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AlgebraQuestion and Answers: Page 29
Question Number 202984 Answers: 2 Comments: 2
$$\mathrm{If}\:\:\:\mathrm{x}^{\boldsymbol{\mathrm{x}}^{\mathrm{3}} } \:=\:\mathrm{3} \\ $$$$\mathrm{Find}\:\:\:\mathrm{x}\:=\:? \\ $$
Question Number 202964 Answers: 0 Comments: 0
$${just}\:{the}\:{exact}\:{real}\:{root}\:{please}: \\ $$$$\left({x}^{\mathrm{6}} −\mathrm{1}\right)\left({x}^{\mathrm{4}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{3}\right)+\mathrm{2}{x}\left(\mathrm{3}{x}^{\mathrm{4}} −\mathrm{1}\right)=\mathrm{0} \\ $$
Question Number 202873 Answers: 0 Comments: 1
$${if}\:{x}^{\mathrm{4}} −{x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{S}\:\&\:\mathrm{P}\:{are}\:+\:\&\:×\:{real}\:{roots} \\ $$$${what}\:{is}\:{value}\:{of}\:\mathrm{4}{P}^{\mathrm{2}} \:+{S}^{\mathrm{2}} ? \\ $$
Question Number 202868 Answers: 1 Comments: 2
Question Number 202759 Answers: 0 Comments: 0
$${Let}\:{A}=\left\{{x}\mid{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{3}<\mathrm{0},{x}\in{R}\right\},{B}=\left\{{x}\mid\mathrm{2}^{\mathrm{1}−{x}} +{a}\leqslant\mathrm{0},{x}^{\mathrm{2}} −\mathrm{2}\left({a}+\mathrm{7}\right){x}+\mathrm{5}\leqslant\mathrm{0},{x}\in{R}\right\} \\ $$$${If}\:{A}\subseteq{B}\:{Find}\:{the}\:{range}\:{of}\:{real}\:{number}\:{a} \\ $$
Question Number 202715 Answers: 2 Comments: 0
$$\begin{array}{|c|}{\:\:\underset{\mathrm{is}\:\mathrm{2025}} {\mathrm{Square}\:\mathrm{of}\:\mathrm{mean}\:\mathrm{of}\:\mathrm{two}\:\mathrm{numbers}}\:\:}\\\hline\end{array}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\& \\ $$$$\begin{array}{|c|}{\:\:\underset{\:\mathrm{is}\:\mathrm{2026}} {\mathrm{mean}\:\mathrm{of}\:\mathrm{squares}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}}\:\:}\\\hline\end{array} \\ $$$$\:\:\:\:\: \\ $$$$\begin{array}{|c|}{\mathrm{Find}\:\mathrm{the}\:\mathrm{product}\:\mathrm{of}\:\mathrm{the}\:\mathrm{numbers}.}\\\hline\end{array}\: \\ $$
Question Number 202726 Answers: 0 Comments: 4
$${pk}={b} \\ $$$${pq}+{sk}={c} \\ $$$${sq}={d} \\ $$$$\sqrt{{t}}=\frac{{t}−{q}}{{k}}=\frac{{t}−{s}}{{p}} \\ $$$${Given}\:{are}\:{b},\:{c},\:{d}.\:{Find}\:{t}. \\ $$
Question Number 202701 Answers: 0 Comments: 1
Question Number 202651 Answers: 1 Comments: 1
Question Number 202638 Answers: 1 Comments: 4
$$\sqrt{{x}−\frac{\mathrm{1}}{{x}}}−\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{x}}}=\mathrm{1}−\frac{\mathrm{1}}{{x}} \\ $$
Question Number 202633 Answers: 1 Comments: 0
Question Number 202616 Answers: 2 Comments: 0
$$\mathrm{If}\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{a}\:+\:\mathrm{b}}\:−\:\frac{\mathrm{1}\:−\:\mathrm{a}}{\mathrm{a}\:−\:\mathrm{b}}\:=\:\mathrm{x} \\ $$$$\mathrm{Find}\:\:\:\:\:\frac{\mathrm{1}\:−\:\mathrm{b}}{\mathrm{a}\:−\:\mathrm{b}}\:+\:\frac{\mathrm{b}}{\mathrm{a}\:+\:\mathrm{b}}\:=\:? \\ $$
Question Number 202605 Answers: 1 Comments: 3
Question Number 202604 Answers: 1 Comments: 0
