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AlgebraQuestion and Answers: Page 348

Question Number 16519 Answers: 2 Comments: 0

Solve: ∣2 − x∣ − 2 ∣x + 1∣ < 1

$$\mathrm{Solve}: \\ $$$$\mid\mathrm{2}\:−\:\mathrm{x}\mid\:−\:\mathrm{2}\:\mid\mathrm{x}\:+\:\mathrm{1}\mid\:<\:\mathrm{1} \\ $$

Question Number 16373 Answers: 1 Comments: 1

if Σ_(k=0) ^(200) i^k +Π_(p=1) ^(50) i^p =x+iy then..(x,y)is... a. (0,1) b. (1,−1) c. (2,3) d. (4,8)

$${if}\:\underset{{k}=\mathrm{0}} {\overset{\mathrm{200}} {\sum}}{i}^{{k}} +\underset{{p}=\mathrm{1}} {\overset{\mathrm{50}} {\prod}}{i}^{{p}} ={x}+{iy}\:{then}..\left({x},{y}\right){is}... \\ $$$${a}.\:\left(\mathrm{0},\mathrm{1}\right) \\ $$$${b}.\:\left(\mathrm{1},−\mathrm{1}\right) \\ $$$${c}.\:\left(\mathrm{2},\mathrm{3}\right) \\ $$$${d}.\:\left(\mathrm{4},\mathrm{8}\right) \\ $$

Question Number 16238 Answers: 0 Comments: 0

we have a^5 +b^5 =1 and u^5 +v^5 =1 find value a^3 u^5 +b^3 v^5 =?

$${we}\:{have}\:{a}^{\mathrm{5}} +{b}^{\mathrm{5}} =\mathrm{1}\:{and}\:{u}^{\mathrm{5}} +{v}^{\mathrm{5}} =\mathrm{1} \\ $$$${find}\:{value}\:\:{a}^{\mathrm{3}} {u}^{\mathrm{5}} +{b}^{\mathrm{3}} {v}^{\mathrm{5}} =? \\ $$

Question Number 16116 Answers: 0 Comments: 0

Question Number 16115 Answers: 0 Comments: 0

Question Number 16374 Answers: 1 Comments: 0

The Value of the sum.. Σ_(n=1) ^(13) (i^n +i^(n+1) ), Where i=(√(−1 )) is.. (a.) i (b.) i−1 (c.) −i (d.) 0

$${The}\:{Value}\:{of}\:{the}\:{sum}..\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{13}} {\sum}}\left({i}^{{n}} +{i}^{{n}+\mathrm{1}} \right),\:{Where}\:{i}=\sqrt{−\mathrm{1}\:\:} \\ $$$${is}.. \\ $$$$\left({a}.\right)\:{i} \\ $$$$\left({b}.\right)\:{i}−\mathrm{1} \\ $$$$\left({c}.\right)\:−{i} \\ $$$$\left({d}.\right)\:\mathrm{0} \\ $$

Question Number 16047 Answers: 0 Comments: 1

number of real solution of equation a^(2008) + b^(2008) = 2008ab−2006

$${number}\:{of}\:{real}\:{solution}\:{of}\:{equation}\:\:\:\:{a}^{\mathrm{2008}} \:+\:\:{b}^{\mathrm{2008}} \:\:=\:\mathrm{2008}{ab}−\mathrm{2006}\:\:\:\:\:\:\:\:\: \\ $$

Question Number 16029 Answers: 0 Comments: 1

Question Number 16000 Answers: 2 Comments: 0

Question Number 15999 Answers: 1 Comments: 0

Solve simultaneously 2xy = x + y 5xz = 6z − 2x 3yz = 3y + 4z

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$\mathrm{2xy}\:=\:\mathrm{x}\:+\:\mathrm{y} \\ $$$$\mathrm{5xz}\:=\:\mathrm{6z}\:−\:\mathrm{2x} \\ $$$$\mathrm{3yz}\:=\:\mathrm{3y}\:+\:\mathrm{4z} \\ $$

