Question and Answers Forum

I think of a two digit number.The sum of the digits is 9. When the number is reversed and subtracted from the original, the result is 45. Find the original number

$$\mathrm{I}\:\mathrm{think}\:\mathrm{of}\:\mathrm{a}\:\mathrm{two}\:\mathrm{digit}\:\mathrm{number}.\mathrm{The}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{digits}\:\mathrm{is}\:\mathrm{9}. \\ $$$$\mathrm{When}\:\mathrm{the}\:\mathrm{number}\:\mathrm{is}\:\mathrm{reversed}\:\mathrm{and}\:\mathrm{subtracted}\:\mathrm{from}\:\mathrm{the} \\ $$$$\mathrm{original},\:\mathrm{the}\:\mathrm{result}\:\mathrm{is}\:\mathrm{45}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{original}\:\mathrm{number} \\ $$

Question Number 26153 Answers: 1 Comments: 0

ratio of income of two persons is 9 is to 7.ratio of their expenses is 4 is to 3 .every person saves rupees 200. find income of each.

$${ratio}\:{of}\:{income}\:{of}\:{two}\:{persons}\:{is} \\ $$$$\mathrm{9}\:{is}\:{to}\:\mathrm{7}.{ratio}\:{of}\:{their}\:{expenses} \\ $$$${is}\:\mathrm{4}\:{is}\:{to}\:\mathrm{3}\:.{every}\:{person}\:{saves}\: \\ $$$${rupees}\:\mathrm{200}.\:{find}\:{income}\:{of}\:{each}. \\ $$

Question Number 26147 Answers: 1 Comments: 0

There are 5 more girls than boys in a class. If 2 boys join the class, the ratio of girls to boys will be 5:4. Find the number of of girls in the class.

$$\mathrm{There}\:\mathrm{are}\:\mathrm{5}\:\mathrm{more}\:\mathrm{girls}\:\mathrm{than}\:\mathrm{boys}\:\mathrm{in}\:\mathrm{a}\:\mathrm{class}.\:\mathrm{If}\:\mathrm{2}\:\mathrm{boys}\:\mathrm{join} \\ $$$$\mathrm{the}\:\mathrm{class},\:\mathrm{the}\:\mathrm{ratio}\:\mathrm{of}\:\mathrm{girls}\:\mathrm{to}\:\mathrm{boys}\:\mathrm{will}\:\mathrm{be}\:\mathrm{5}:\mathrm{4}.\:\mathrm{Find}\:\mathrm{the} \\ $$$$\mathrm{number}\:\mathrm{of}\:\mathrm{of}\:\mathrm{girls}\:\mathrm{in}\:\mathrm{the}\:\mathrm{class}. \\ $$

Question Number 26143 Answers: 2 Comments: 1

2000^(3000) vs 3000^(2000) who is stronger ?

$$\mathrm{2000}^{\mathrm{3000}} \:\:\boldsymbol{{vs}}\:\mathrm{3000}^{\mathrm{2000}} \\ $$$$ \\ $$$$\:\boldsymbol{{who}}\:\boldsymbol{{is}}\:\boldsymbol{{stronger}}\:? \\ $$

Question Number 26133 Answers: 1 Comments: 1

find the value of (C_n ^(0 ) )^2 +(C_n ^1 )^2 +(C_n ^2 )^2 +...(C_n ^n )^2 .

$${find}\:{the}\:{value}\:{of}\:\:\left({C}_{{n}} ^{\mathrm{0}\:\:} \right)^{\mathrm{2}} \:+\left({C}_{{n}} ^{\mathrm{1}} \right)^{\mathrm{2}} \:+\left({C}_{{n}} ^{\mathrm{2}} \right)^{\mathrm{2}} \:+...\left({C}_{{n}} ^{{n}} \right)^{\mathrm{2}} . \\ $$

Question Number 26117 Answers: 1 Comments: 0

(1/6)(√((3log1728)/(1+(1/2)log36+(1/3)log8))) simplify the question above

$$\frac{\mathrm{1}}{\mathrm{6}}\sqrt{\frac{\mathrm{3}{log}\mathrm{1728}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{log}\mathrm{36}+\frac{\mathrm{1}}{\mathrm{3}}{log}\mathrm{8}}} \\ $$$${simplify}\:{the}\:{question}\:{above} \\ $$

