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Question Number 202761 Answers: 1 Comments: 0
Question Number 202760 Answers: 1 Comments: 0
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 202750 Answers: 2 Comments: 1
$${place}\left(\mathrm{1}\rightarrow\mathrm{25}\right) \\ $$$${in}\:{table}\left(\mathrm{5}×\mathrm{5}\right){in}\:{such}\:{away}\:{that} \\ $$$${the}\:{sum}\:{is}\:{constant}\:{in}\:{all}\:{directions}. \\ $$$$\left[\:\mid\underset{−} {\overset{−} {×}}\mid\:\right]\rightarrow\left({direction}\right) \\ $$
Question Number 202749 Answers: 1 Comments: 3
Question Number 202747 Answers: 1 Comments: 0
$$\mathrm{Series}\:\:\Sigma\:\frac{\left(−\mathrm{1}\right)^{{h}−\mathrm{1}} }{{p}_{{h}} }\:\mathrm{is}\:\mathrm{Converge}??\:{h}\in\mathbb{Z}^{+} \:,\:{p}_{{h}} \in\mathbb{P}^{+} \\ $$$$\mathbb{P}^{+} =\:\mathrm{2}\:,\:\mathrm{3}\:,\:\mathrm{5}\:,\:\mathrm{7}\:,\:\mathrm{11}\:,\:\mathrm{13}\:,\:....\: \\ $$
Question Number 202740 Answers: 0 Comments: 0
Question Number 202737 Answers: 1 Comments: 0
$$\:\:\:\:\:\:\:\:\:\begin{array}{|c|}{\underset{\overset{\overset{\mathrm{2}} {\mathrm{0}}} {\mathrm{24}}} {\mathbb{H}^{\mathbb{A}^{\mathbb{P}^{\mathbb{P}^{\mathbb{Y}^{\mathbb{N}^{\mathbb{E}} \mathbb{W}} \mathbb{Y}} \mathbb{E}} \mathbb{A}} \mathbb{R}} \:!}}\\\hline\end{array} \\ $$
Question Number 202723 Answers: 0 Comments: 0
Question Number 202721 Answers: 0 Comments: 0
$$\:\:\:\begin{array}{|c|}{\underset{\mathcal{HAPPY}\:\mathcal{NEW}\:\mathcal{YEAR}!} {\mathrm{2}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} +\mathrm{6}^{\mathrm{2}} +\centerdot\centerdot\centerdot\centerdot\centerdot+\mathrm{22}^{\mathrm{2}} =\mathrm{2024}}}\\\hline\end{array} \\ $$$$\begin{array}{|c|}{\:\:\:\mathrm{May}\:\mathrm{2024}\:\mathrm{be}\:\begin{cases}{\mathrm{war}-\mathrm{free}!}\\{\&}\\{{peaceful}!}\end{cases}\:\:\:}\\\hline\end{array} \\ $$$$ \\ $$
Question Number 202716 Answers: 1 Comments: 0
$$\overline {\:\:{abcd}\:\:}{is}\:{a}\:{four}\:{digit}\:{number} \\ $$$${such}\:{that}\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} =\overline {\:{cd}\:} \\ $$$${and}\:\overline {\:{cd}\:}−\overline {\:{d}\:}=\overline {\:{ab}\:}. \\ $$$$\mathcal{F}{ind}\:{the}\:{number}. \\ $$
Question Number 202731 Answers: 1 Comments: 0
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({x}\right){ln}\left(\mathrm{1}−{x}\right)}{\:{x}\sqrt{\mathrm{1}−{x}}}{dx} \\ $$
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 202698 Answers: 1 Comments: 0
$${determine}\:{le}\:{reste}\:{de}\:{la}\:{division}\:{eucludienne}\:{de}\:\mathrm{2023}^{\mathrm{2019}} {par}\:\mathrm{13} \\ $$
Question Number 202694 Answers: 3 Comments: 0
Question Number 202684 Answers: 1 Comments: 0
Question Number 202679 Answers: 1 Comments: 0
Question Number 202677 Answers: 2 Comments: 0
$$ \\ $$$$\mathrm{If}\:{x}^{\mathrm{2}} \:−\:{y}^{\mathrm{2}} \:=\:\mathrm{2023}^{\mathrm{2024}} \:\&\:{x},\:{y}\:\in\:\boldsymbol{\mathrm{N}} \\ $$$$\mathrm{how}\:\mathrm{many}\:\mathrm{set}\left\{{x},\:{y}\right\} \\ $$
Question Number 202672 Answers: 3 Comments: 0
$$\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{divisors}\:\mathrm{of}\:\mathrm{24}\:\mathrm{is}\:\mathrm{60}. \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{the}\:\mathrm{divisors}\:\mathrm{of}\:\mathrm{24}×\mathrm{79}. \\ $$
