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AlgebraQuestion and Answers: Page 1

Question Number 222889 Answers: 1 Comments: 10

Question Number 222885 Answers: 3 Comments: 0

Question Number 222856 Answers: 1 Comments: 0

find the possible root of x^3 −2x^2 −5x+6=0 using the fixed point iteration method?

$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{possible}}\:\boldsymbol{{root}}\:\boldsymbol{{of}}\:\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{5}\boldsymbol{{x}}+\mathrm{6}=\mathrm{0} \\ $$$$\boldsymbol{{using}}\:\boldsymbol{{the}}\:\boldsymbol{{fixed}}\:\boldsymbol{{point}}\:\boldsymbol{{iteration}}\:\boldsymbol{{method}}? \\ $$

Question Number 222847 Answers: 0 Comments: 0

let f(x)=1.013x^5 −5.262x^3 −0.01732x^2 +0.8389x −1.912. Evaluate f(2.279) by first calculating (2.279)^2 ,(2.279)^3 ,(2.279)^4 and(2.279)^5 using four−digit round arithmetic. hence,compute the absolute and relative errors.

$$\boldsymbol{{let}}\:\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\mathrm{1}.\mathrm{013}\boldsymbol{{x}}^{\mathrm{5}} −\mathrm{5}.\mathrm{262}\boldsymbol{{x}}^{\mathrm{3}} −\mathrm{0}.\mathrm{01732}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{0}.\mathrm{8389}\boldsymbol{{x}} \\ $$$$−\mathrm{1}.\mathrm{912}.\:\boldsymbol{{Evaluate}}\:\boldsymbol{{f}}\left(\mathrm{2}.\mathrm{279}\right)\:\boldsymbol{{by}}\:\boldsymbol{{first}}\:\boldsymbol{{calculating}} \\ $$$$\left(\mathrm{2}.\mathrm{279}\right)^{\mathrm{2}} ,\left(\mathrm{2}.\mathrm{279}\right)^{\mathrm{3}} ,\left(\mathrm{2}.\mathrm{279}\right)^{\mathrm{4}} \boldsymbol{{and}}\left(\mathrm{2}.\mathrm{279}\right)^{\mathrm{5}} \:\boldsymbol{{using}} \\ $$$$\boldsymbol{{four}}−\boldsymbol{{digit}}\:\boldsymbol{{round}}\:\boldsymbol{{arithmetic}}.\:\boldsymbol{{hence}},\boldsymbol{{compute}} \\ $$$$\boldsymbol{{the}}\:\boldsymbol{{absolute}}\:\boldsymbol{{and}}\:\boldsymbol{{relative}}\:\boldsymbol{{errors}}. \\ $$

Question Number 222829 Answers: 1 Comments: 0

If f(x) = ((3x + [x])/(2x)) Find lim_(x→−5^+ ) f(x) − lim_(x→−5^− ) f(x) = ?

$$\mathrm{If}\:\:\:\mathrm{f}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{3x}\:+\:\left[\mathrm{x}\right]}{\mathrm{2x}} \\ $$$$\mathrm{Find}\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow−\mathrm{5}^{+} } {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\:−\:\underset{\boldsymbol{\mathrm{x}}\rightarrow−\mathrm{5}^{−} } {\mathrm{lim}}\:\mathrm{f}\left(\mathrm{x}\right)\:=\:? \\ $$

Question Number 222801 Answers: 2 Comments: 0

Question Number 222777 Answers: 1 Comments: 0

Simplify: (((cos214° + i sin146°)∙(cos10° + i sin10°))/((cos66° − i sin246°))) = ?

$$\mathrm{Simplify}: \\ $$$$\frac{\left(\mathrm{cos214}°\:+\:\boldsymbol{\mathrm{i}}\:\mathrm{sin146}°\right)\centerdot\left(\mathrm{cos10}°\:+\:\boldsymbol{\mathrm{i}}\:\mathrm{sin10}°\right)}{\left(\mathrm{cos66}°\:−\:\boldsymbol{\mathrm{i}}\:\mathrm{sin246}°\right)}\:=\:? \\ $$

Question Number 222754 Answers: 1 Comments: 1

Question Number 222743 Answers: 1 Comments: 0

Compare: a = arcctg (√2) b = arccos ((√2)/2) c = arctg (√2)

$$\mathrm{Compare}: \\ $$$$\boldsymbol{\mathrm{a}}\:=\:\mathrm{arcctg}\:\sqrt{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{b}}\:=\:\mathrm{arccos}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{c}}\:=\:\mathrm{arctg}\:\sqrt{\mathrm{2}} \\ $$

