Recent content by dragonblood
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Geometry Problem: Solving w/o Extra Variable?
Thanks for making an effort! It's always rewarding even if we don't solve exactly everything :)- dragonblood
- Post #16
- Forum: Precalculus Mathematics Homework Help
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Geometry Problem: Solving w/o Extra Variable?
Use the green triangle how? I don't know the short side or the hypothenus of it. I can't know how to aim the angle to the zero point 'inside' the mountain...- dragonblood
- Post #14
- Forum: Precalculus Mathematics Homework Help
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Geometry Problem: Solving w/o Extra Variable?
I would very much like you to explain this in detail (with drawings). Else I'm not quite sure I understand your method :P- dragonblood
- Post #12
- Forum: Precalculus Mathematics Homework Help
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Geometry Problem: Solving w/o Extra Variable?
No, it's not that easy unfortunatel. I don't know the distance to the zero point. In fact, I don't know any distances at all. Only the height of the two mountains. And the point was to use angles along with the mountain heights to solve the problem...- dragonblood
- Post #10
- Forum: Precalculus Mathematics Homework Help
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Geometry Problem: Solving w/o Extra Variable?
The point of the problem is this: Let's say you outside taking a hike somewhere. You can see the summit of to mountains (which you know the height of), and you're wondering how high above sea-level you are right now. The only tools you have is some kind of instrument making you able to measure...- dragonblood
- Post #8
- Forum: Precalculus Mathematics Homework Help
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Geometry Problem: Solving w/o Extra Variable?
I'm sorry. You're confusing me...- dragonblood
- Post #6
- Forum: Precalculus Mathematics Homework Help
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Geometry Problem: Solving w/o Extra Variable?
I'm leaning towards there not being a solution. However, if you perform two measurements, with the distance between measuring points known, the angles could tell you the distance to either mountain. Then it would be possible!- dragonblood
- Post #4
- Forum: Precalculus Mathematics Homework Help
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Geometry Problem: Solving w/o Extra Variable?
I can measure the angle between top A and B, but that's about it. I don't know any other length in that triangle. So I would say no. But I'm not sure...- dragonblood
- Post #3
- Forum: Precalculus Mathematics Homework Help
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Geometry Problem: Solving w/o Extra Variable?
Homework Statement I want to know if the problem needs another variable, or if it is possible to solve as it is. Homework Equations See figure in image included: You are a point x, some height h above zero-level. The two mountains have known height 500m and 400m. But you don't know how far...- dragonblood
- Thread
- Geometry
- Replies: 15
- Forum: Precalculus Mathematics Homework Help
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Exponential equation - how to use ln?
Aha! So i get a simple second degree equation: 64x^{2}-50x+4=0 Thanks :) But I'm thinking now that these numbers were 'convenient'...are there other ways to solve this if the numbers don't play along this nicely?- dragonblood
- Post #3
- Forum: Calculus and Beyond Homework Help
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Exponential equation - how to use ln?
Homework Statement Problem: find t in the following equationHomework Equations 64000e^{-1600t}+4000e^{-400t}=50000e^{-1000t} The Attempt at a Solution I know the answer: t=6.17\cdot\;10^{-4}s. But I'm struggling with how to get there. This is my attempt: Factorizing down to...- dragonblood
- Thread
- Exponential Ln
- Replies: 3
- Forum: Calculus and Beyond Homework Help
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Undergrad Struggling with Calculating an Integral: Can Someone Help?
Ah, yes. That is much easier :) Thanks- dragonblood
- Post #6
- Forum: Calculus
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Undergrad Struggling with Calculating an Integral: Can Someone Help?
Nevermind. I did it now. It was a simple mistake on my part. The integral should look like: \int_0^1{\frac{2u}{1+u}du} after substitution and then it should be calculated with integration by parts from there.- dragonblood
- Post #2
- Forum: Calculus
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Undergrad Struggling with Calculating an Integral: Can Someone Help?
I'm having trouble calculating an integral: \int_0^1{\frac{1}{1+\sqrt{x}}dx} I decided to do a substitution: u=\sqrt{x} du=\frac{1}{2\sqrt{x}}dx=\frac{1}{2u}dx thus making the integral look like this: \int_0^1{\frac{1}{2u}\cdot\frac{1}{1+u}du} I transformed this integral to...- dragonblood
- Thread
- Integral
- Replies: 5
- Forum: Calculus
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Undergrad First order DE with exponential
Thank you very much!- dragonblood
- Post #3
- Forum: Differential Equations