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AlgebraQuestion and Answers: Page 247

Question Number 103151 Answers: 0 Comments: 1

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Question Number 103120 Answers: 0 Comments: 0

Question Number 103094 Answers: 0 Comments: 4

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Question Number 103093 Answers: 5 Comments: 0

solve for x,y∈N (1/x)+(1/y)=(3/(202))

$${solve}\:{for}\:{x},{y}\in\mathbb{N} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{3}}{\mathrm{202}} \\ $$

Question Number 102975 Answers: 0 Comments: 1

A student can recall 6 digits of a 9 digit number. In how many ways can he get the complete number?

$$\mathrm{A}\:\mathrm{student}\:\mathrm{can}\:\mathrm{recall}\:\mathrm{6}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{a}\:\mathrm{9}\:\mathrm{digit}\:\mathrm{number}. \\ $$$$\mathrm{In}\:\mathrm{how}\:\mathrm{many}\:\mathrm{ways}\:\mathrm{can}\:\mathrm{he}\:\mathrm{get}\:\mathrm{the}\:\mathrm{complete}\:\mathrm{number}? \\ $$

Question Number 102967 Answers: 1 Comments: 0

Question Number 102940 Answers: 0 Comments: 1

Question Number 102927 Answers: 3 Comments: 1

If (x/(x^2 + x + 1)) = (1/4) Find x + (1/x)

$$\mathrm{If}\:\:\:\frac{\mathrm{x}}{\mathrm{x}^{\mathrm{2}} \:\:+\:\:\mathrm{x}\:\:+\:\:\mathrm{1}}\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{Find}\:\:\:\:\:\mathrm{x}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{x}} \\ $$

Question Number 102548 Answers: 0 Comments: 0

if n is a real number,find the least possible number of n,if n^2 −3n+(√(n−3)) −(√(n+3))

$${if}\:{n}\:{is}\:{a}\:{real}\:{number},{find}\:{the}\:{least}\:{possible}\:{number}\:{of}\:{n},{if} \\ $$$${n}^{\mathrm{2}} −\mathrm{3}{n}+\sqrt{{n}−\mathrm{3}}\:−\sqrt{{n}+\mathrm{3}} \\ $$

Question Number 102490 Answers: 1 Comments: 0

x^3 −bx−c=0 ; b, c >0 ; ((b/3))^3 >((c/2))^2 To find the three real roots without the use of trigonometric solution to cubic polynomial...

$${x}^{\mathrm{3}} −{bx}−{c}=\mathrm{0}\:\:\:\:\:;\:\:{b},\:{c}\:>\mathrm{0}\:;\:\:\left(\frac{{b}}{\mathrm{3}}\right)^{\mathrm{3}} >\left(\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${To}\:{find}\:{the}\:{three}\:{real}\:{roots}\:{without} \\ $$$${the}\:{use}\:{of}\:{trigonometric}\:{solution} \\ $$$${to}\:{cubic}\:{polynomial}... \\ $$

Question Number 102418 Answers: 1 Comments: 0

Question Number 102278 Answers: 3 Comments: 4

Find the greatest coefficient in the following without actually expand. (i) (5 − 3x)^(10) (ii) (5 + 3x)^(− 10)

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{without} \\ $$$$\mathrm{actually}\:\mathrm{expand}. \\ $$$$\left(\mathrm{i}\right)\:\:\:\:\:\:\:\:\left(\mathrm{5}\:\:−\:\:\mathrm{3x}\right)^{\mathrm{10}} \\ $$$$\left(\mathrm{ii}\right)\:\:\:\:\:\:\:\:\left(\mathrm{5}\:\:+\:\:\mathrm{3x}\right)^{−\:\mathrm{10}} \\ $$

Question Number 102201 Answers: 0 Comments: 0

Question Number 102186 Answers: 0 Comments: 0

Question Number 102179 Answers: 1 Comments: 0

Question Number 102178 Answers: 1 Comments: 0

Question Number 102109 Answers: 0 Comments: 0

Question Number 102078 Answers: 5 Comments: 0

Question Number 102002 Answers: 0 Comments: 6

Question Number 101971 Answers: 0 Comments: 2

Question Number 101961 Answers: 0 Comments: 4

Question Number 101951 Answers: 0 Comments: 3

Question Number 101921 Answers: 1 Comments: 1

Please, help i cannot solve problems such as: x^4 =1, x^5 =1, or x^n =1 n∈N...

$$\boldsymbol{{Please}},\:\boldsymbol{{help}}\:\:\boldsymbol{{i}}\:\:\boldsymbol{{cannot}}\:\boldsymbol{{solve}}\:\boldsymbol{{problems}} \\ $$$$\boldsymbol{{such}}\:\boldsymbol{{as}}:\:\boldsymbol{{x}}^{\mathrm{4}} =\mathrm{1},\:\boldsymbol{{x}}^{\mathrm{5}} =\mathrm{1},\:\boldsymbol{{or}}\:\boldsymbol{{x}}^{\boldsymbol{{n}}} =\mathrm{1}\:\boldsymbol{{n}}\in\mathbb{N}... \\ $$

Question Number 101803 Answers: 1 Comments: 2

((√2)−1)^x +((√2)+1)^x =((√6))^x

$$\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{{x}} +\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{{x}} =\left(\sqrt{\mathrm{6}}\right)^{{x}} \\ $$

Question Number 101800 Answers: 1 Comments: 3

Question Number 101742 Answers: 3 Comments: 0

{ ((ab+a+b = 5)),((bc + b+c = 14)),((ac + a+c = 9)) :} find a+b+c = ___

$$\begin{cases}{{ab}+{a}+{b}\:=\:\mathrm{5}}\\{{bc}\:+\:{b}+{c}\:=\:\mathrm{14}}\\{{ac}\:+\:{a}+{c}\:=\:\mathrm{9}}\end{cases} \\ $$$$\mathrm{find}\:{a}+{b}+{c}\:=\:\_\_\_ \\ $$

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