Question Number 15927 Answers: 0 Comments: 5

Question Number 15916 Answers: 2 Comments: 2

Question Number 15869 Answers: 1 Comments: 1

If ax^2 +(b/x)≥c for all x,a,b>0 then prove minimum value of 27ab^2 is 4c^2 .

$$\mathrm{If}\:{ax}^{\mathrm{2}} +\frac{{b}}{{x}}\geqslant{c}\:\mathrm{for}\:\mathrm{all}\:{x},{a},{b}>\mathrm{0}\:\mathrm{then} \\ $$$$\mathrm{prove}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\mathrm{27}{ab}^{\mathrm{2}} \:\mathrm{is}\:\mathrm{4}{c}^{\mathrm{2}} . \\ $$

Question Number 15865 Answers: 2 Comments: 0

a,b,c∈R^(+ ) andIf a+b+c=18 then maximum value of a^2 b^3 c^4 is

$${a},{b},{c}\in\mathbb{R}^{+\:} \mathrm{andIf}\:{a}+{b}+{c}=\mathrm{18}\:\mathrm{then}\:\mathrm{maximum}\:\mathrm{value} \\ $$$$\mathrm{of}\:{a}^{\mathrm{2}} {b}^{\mathrm{3}} {c}^{\mathrm{4}} \:\mathrm{is} \\ $$

Question Number 15721 Answers: 1 Comments: 0

If x+y+z=1 with 0<x, y, z <(1/2) then find tbe range of values of (1/(x+y))+(1/(y+z))+(1/(z+x)) .

$${If}\:\:\:\:\boldsymbol{{x}}+\boldsymbol{{y}}+\boldsymbol{{z}}=\mathrm{1} \\ $$$$\:\:\:\:\:{with}\:\mathrm{0}<{x},\:{y},\:{z}\:<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:{then}\:{find}\:{tbe}\:\:{range}\:{of}\:{values}\:{of} \\ $$$$\:\:\:\frac{\mathrm{1}}{\boldsymbol{{x}}+\boldsymbol{{y}}}+\frac{\mathrm{1}}{\boldsymbol{{y}}+\boldsymbol{{z}}}+\frac{\mathrm{1}}{\boldsymbol{{z}}+\boldsymbol{{x}}}\:. \\ $$

Question Number 15656 Answers: 1 Comments: 0

Solve : 0 ≤ x^2 − 5x + 7 < 1

$$\mathrm{Solve}\::\:\mathrm{0}\:\leqslant\:{x}^{\mathrm{2}} \:−\:\mathrm{5}{x}\:+\:\mathrm{7}\:<\:\mathrm{1} \\ $$

Question Number 15570 Answers: 2 Comments: 1

Solve: 2^x = x^4

$$\mathrm{Solve}:\:\:\:\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{x}^{\mathrm{4}} \\ $$

Question Number 15504 Answers: 1 Comments: 3

Solve ⌈x^2 ⌉=(⌊x⌋)^2 +2x

$$\mathrm{Solve}\:\lceil{x}^{\mathrm{2}} \rceil=\left(\lfloor{x}\rfloor\right)^{\mathrm{2}} +\mathrm{2}{x} \\ $$

Question Number 15501 Answers: 1 Comments: 0

Solve x^3 −⌊x⌋=3

$$\mathrm{Solve} \\ $$$${x}^{\mathrm{3}} −\lfloor{x}\rfloor=\mathrm{3} \\ $$

Question Number 15497 Answers: 1 Comments: 5

P=Σ_(n∈P) ^∞ n Q=Σ_(n∉P) ^∞ n P=2+3+5+7+... Q=1+4+6+8+... Is P>Q? Is Q>P?