Question Number 222616 Answers: 1 Comments: 0

Question Number 222585 Answers: 3 Comments: 2

Question Number 222584 Answers: 0 Comments: 0

find the correct const. to preconst. “ x^n −2=−x ”

$${find}\:{the}\:{correct}\:{const}.\:{to}\:{preconst}.\:``\:{x}^{{n}} −\mathrm{2}=−{x}\:'' \\ $$

Question Number 222582 Answers: 2 Comments: 0

If: a_i > 0 , b_i > 0 , i = 1,...,n^(−) Prove that: (√(a_1 ^2 + b_1 ^2 )) + (√(a_2 ^2 + b_2 ^2 )) +...+ (√(a_n ^2 + b_n ^2 )) ≥ ≥ (√((a_1 +a_2 +...+a_n )^2 + (b_1 +b_2 +...+b_n )^2 ))

$$\mathrm{If}:\:\:\:\mathrm{a}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{b}_{\boldsymbol{\mathrm{i}}} \:>\:\mathrm{0}\:\:\:,\:\:\:\mathrm{i}\:=\:\overline {\mathrm{1},...,\mathrm{n}} \\ $$$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\sqrt{\mathrm{a}_{\mathrm{1}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{1}} ^{\mathrm{2}} }\:+\:\sqrt{\mathrm{a}_{\mathrm{2}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\mathrm{2}} ^{\mathrm{2}} }\:+...+\:\sqrt{\mathrm{a}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} \:+\:\mathrm{b}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} }\:\:\:\geqslant \\ $$$$\geqslant\:\sqrt{\left(\mathrm{a}_{\mathrm{1}} +\mathrm{a}_{\mathrm{2}} +...+\mathrm{a}_{\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} \:+\:\left(\mathrm{b}_{\mathrm{1}} +\mathrm{b}_{\mathrm{2}} +...+\mathrm{b}_{\boldsymbol{\mathrm{n}}} \right)^{\mathrm{2}} } \\ $$

Question Number 222552 Answers: 1 Comments: 0

x^x =64 , x=?

$${x}^{{x}} =\mathrm{64}\:\:\:\:\:\:,\:\:{x}=? \\ $$

Question Number 222523 Answers: 2 Comments: 0

x+V−J(x)((ustx)/(zac^2 x))=x−((ustx_0 )/( (√x)ψ+ζ(((−1+v%)/(π+2))+L_(l(x→0)) ^( v%) X_(2x^2 ) ^( 1) )))

$${x}+{V}−{J}\left({x}\right)\frac{{ustx}}{{zac}^{\mathrm{2}} {x}}={x}−\frac{{ustx}_{\mathrm{0}} }{\:\sqrt{{x}}\psi+\zeta\left(\frac{−\mathrm{1}+{v\%}}{\pi+\mathrm{2}}+\mathscr{L}_{{l}\left({x}\rightarrow\mathrm{0}\right)} ^{\:\:{v\%}} {X}_{\mathrm{2}{x}^{\mathrm{2}} } ^{\:\mathrm{1}} \right)} \\ $$

Question Number 222521 Answers: 1 Comments: 1

−3ix+(π/2)+((ix−(√(5π)))/6)=0 x=?

$$−\mathrm{3}{ix}+\frac{\pi}{\mathrm{2}}+\frac{{ix}−\sqrt{\mathrm{5}\pi}}{\mathrm{6}}=\mathrm{0} \\ $$$${x}=?\: \\ $$

Question Number 222511 Answers: 0 Comments: 0

Is the statement correct? in const.“ x+a−bx^n ={0} ” x= { ((−a±((2b)/(n+1)) , n>0)),((N_b ^( a) ∫_b ^( a) G(n−1), n≤0)) :}

$${Is}\:{the}\:{statement}\:{correct}? \\ $$$${in}\:{const}.``\:{x}+{a}−{bx}^{{n}} =\left\{\mathrm{0}\right\}\:'' \\ $$$${x}=\begin{cases}{−{a}\pm\frac{\mathrm{2}{b}}{{n}+\mathrm{1}}\:,\:{n}>\mathrm{0}}\\{{N}_{{b}} ^{\:{a}} \:\int_{{b}} ^{\:{a}} {G}\left({n}−\mathrm{1}\right),\:{n}\leqslant\mathrm{0}}\end{cases} \\ $$

Question Number 222501 Answers: 1 Comments: 0

This is VERY HARD { ( { ((x+y=0)),((l(y)=1)) :}),( { ((x∈N)),((−y= { ((v%, for x↺γ(1))),((−v%, for x↬θ(∮_(−x) ^( 0) (c/7)))) :})) :}) :} x=?, y=?

