, you can multiply this equation by any number a without changing the circle:
ax^2 + ay^2 + (-2ah)x +(-2ak)y +(ah^2 + ak^2 - ar^2) = 0
Setting A = a, B = 0, C = a, D = -2ah, E = -2ak, and F = (ah^2 + ak^2 - ar^2) gives:
Ax^2 + Cy^2 + Dx + Ey + F = 0
Ax^2 + Ay^2 + Dx + Ey + F = 0
So indeed you are correct for the circle. Adding Bxy with B ≠ 0 would be incorrect though.
I'm not entirely sure about the other equations, but they're likely correct. If I recall and like you mentioned, the Bxy term affects the slant of the conic (circles don't have slants which makes the case above easier xP), however, you can verify the equations of unslanted parabolas, hyperbolas, and ellipses by expanding out the following equations:
Parabolas: y = (x-h)^2 + k and x = (y - h)^2 + k
Ellipses: (x - h)^2/a^2 + (y - k)^2/b^2 = 1
Hyperbolas: (x - h)^2/a^2 - (y - k)^2/b^2 = 1
None of them should include a Bxy term because these are standard unslanted formulas for parabolas, ellipses, and hyperbolas.
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I have a question if anyone can answer it!
Is it true that for every a > b > 1 that ln(a) - ln(b) ≤ a - b? If someone could find a source of this fact (or a counter example showing it is false) I'd greatly appreciate it!
Found that the answer is yes xP.
Any circle in the plane can be factored in the form (x - h)^2 + (y - k)^2 = r^2 where (h, k) is the center of the circle and r is the radius. If you expand this formula out you get
x^2 - 2hx + h^2 + y^2 - 2ky + k^2 = r^2
x^2 + y^2 + (-2h)x + (-2k)y + (h^2 + k^2 - r^2) = 0
Of course, you can multiply this equation by any number a without changing the circle:
ax^2 + ay^2 + (-2ah)x +(-2ak)y +(ah^2 + ak^2 - ar^2) = 0
Setting A = a, B = 0, C = a, D = -2ah, E = -2ak, and F = (ah^2 + ak^2 - ar^2) gives:
Ax^2 + Cy^2 + Dx + Ey + F = 0
Ax^2 + Ay^2 + Dx + Ey + F = 0
So indeed you are correct for the circle
. Adding Bxy with B ≠ 0 would be incorrect though.I'm not entirely sure about the other equations, but they're likely correct. If I recall and like you mentioned, the Bxy term affects the slant of the conic (circles don't have slants which makes the case above easier xP), however, you can verify the equations of unslanted parabolas, hyperbolas, and ellipses by expanding out the following equations:
Parabolas: y = (x-h)^2 + k and x = (y - h)^2 + k
Ellipses: (x - h)^2/a^2 + (y - k)^2/b^2 = 1
Hyperbolas: (x - h)^2/a^2 - (y - k)^2/b^2 = 1
None of them should include a Bxy term because these are standard unslanted formulas for parabolas, ellipses, and hyperbolas.
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I have a question if anyone can answer it!
Is it true that for every a > b > 1 that ln(a) - ln(b) ≤ a - b? If someone could find a source of this fact (or a counter example showing it is false) I'd greatly appreciate it
!Found that the answer is yes xP.
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What's 0/0? It could be "not defined", since anything divided by a zero does not have a defined value. On the other hand, it also follows a 1:1 ratio. Just a couple possibilities.
I know the answer, just feel like knowing what you guys think c:
- - - Post Merge - - -
Oh, by the way, try asking Siri that.
I know the answer, just feel like knowing what you guys think c:
- - - Post Merge - - -
Oh, by the way, try asking Siri that.
Often times 0/0 is said to be undefined because defining 0/0 would break math xD.
That said, there are some contexts in which 0/0 can be defined and is locally consistent, but in those cases, 0/0 is often treated more as a symbol representation of a concept similarly to how infinity is not a number but rather just a concept of an upper bound to all real numbers.
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