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Gnesinka [82]
3 years ago
6

Triangle ABC is similar to triangle DEF. What is the value of x ?

Mathematics
1 answer:
Delicious77 [7]3 years ago
5 0
48382939392838338383839
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Read 2 more answers
Use series to verify that<br><br> <img src="https://tex.z-dn.net/?f=y%3De%5E%7Bx%7D" id="TexFormula1" title="y=e^{x}" alt="y=e^{
SVETLANKA909090 [29]

y = e^x\\\\\displaystyle y = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y= 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \frac{d}{dx}\left( 1+x+\frac{x^2}{2!} + \frac{x^3}{3!}+\frac{x^4}{4!}+\ldots\right)\\\\

\displaystyle y' = \frac{d}{dx}\left(1\right)+\frac{d}{dx}\left(x\right)+\frac{d}{dx}\left(\frac{x^2}{2!}\right) + \frac{d}{dx}\left(\frac{x^3}{3!}\right) + \frac{d}{dx}\left(\frac{x^4}{4!}\right)+\ldots\\\\\displaystyle y' = 0+1+\frac{2x^1}{2*1} + \frac{3x^2}{3*2!} + \frac{4x^3}{4*3!}+\ldots\\\\\displaystyle y' = 1 + x + \frac{x^2}{2!}+ \frac{x^3}{3!}+\ldots\\\\\displaystyle y' = \sum_{k=1}^{\infty}\frac{x^k}{k!}\\\\\displaystyle y' = e^{x}\\\\

This shows that y' = y is true when y = e^x

-----------------------

  • Note 1: A more general solution is y = Ce^x for some constant C.
  • Note 2: It might be tempting to say the general solution is y = e^x+C, but that is not the case because y = e^x+C \to y' = e^x+0 = e^x and we can see that y' = y would only be true for C = 0, so that is why y = e^x+C does not work.
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3 years ago
tensor-on-tensor regression: riemannian optimization, over-parameterization, statistical-computational gap and their interplay
Contact [7]

The purpose of the tensor-on-tensor regression, which we examine, is to relate tensor responses to tensor covariates with a low Tucker rank parameter tensor/matrix without being aware of its intrinsic rank beforehand.

By examining the impact of rank over-parameterization, we suggest the Riemannian Gradient Descent (RGD) and Riemannian Gauss-Newton (RGN) methods to address the problem of unknown rank. By demonstrating that RGD and RGN, respectively, converge linearly and quadratically to a statistically optimal estimate in both rank correctly-parameterized and over-parameterized scenarios, we offer the first convergence guarantee for the generic tensor-on-tensor regression. According to our theory, Riemannian optimization techniques automatically adjust to over-parameterization without requiring implementation changes.

Learn more about tensor-on-tensor here

brainly.com/question/16382372

#SPJ4

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