$${This}\:{is}\:{VERY}\:{HARD} \\ $$$$\begin{cases}{\begin{cases}{{x}+{y}=\mathrm{0}}\\{{l}\left({y}\right)=\mathrm{1}}\end{cases}}\\{\begin{cases}{{x}\in\mathbb{N}}\\{−{y}=\begin{cases}{{v\%},\:\:{for}\:{x}\circlearrowleft\gamma\left(\mathrm{1}\right)}\\{−{v\%},\:\:{for}\:{x}\looparrowright\theta\left(\oint_{−{x}} ^{\:\mathrm{0}} \frac{{c}}{\mathrm{7}}\right)}\end{cases}}\end{cases}}\end{cases} \\ $$$${x}=?,\:{y}=? \\ $$

Question Number 222583 Answers: 0 Comments: 0

solve for p,q,s in terms of c. • (((qs)/(q−sp)))^2 −s(((qs)/(q−sp)))+p=0 • (((q+c)/(p+1)))^2 =sp−q • (q−cp)(p+1)^2 =(q+c)^3 I have to find non zero real x=−(((q+c)/(p+1))) .

$${solve}\:{for}\:{p},{q},{s}\:{in}\:{terms}\:{of}\:{c}. \\ $$$$\bullet\:\left(\frac{{qs}}{{q}−{sp}}\right)^{\mathrm{2}} −{s}\left(\frac{{qs}}{{q}−{sp}}\right)+{p}=\mathrm{0} \\ $$$$\bullet\:\left(\frac{{q}+{c}}{{p}+\mathrm{1}}\right)^{\mathrm{2}} ={sp}−{q} \\ $$$$\bullet\:\left({q}−{cp}\right)\left({p}+\mathrm{1}\right)^{\mathrm{2}} =\left({q}+{c}\right)^{\mathrm{3}} \\ $$$${I}\:{have}\:{to}\:{find}\:{non}\:{zero}\:{real}\:{x}=−\left(\frac{{q}+{c}}{{p}+\mathrm{1}}\right)\:. \\ $$

Question Number 222568 Answers: 0 Comments: 0

f(x)=((√(x−2)))^0 and g(x)=(√((x−2)^0 )) dom f(x)=? , dom g(x)=?

$${f}\left({x}\right)=\left(\sqrt{{x}−\mathrm{2}}\right)^{\mathrm{0}} \:\:{and}\:{g}\left({x}\right)=\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{0}} } \\ $$$${dom}\:{f}\left({x}\right)=?\:,\:{dom}\:{g}\left({x}\right)=? \\ $$

Question Number 222404 Answers: 1 Comments: 0

Question Number 222353 Answers: 1 Comments: 1

solve; ((11)/(29)) = ? , no fraction and no decimal

$$ \\ $$$$\:\:\:\:\:\mathrm{solve};\:\:\frac{\mathrm{11}}{\mathrm{29}}\:=\:?\:,\:\mathrm{no}\:\mathrm{fraction}\:\mathrm{and}\:\mathrm{no}\:\mathrm{decimal}\:\:\:\:\: \\ $$$$ \\ $$

Question Number 222339 Answers: 2 Comments: 0

Solve: ((36))^(1/x) + ((24))^(1/x) = ((16))^(1/x)

$$\mathrm{Solve}:\:\:\:\:\:\:\sqrt[{\mathrm{x}}]{\mathrm{36}}\:\:\:+\:\:\:\sqrt[{\mathrm{x}}]{\mathrm{24}}\:\:\:\:=\:\:\:\sqrt[{\mathrm{x}}]{\mathrm{16}} \\ $$

Question Number 222323 Answers: 1 Comments: 0

a+3^b =b,3^b ∙a^(b+1) max=?

$${a}+\mathrm{3}^{{b}} ={b},\mathrm{3}^{{b}} \centerdot{a}^{{b}+\mathrm{1}} \:\mathrm{max}=? \\ $$

Question Number 222309 Answers: 2 Comments: 0

determinant (((4^x +6^x =9^x )))

$$\begin{array}{|c|}{\mathrm{4}^{{x}} +\mathrm{6}^{{x}} =\mathrm{9}^{{x}} }\\\hline\end{array} \\ $$

Question Number 222299 Answers: 1 Comments: 0

For what value of k the roots of the equation ((x^2 −2x)/(4x−1))=((k−1)/(k+1)) will have same value but with opposite symbol(like x=a and −a) i mean the two valuea of x will be this type x=2 and −2(both 2 but opposite symbols)

$${For}\:{what}\:{value}\:{of}\:\:{k}\:\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{4}{x}−\mathrm{1}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${will}\:{have}\:{same}\:{value}\:{but}\:\:{with}\:{opposite}\:{symbol}\left({like}\:{x}={a}\:{and}\:−{a}\right) \\ $$$${i}\:{mean}\:{the}\:{two}\:{valuea}\:{of}\:{x}\:{will}\:{be}\:{this}\:{type} \\ $$$${x}=\mathrm{2}\:{and}\:−\mathrm{2}\left({both}\:\mathrm{2}\:{but}\:{opposite}\:{symbols}\right) \\ $$

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