Question

Finding inverse function of $y=\sqrt{x+\sqrt{2x-1}}$

score 90 · Answer 1 · 2026-04-15 12:23:15

90

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Finding inverse function of $y=\sqrt{x+\sqrt{2x-1}}$

Published on 15 Apr 2026 - 12:23

score 242 · Answer 2 · 2026-04-16 16:00:11

242

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Image of the function $y=\sqrt x$

Published on 16 Apr 2026 - 16:00

score 109 · Answer 3 · 2026-04-15 19:40:33

109

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Solve the following equation: $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$

Published on 15 Apr 2026 - 19:40

score 66 · Answer 4 · 2026-04-17 06:44:30

66

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How does $\frac{1}{z}-\frac{z}{\sqrt{L^2+z^2}\left(L+\sqrt{L^2+z^2}\right)}$ become $\frac{L}{z\sqrt{L^2+z^2}}$?

Published on 17 Apr 2026 - 6:44

score 134 · Answer 5 · 2026-04-15 20:32:54

134

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What is bigger, $\sqrt[3]{3}+\sqrt[3]{24}$ or $\sqrt[3]{2}+\sqrt[3]{54}\;$?

Published on 15 Apr 2026 - 20:32

score 102 · Answer 6 · 2026-04-17 14:49:58

102

Views

Simplest expression for cos(1°) predating calculus

Published on 17 Apr 2026 - 14:49

score 133 · Answer 7 · 2026-04-16 19:03:43

133

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Showing $ 2\sqrt{\frac{x+3}{x}}+8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{(x+1)^{3/2}(x+3)}{(x-1)^{5/2}}\right)\geq 10 $ for $x\geq7$

Published on 16 Apr 2026 - 19:03

score 109 · Answer 8 · 2026-03-25 06:01:25

109

Views

Solving $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$

Published on 25 Mar 2026 - 6:01

score 329 · Answer 9 · 2026-04-09 20:36:21

329

Views

What is $\sqrt{-36}?$

Published on 09 Apr 2026 - 20:36

score 73 · Answer 10 · 2026-03-25 19:02:38

73

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How to determine the sign of a square root in Euler's method?

Published on 25 Mar 2026 - 19:02

score 69 · Answer 11 · 2026-04-15 05:27:49

69

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Prove that if $a, b \in \mathbb{R}$ are positive, then $a \leq b$ implies $\sqrt a \leq \sqrt b$.

Published on 15 Apr 2026 - 5:27

score 55 · Answer 12 · 2026-04-12 08:43:24

55

Views

Are there always 2 solutions to the square root of a complex number?

Published on 12 Apr 2026 - 8:43

score 758 · Answer 13 · 2026-03-25 06:06:47

758

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If $\sqrt{a}\sqrt{b}=\sqrt{ab}$ only holds for positive real $a$ & $b$, then why can we say $\sqrt{-a}=\sqrt{-1\cdot a}=\sqrt{-1}\sqrt{a}=i\sqrt{a}$?

Published on 25 Mar 2026 - 6:06

score 34 · Answer 14 · 2026-04-14 19:22:10

"If $x^2>a$, then $(x-h)^2>a$ for some $h>0$" How should I modify this sentence? (Munkres "Topology 2nd Edition" Exercise 10(b) Section 4)

score 55 · Answer 15 · 2026-04-12 14:09:36

55

Views

Transforming graphs using matrices

Published on 12 Apr 2026 - 14:09

List Question

Finding inverse function of $y=\sqrt{x+\sqrt{2x-1}}$

Image of the function $y=\sqrt x$

Solve the following equation: $\sqrt{x^2+x}-\sqrt{x^2-x}=x+1$

How does $\frac{1}{z}-\frac{z}{\sqrt{L^2+z^2}\left(L+\sqrt{L^2+z^2}\right)}$ become $\frac{L}{z\sqrt{L^2+z^2}}$?

What is bigger, $\sqrt[3]{3}+\sqrt[3]{24}$ or $\sqrt[3]{2}+\sqrt[3]{54}\;$?

Simplest expression for cos(1°) predating calculus

Showing $ 2\sqrt{\frac{x+3}{x}}+8\sqrt{\frac{x+1}{x}}-\ln\left(\frac{(x+1)^{3/2}(x+3)}{(x-1)^{5/2}}\right)\geq 10 $ for $x\geq7$

Solving $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$

What is $\sqrt{-36}?$

How to determine the sign of a square root in Euler's method?

Prove that if $a, b \in \mathbb{R}$ are positive, then $a \leq b$ implies $\sqrt a \leq \sqrt b$.

Are there always 2 solutions to the square root of a complex number?

If $\sqrt{a}\sqrt{b}=\sqrt{ab}$ only holds for positive real $a$ & $b$, then why can we say $\sqrt{-a}=\sqrt{-1\cdot a}=\sqrt{-1}\sqrt{a}=i\sqrt{a}$?

"If $x^2>a$, then $(x-h)^2>a$ for some $h>0$" How should I modify this sentence? (Munkres "Topology 2nd Edition" Exercise 10(b) Section 4)

Transforming graphs using matrices

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