$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{and}\: \\ $$$${y}\:=\:\frac{\sqrt{{a}\:+\:\mathrm{1}}\:−\:\sqrt{{a}\:−\:\mathrm{1}}}{\:\sqrt{{a}\:+\:\mathrm{1}}\:+\:\sqrt{{a}\:−\:\mathrm{1}}}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$$\frac{{x}^{\mathrm{2}} \:−\:{xy}\:+\:{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:+\:{xy}\:+\:{y}^{\mathrm{2}} }\:=\:\frac{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{3}}{\mathrm{4}{a}^{\mathrm{2}} \:−\:\mathrm{1}}\:. \\ $$
Question Number 202551 Answers: 0 Comments: 1
$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{i}} =\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{1}} +\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{2}} +...\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}_{\mathrm{1}} +\mathrm{y}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{y}_{\mathrm{2}} +...+\mathrm{x}_{\mathrm{n}} +\mathrm{y}_{\mathrm{n}} \: \\ $$$$\mathrm{please}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}\:? \\ $$$$ \\ $$
Question Number 202540 Answers: 1 Comments: 0
$$\mathrm{If}\:\frac{{b}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{3}{a}\:−\:{b}\:−\:{c}}{\mathrm{2}{b}\:+\:{c}\:−\:{a}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{\mathrm{2}{a}\:−\:{b}\:+\:\mathrm{3}{c}}\: \\ $$$$\left[{a}\:+\:{b}\:+\:{c}\:\neq\:\mathrm{0}\right]\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:{a}\:=\:{b}\:=\:{c}. \\ $$
Question Number 202535 Answers: 1 Comments: 0
$$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5}}\:=\mathrm{2} \\ $$$$\left(\mathrm{An}\:\mathrm{alteration}\:\mathrm{of}\:\mathrm{Q}#\mathrm{202500}\right) \\ $$
Question Number 202532 Answers: 0 Comments: 1
$${please}\:{help} \\ $$
Question Number 202500 Answers: 4 Comments: 0
Question Number 202497 Answers: 3 Comments: 0
$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{qx}\:+\:{p}\:=\:\mathrm{0}\:\left[{p}\:\neq\:{q}\right]\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{p}\:+\:{q}\:+\:\mathrm{1}\:=\:\mathrm{0}. \\ $$
Question Number 202477 Answers: 3 Comments: 0
$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0}. \\ $$
Question Number 202473 Answers: 0 Comments: 0
$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}^{\boldsymbol{\mathrm{n}}} \:\centerdot\:\mathrm{n}!}{\left(\mathrm{2n}\right)!}\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{x}\:+\:\mathrm{5}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\centerdot\:\mathrm{n}\:\centerdot\:\mathrm{ln}\left(\mathrm{n}\right)} \\ $$
Question Number 202468 Answers: 1 Comments: 0
$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\mathrm{16}}{\mathrm{16n}^{\mathrm{2}} \:−\:\mathrm{8n}\:−\:\mathrm{3}}\:=\:?\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{2n}^{\mathrm{3}} }\:=\:? \\ $$
Question Number 202464 Answers: 0 Comments: 0
Question Number 202459 Answers: 3 Comments: 0
$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:{lx}\:+\:{m}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${l}^{\mathrm{2}} \:+\:\mathrm{4}{m}^{\mathrm{2}} \:=\:\left(\mathrm{1}\:+\:\mathrm{2}{m}\right)^{\mathrm{2}} \:. \\ $$
Question Number 202449 Answers: 0 Comments: 0
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