$${P}=\underset{{n}\in\mathbb{P}} {\overset{\infty} {\sum}}{n}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Q}=\underset{{n}\notin\mathbb{P}} {\overset{\infty} {\sum}}{n} \\ $$$${P}=\mathrm{2}+\mathrm{3}+\mathrm{5}+\mathrm{7}+... \\ $$$${Q}=\mathrm{1}+\mathrm{4}+\mathrm{6}+\mathrm{8}+... \\ $$$$\: \\ $$$$\mathrm{Is}\:{P}>{Q}?\:\:\:\mathrm{Is}\:{Q}>{P}? \\ $$

Question Number 15234 Answers: 0 Comments: 2

A question related to Q.15184 Find the maximum of f(x)=(ln x)^(1/x)

$$\mathrm{A}\:\mathrm{question}\:\mathrm{related}\:\mathrm{to}\:\mathrm{Q}.\mathrm{15184} \\ $$$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{of}\:\mathrm{f}\left(\mathrm{x}\right)=\left(\mathrm{ln}\:\mathrm{x}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$

Question Number 15051 Answers: 1 Comments: 0

Solve simultaneously x + y + z = 6 ............ equation (i) x^3 + y^3 + z^3 = 92 .......... equation (ii) x − y = z ........... equation (iii)

$$\mathrm{Solve}\:\mathrm{simultaneously} \\ $$$$ \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:............\:\mathrm{equation}\:\left(\mathrm{i}\right) \\ $$$$\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{y}^{\mathrm{3}} \:+\:\mathrm{z}^{\mathrm{3}} \:=\:\mathrm{92}\:\:\:\:\:\:\:\:\:..........\:\mathrm{equation}\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{x}\:−\:\mathrm{y}\:=\:\mathrm{z}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...........\:\mathrm{equation}\:\left(\mathrm{iii}\right) \\ $$

Question Number 14988 Answers: 0 Comments: 2

Solve on Z_4 ax+b=[0]_4 a,b∈Z_4 ax^2 +bx+c=[0]_4 a,b,c∈Z_4

$${Solve}\:{on}\:\mathbb{Z}_{\mathrm{4}} \: \\ $$$${ax}+{b}=\left[\mathrm{0}\right]_{\mathrm{4}} \:\:{a},{b}\in\mathbb{Z}_{\mathrm{4}} \\ $$$${ax}^{\mathrm{2}} +{bx}+{c}=\left[\mathrm{0}\right]_{\mathrm{4}} \:\:{a},{b},{c}\in\mathbb{Z}_{\mathrm{4}} \\ $$

Question Number 14811 Answers: 1 Comments: 0

why (√x^2 )=∣x∣ ?

$${why}\:\:\:\:\:\sqrt{{x}^{\mathrm{2}} }=\mid{x}\mid\:\:\:\:?\: \\ $$

Question Number 14775 Answers: 1 Comments: 0

Using the remainder theorem to factorize completely the expression x^3 (y − z) + y^3 (z − x) + z^3 (x − y)

$$\mathrm{Using}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{theorem}\:\mathrm{to}\:\mathrm{factorize}\:\mathrm{completely}\:\mathrm{the}\:\mathrm{expression}\: \\ $$$$\mathrm{x}^{\mathrm{3}} \left(\mathrm{y}\:−\:\mathrm{z}\right)\:+\:\mathrm{y}^{\mathrm{3}} \left(\mathrm{z}\:−\:\mathrm{x}\right)\:+\:\mathrm{z}^{\mathrm{3}} \left(\mathrm{x}\:−\:\mathrm{y}\right)\: \\ $$

Question Number 14747 Answers: 1 Comments: 0

ε>0 6−ε≤xy≤6+ε 5−ε≤x+y≤5+ε Find x & y

$$\epsilon>\mathrm{0} \\ $$$$\mathrm{6}−\epsilon\leqslant{xy}\leqslant\mathrm{6}+\epsilon \\ $$$$\mathrm{5}−\epsilon\leqslant{x}+{y}\leqslant\mathrm{5}+\epsilon \\ $$$${Find}\:{x}\:\&\:{y} \\ $$

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