Question Number 26115 Answers: 1 Comments: 0

x^2 +(1/x^2 )=3 thrn find the valu of( x−(1/x))^2

$$\mathrm{x}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{3}\:\mathrm{thrn}\:\mathrm{find}\:\mathrm{the}\:\mathrm{valu}\:\mathrm{of}\left(\:\mathrm{x}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} \\ $$

Question Number 26067 Answers: 2 Comments: 0

Find the value of ((2 + 3^2 )/(1! + 2! + 3! + 4!)) + ((3 + 4^2 )/(2! + 3! + 4! + 5!)) + ... + ((2013 + 2014^2 )/(2012! + 2013! + 2014! + 2015!))

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{2}\:+\:\mathrm{3}^{\mathrm{2}} }{\mathrm{1}!\:+\:\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!}\:+\:\frac{\mathrm{3}\:+\:\mathrm{4}^{\mathrm{2}} }{\mathrm{2}!\:+\:\mathrm{3}!\:+\:\mathrm{4}!\:+\:\mathrm{5}!}\:+\:...\:+\:\frac{\mathrm{2013}\:+\:\mathrm{2014}^{\mathrm{2}} }{\mathrm{2012}!\:+\:\mathrm{2013}!\:+\:\mathrm{2014}!\:+\:\mathrm{2015}!} \\ $$

Question Number 26053 Answers: 1 Comments: 0

If x^2 + 9x + 2 = 0 and x^2 + kx + 5 = 0 have a common root, show that 2k^2 + 63k − 414 = 0 , hence find the value of k such that k > 9.3

$$\mathrm{If}\:\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{9x}\:+\:\mathrm{2}\:=\:\mathrm{0}\:\:\mathrm{and}\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{kx}\:+\:\mathrm{5}\:=\:\mathrm{0}\:\:\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root},\:\mathrm{show}\:\mathrm{that}\: \\ $$$$\mathrm{2k}^{\mathrm{2}} \:+\:\mathrm{63k}\:−\:\mathrm{414}\:=\:\mathrm{0}\:,\:\:\mathrm{hence}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\:\mathrm{k}\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{k}\:>\:\mathrm{9}.\mathrm{3} \\ $$

Question Number 26042 Answers: 0 Comments: 2

ax^2 −bx=0

$${ax}^{\mathrm{2}} −{bx}=\mathrm{0} \\ $$

Question Number 25971 Answers: 1 Comments: 2

In finding the equations of the bisectors of the angles between two lines a_1 x+b_1 y+c_1 =0 and a_2 x+b_2 y+c_2 =0, why we observe a_1 a_2 +b_1 b_2 >0 or <0 for obtuse and acute angle bisectors?

$${In}\:{finding}\:{the}\:{equations}\:{of}\:{the} \\ $$$${bisectors}\:{of}\:{the}\:{angles}\:{between}\:{two} \\ $$$${lines}\:{a}_{\mathrm{1}} {x}+{b}_{\mathrm{1}} {y}+{c}_{\mathrm{1}} =\mathrm{0}\:{and}\:{a}_{\mathrm{2}} {x}+{b}_{\mathrm{2}} {y}+{c}_{\mathrm{2}} =\mathrm{0}, \\ $$$${why}\:{we}\:{observe}\:{a}_{\mathrm{1}} {a}_{\mathrm{2}} +{b}_{\mathrm{1}} {b}_{\mathrm{2}} >\mathrm{0}\:{or}\:<\mathrm{0} \\ $$$${for}\:{obtuse}\:{and}\:{acute}\:{angle}\:{bisectors}? \\ $$

Question Number 25961 Answers: 0 Comments: 0

8cos^4 x−8cos^2 x+1=0 solution:8cos^2 x(cos^2 x−1)+1=0⇒ −8cos^2 xsin^2 x=−1⇒ sin^2 2x=(1/2)⇒sinx=+_− ((√2)/2)⇒ { ((2x=2kπ+(π/4))),((2x=2kπ+((3π)/4))) :} { ((2x=2kπ−(π/4))),((2x=2kπ+((5π)/4))) :} and so we have { ((x=kπ+(π/8))),((x=kπ+((3π)/8))) :} { ((x=kπ−(π/8))),((x=kπ+((5π)/8))) :} why x=((kπ)/4)+(π/8)? can we solve by another way?

