Question-228417

Question Number 228417 by s12345 last updated on 12/Apr/26

Commented by TonyCWX last updated on 12/Apr/26

First of all, could you translate it to English so that more users can attempt it? Secondly, this is not a homework solving forum.

$$\boldsymbol{\mathrm{First}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{all}},\:\boldsymbol{\mathrm{could}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{translate}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{English}}\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{that}}\:\boldsymbol{\mathrm{more}}\:\boldsymbol{\mathrm{users}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{attempt}}\:\boldsymbol{\mathrm{it}}? \\ $$$$ \\ $$$$\boldsymbol{\mathrm{Secondly}},\:\boldsymbol{\mathrm{this}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{not}}\:\boldsymbol{\mathrm{a}}\:\boldsymbol{\mathrm{homework}}\:\boldsymbol{\mathrm{solving}}\:\boldsymbol{\mathrm{forum}}. \\ $$

Commented by fantastic2 last updated on 12/Apr/26

$${i}\:{couldnt}\:{understand}\:{the}\:\mathrm{5}{th}\:{q} \\ $$$$ \\ $$

Commented by who133 last updated on 17/Apr/26

$${pls} \\ $$

Commented by who133 last updated on 17/Apr/26

$${plsineedatoughproblemonlogarithm} \\ $$

Answered by fantastic2 last updated on 12/Apr/26

1) highest power=2 general form⇒ax^2 +bx+c=0(a≠0) 2)let =p x=((p×x×x)/(100))⇒p=((100)/x) 3) (a/2)=(l/3)=(c/4)=((2a−3l+4c)/p)=k ∴a=2k . l=3k . c=4k ((2a−3l+4c)/p)=k ⇒((4k−9k+16k)/p)=k⇒p=11 4) p+q=(√(13)) ,p−q=(√5) ⇒(p+q)^2 −(p−q)^2 =13−5 ⇒4pq=8 ⇒pq=2 5) 6) r% is the compund interest rate 441=p(1+(r/(100)))^2 441=400(1+(r/(100)))^2 1+(r/(100))=((21)/(20)) (r/(100))=(1/(20)) ∴r=5% 7)x^2 −22x+105=0 ∴−(α+β)=−22⇒α+β=22 αβ=105 (√((α+β)^2 −4αβ))=α−β (√(564−420))=(√(144))=±12 ⇒α−β=±12 8)(1/( (√2)+1))+(1/( (√3)+(√2)))+(1/( (√4)+(√3))) =(√2)−1+(√3)−(√2)+(√4)−(√3) =(√4)−1 =1

$$\left.\mathrm{1}\right)\:{highest}\:{power}=\mathrm{2} \\ $$$${general}\:{form}\Rightarrow{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\left({a}\neq\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right){let}\: ={p}\: \\ $$$${x}=\frac{{p}×{x}×{x}}{\mathrm{100}}\Rightarrow{p}=\frac{\mathrm{100}}{{x}} \\ $$$$\left.\mathrm{3}\right)\:\frac{{a}}{\mathrm{2}}=\frac{{l}}{\mathrm{3}}=\frac{{c}}{\mathrm{4}}=\frac{\mathrm{2}{a}−\mathrm{3}{l}+\mathrm{4}{c}}{{p}}={k} \\ $$$$\therefore{a}=\mathrm{2}{k}\:.\:{l}=\mathrm{3}{k}\:.\:{c}=\mathrm{4}{k} \\ $$$$\frac{\mathrm{2}{a}−\mathrm{3}{l}+\mathrm{4}{c}}{{p}}={k} \\ $$$$\Rightarrow\frac{\mathrm{4}{k}−\mathrm{9}{k}+\mathrm{16}{k}}{{p}}={k}\Rightarrow{p}=\mathrm{11} \\ $$$$\left.\mathrm{4}\right)\:{p}+{q}=\sqrt{\mathrm{13}}\:,{p}−{q}=\sqrt{\mathrm{5}} \\ $$$$\Rightarrow\left({p}+{q}\right)^{\mathrm{2}} −\left({p}−{q}\right)^{\mathrm{2}} =\mathrm{13}−\mathrm{5} \\ $$$$\Rightarrow\mathrm{4}{pq}=\mathrm{8} \\ $$$$\Rightarrow{pq}=\mathrm{2} \\ $$$$\left.\mathrm{5}\right) \\ $$$$\left.\mathrm{6}\right)\:{r\%}\:{is}\:{the}\:{compund}\:{interest}\:{rate} \\ $$$$\mathrm{441}={p}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{2}} \\ $$$$\mathrm{441}=\mathrm{400}\left(\mathrm{1}+\frac{{r}}{\mathrm{100}}\right)^{\mathrm{2}} \\ $$$$\mathrm{1}+\frac{{r}}{\mathrm{100}}=\frac{\mathrm{21}}{\mathrm{20}} \\ $$$$\frac{{r}}{\mathrm{100}}=\frac{\mathrm{1}}{\mathrm{20}} \\ $$$$\therefore{r}=\mathrm{5\%} \\ $$$$\left.\mathrm{7}\right){x}^{\mathrm{2}} −\mathrm{22}{x}+\mathrm{105}=\mathrm{0} \\ $$$$\therefore−\left(\alpha+\beta\right)=−\mathrm{22}\Rightarrow\alpha+\beta=\mathrm{22} \\ $$$$\alpha\beta=\mathrm{105} \\ $$$$\sqrt{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta}=\alpha−\beta \\ $$$$\sqrt{\mathrm{564}−\mathrm{420}}=\sqrt{\mathrm{144}}=\pm\mathrm{12} \\ $$$$\Rightarrow\alpha−\beta=\pm\mathrm{12} \\ $$$$\left.\mathrm{8}\right)\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{1}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}}+\sqrt{\mathrm{3}}} \\ $$$$=\sqrt{\mathrm{2}}−\mathrm{1}+\sqrt{\mathrm{3}}−\sqrt{\mathrm{2}}+\sqrt{\mathrm{4}}−\sqrt{\mathrm{3}} \\ $$$$=\sqrt{\mathrm{4}}−\mathrm{1} \\ $$$$=\mathrm{1} \\ $$

Commented by AgniMath last updated on 15/Apr/26

a+(a/2)=((3a)/2) Prev area = 6a^2 ⇒ Final area = 6a^2 .(9/4) Percentage increase = ((6a^2 ((9/4)−1))/(6a^2 )) × 100% = 125%

$${a}+\frac{{a}}{\mathrm{2}}=\frac{\mathrm{3}{a}}{\mathrm{2}} \\ $$$${Prev}\:{area}\:=\:\mathrm{6}{a}^{\mathrm{2}} \\ $$$$\Rightarrow\:{Final}\:{area}\:=\:\mathrm{6}{a}^{\mathrm{2}} .\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${Percentage}\:{increase}\:= \\ $$$$\frac{\mathrm{6}{a}^{\mathrm{2}} \left(\frac{\mathrm{9}}{\mathrm{4}}−\mathrm{1}\right)}{\mathrm{6}{a}^{\mathrm{2}} }\:×\:\mathrm{100\%}\:=\:\mathrm{125\%} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply