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algebra precalculus - Evaluating $\frac {1} {a^ {2025}}+\frac {1} {b . . .
Well, the image equation is a different equation? One has $\frac1 {2024}$ on the right, and the other has $2024$ on the right?
Evaluating $ \\lim_{x \\to 0} \\frac{e - (1 + 2x)^{1 2x}}{x} $ without . . .
The following is a question from the Joint Entrance Examination (Main) from the 09 April 2024 evening shift: $$ \lim_ {x \to 0} \frac {e - (1 + 2x)^ {1 2x}} {x} $$ is equal to: (A) $0$ (B) $\frac {-2} {
Evaluating $\iiint_B (x^2+y^2+z^2)dV$ where $B$ is the ball of radius . . .
The question asks to use spherical coords My answer is coming out wrong and symbolab is saying I'm evaluating the integrals correctly so my set up must be wrong Since $\\rho$ is the distance from
integration - Evaluating $\iiint z (x^2+y^2+z^2)^ {−3 2}\,dx\,dy\,dz . . .
Spherical Coordinate Homework Question Evaluate the triple integral of $f (x,y,z)=z (x^2+y^2+z^2)^ {−3 2}$ over the part of the ball $x^2+y^2+z^2\le 81$ defined by
Evaluating $\\prod_{n=1}^{\\infty}\\left(1+\\frac{1}{2^n}\\right)$
Compute:$$\prod_ {n=1}^ {\infty}\left (1+\frac {1} {2^n}\right)$$ I and my friend came across this product Is the product till infinity equal to $1$? If no, what is the answer?
Evaluating $\lim\limits_ {n\to\infty} e^ {-n} \sum\limits_ {k=0}^ {n . . .
I'm supposed to calculate: $$\\lim_{n\\to\\infty} e^{-n} \\sum_{k=0}^{n} \\frac{n^k}{k!}$$ By using WolframAlpha, I might guess that the limit is $\\frac{1}{2