Question Number 202653 Answers: 0 Comments: 0
Question Number 202651 Answers: 1 Comments: 1
Question Number 202650 Answers: 1 Comments: 0
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 202636 Answers: 0 Comments: 0
$$\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\:^{\mathrm{3}} \sqrt{\mathrm{4}^{\mathrm{5}−\mathrm{x}} }}{\int_{\mathrm{4}} ^{\mathrm{6}} \left(\mathrm{x}−\mathrm{1}\right){dx}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} }\:,\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underline{\mathrm{Solution}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{4}^{\frac{\mathrm{5}−\mathrm{x}}{\mathrm{3}}} }{\int_{\mathrm{4}} ^{\mathrm{6}} \left(\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{x}+\mathrm{k}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\mathrm{2}\bullet\frac{\mathrm{5}−\mathrm{x}}{\mathrm{3}}} }{\left(\frac{\mathrm{6}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{6}+\mathrm{k}\right)−\left(\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{2}}−\mathrm{4}+\mathrm{k}\right)}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\frac{\mathrm{36}}{\mathrm{2}}−\mathrm{6}+\cancel{\mathrm{k}}−\frac{\mathrm{16}}{\mathrm{2}}+\mathrm{4}−\cancel{\mathrm{k}}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\mathrm{18}−\mathrm{6}−\mathrm{8}+\mathrm{4}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} }{\mathrm{8}}\:=\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2x}−\mathrm{1}} }\:\:\left(\mathrm{Cross}\:\mathrm{Multiply}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}} ×\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{8}×\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}} ×\mathrm{2}^{\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{2}^{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}−\mathrm{1}+\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}} \:=\:\mathrm{2}^{\mathrm{3}} \:\left(\mathrm{Since},\:\mathrm{the}\:\mathrm{bases}\:\mathrm{are}\:\mathrm{equal}.\:\mathrm{Then},\:\mathrm{we}\:\mathrm{can}\:\mathrm{equate}\:\mathrm{the}\:\mathrm{exponents}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2x}−\mathrm{1}+\frac{\mathrm{10}−\mathrm{2x}}{\mathrm{3}}\:=\:\mathrm{3}\:\left(\mathrm{Multiply}\:\mathrm{each}\:\mathrm{term}\:\mathrm{by}\:\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}\left(\mathrm{2x}\right)−\mathrm{3}\left(\mathrm{1}\right)+\cancel{\mathrm{3}}\left(\frac{\mathrm{10}−\mathrm{2x}}{\cancel{\mathrm{3}}}\right)\:=\:\mathrm{3}\left(\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{6x}−\mathrm{3}+\mathrm{10}−\mathrm{2x}\:=\:\mathrm{9}\:\left(\mathrm{Collect}\:\mathrm{Like}\:\mathrm{Terms}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}+\mathrm{7}\:=\:\mathrm{9} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}\:=\:\mathrm{9}−\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{4x}\:=\:\mathrm{2}\:\left(\mathrm{Divide}\:\mathrm{Both}\:\mathrm{Sides}\:\mathrm{by}\:\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\cancel{\mathrm{4}x}}{\cancel{\mathrm{4}}}\:=\:\frac{\mathrm{2}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\therefore\:\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$
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