$$\mathrm{8cos}^{\mathrm{4}} \mathrm{x}−\mathrm{8cos}^{\mathrm{2}} \mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{solution}:\mathrm{8cos}^{\mathrm{2}} \mathrm{x}\left(\mathrm{cos}^{\mathrm{2}} \mathrm{x}−\mathrm{1}\right)+\mathrm{1}=\mathrm{0}\Rightarrow \\ $$$$−\mathrm{8cos}^{\mathrm{2}} \mathrm{xsin}^{\mathrm{2}} \mathrm{x}=−\mathrm{1}\Rightarrow \\ $$$$\mathrm{sin}^{\mathrm{2}} \mathrm{2x}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{sinx}=\underset{−} {+}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\Rightarrow \\ $$$$\begin{cases}{\mathrm{2x}=\mathrm{2k}\pi+\frac{\pi}{\mathrm{4}}}\\{\mathrm{2x}=\mathrm{2k}\pi+\frac{\mathrm{3}\pi}{\mathrm{4}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{2x}=\mathrm{2k}\pi−\frac{\pi}{\mathrm{4}}}\\{\mathrm{2x}=\mathrm{2k}\pi+\frac{\mathrm{5}\pi}{\mathrm{4}}}\end{cases} \\ $$$$\mathrm{and}\:\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\begin{cases}{\mathrm{x}=\mathrm{k}\pi+\frac{\pi}{\mathrm{8}}}\\{\mathrm{x}=\mathrm{k}\pi+\frac{\mathrm{3}\pi}{\mathrm{8}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{x}=\mathrm{k}\pi−\frac{\pi}{\mathrm{8}}}\\{\mathrm{x}=\mathrm{k}\pi+\frac{\mathrm{5}\pi}{\mathrm{8}}}\end{cases} \\ $$$$\mathrm{why}\:\mathrm{x}=\frac{\mathrm{k}\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{8}}? \\ $$$$ \\ $$$$\mathrm{can}\:\mathrm{we}\:\mathrm{solve}\:\mathrm{by}\:\mathrm{another}\:\mathrm{way}? \\ $$

Question Number 25937 Answers: 1 Comments: 0

For a certain amount of work,Ade takes 6hours less than Bode.if they work together it takes them 13hours 20 minutes.How long will it take Bode alone to complete the work?

$${For}\:{a}\:{certain}\:{amount}\:{of}\:{work},{Ade}\:{takes} \\ $$$$\mathrm{6}{hours}\:{less}\:{than}\:{Bode}.{if}\:{they}\:{work}\:{together} \\ $$$${it}\:{takes}\:{them}\:\mathrm{13}{hours}\:\mathrm{20}\:{minutes}.{How} \\ $$$${long}\:{will}\:{it}\:{take}\:{Bode}\:{alone}\:{to}\:{complete} \\ $$$${the}\:{work}? \\ $$

Question Number 25853 Answers: 0 Comments: 2

simplify: 3^x +4^x =5^x anybody to solve this

$${simplify}:\:\mathrm{3}^{{x}} +\mathrm{4}^{{x}} \:=\mathrm{5}^{{x}} \\ $$$${anybody}\:{to}\:{solve}\:{this} \\ $$

Question Number 25823 Answers: 0 Comments: 1

z_1 =4−3i z_2 =9+2i Find (a) z_1 oz_2 (b) z_1 ×z_2 (c) angle between z_1 and z_2 .

$${z}_{\mathrm{1}} =\mathrm{4}−\mathrm{3}{i} \\ $$$${z}_{\mathrm{2}} =\mathrm{9}+\mathrm{2}{i} \\ $$$${Find}\:\:\left({a}\right)\:{z}_{\mathrm{1}} {oz}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\left({b}\right)\:{z}_{\mathrm{1}} ×{z}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({c}\right)\:{angle}\:{between}\:{z}_{\mathrm{1}} {and}\:{z}_{\mathrm{2}} . \\ $$

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