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AlgebraQuestion and Answers: Page 86

Question Number 187357 Answers: 1 Comments: 0

Question Number 187302 Answers: 0 Comments: 0

f(4)=44, f(m)=52,f(l)=−33 l,m are positive integers such that 4<m<l and f(x) is a polynomial with integer coefficients. Find l+m.

$${f}\left(\mathrm{4}\right)=\mathrm{44},\:{f}\left({m}\right)=\mathrm{52},{f}\left({l}\right)=−\mathrm{33} \\ $$$${l},{m}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integers}\:\mathrm{such}\:\mathrm{that}\:\mathrm{4}<{m}<{l} \\ $$$$\mathrm{and}\:{f}\left({x}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{polynomial}\:\mathrm{with}\:\mathrm{integer}\:\mathrm{coefficients}. \\ $$$$\mathrm{Find}\:{l}+{m}. \\ $$

Question Number 187295 Answers: 0 Comments: 2

solve: ((√(4+(√4))^x )) +[(√(4−(√(4]^x )))) =4^x

$${solve}: \\ $$$$\left(\sqrt{\left.\mathrm{4}+\sqrt{\mathrm{4}}\right)\:^{{x}} }\:+\left[\sqrt{\mathrm{4}−\sqrt{\left.\mathrm{4}\right]^{{x}} }}\:=\mathrm{4}^{{x}} \right.\right. \\ $$

Question Number 187294 Answers: 0 Comments: 1

solve: −1−[x^x −5x+6]^x =1

$${solve}: \\ $$$$−\mathrm{1}−\left[{x}^{{x}} −\mathrm{5}{x}+\mathrm{6}\right]^{{x}} =\mathrm{1} \\ $$

Question Number 187288 Answers: 1 Comments: 0

logx+x!=2 x=?

$${logx}+{x}!=\mathrm{2} \\ $$$${x}=? \\ $$

Question Number 187255 Answers: 1 Comments: 0

Find the directional derivatives of the function f(x,y,z)=2x^2 +3y^2 +z^2 at the point p(2,1,3)

$${Find}\:{the}\:{directional}\:{derivatives}\:{of}\:{the} \\ $$$${function}\: \\ $$$${f}\left({x},\mathrm{y},\mathrm{z}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \:{at}\:{the}\:{point}\:{p}\left(\mathrm{2},\mathrm{1},\mathrm{3}\right) \\ $$

Question Number 187254 Answers: 1 Comments: 0

Question Number 187248 Answers: 0 Comments: 1

what is the cardinality of the set of prime numbers whose base ten digits sums up to 10

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{cardinality}\:\mathrm{of}\:\mathrm{the}\:\mathrm{set}\:\mathrm{of}\:\mathrm{prime}\:\mathrm{numbers}\:\mathrm{whose} \\ $$$$\mathrm{base}\:\mathrm{ten}\:\mathrm{digits}\:\mathrm{sums}\:\mathrm{up}\:\mathrm{to}\:\mathrm{10} \\ $$

Question Number 187233 Answers: 1 Comments: 1

Find x if : x^4 +a=x ∀ (0<a<(2/9))

$${Find}\:{x}\:{if}\::\:\:\:{x}^{\mathrm{4}} +{a}={x}\:\:\:\:\forall\:\left(\mathrm{0}<{a}<\frac{\mathrm{2}}{\mathrm{9}}\right)\:\:\: \\ $$

Question Number 187209 Answers: 0 Comments: 1

Question Number 187166 Answers: 0 Comments: 0

x^3 =x+c let x=((mt)/(1−t)) m^3 t^3 =mt(1−t)^2 +c(1−t)^3 ⇒ (m^3 −m−c)t^3 +(2m−3c)t^2 +(3c−m)t−c=0 t^3 +At^2 +Bt+C=0 let AB=C ⇒ (2m−3c)(m−3c)=c(m^3 −m−c) ⇒ m^3 −(2/c)m^2 −8m+10c=0 ...

$${x}^{\mathrm{3}} ={x}+{c} \\ $$$${let}\:\:{x}=\frac{{mt}}{\mathrm{1}−{t}} \\ $$$${m}^{\mathrm{3}} {t}^{\mathrm{3}} ={mt}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} +{c}\left(\mathrm{1}−{t}\right)^{\mathrm{3}} \\ $$$$\Rightarrow \\ $$$$\left({m}^{\mathrm{3}} −{m}−{c}\right){t}^{\mathrm{3}} +\left(\mathrm{2}{m}−\mathrm{3}{c}\right){t}^{\mathrm{2}} \\ $$$$\:\:\:\:\:+\left(\mathrm{3}{c}−{m}\right){t}−{c}=\mathrm{0} \\ $$$${t}^{\mathrm{3}} +{At}^{\mathrm{2}} +{Bt}+{C}=\mathrm{0} \\ $$$${let}\:\:{AB}={C} \\ $$$$\Rightarrow\:\left(\mathrm{2}{m}−\mathrm{3}{c}\right)\left({m}−\mathrm{3}{c}\right)={c}\left({m}^{\mathrm{3}} −{m}−{c}\right) \\ $$$$\Rightarrow\:{m}^{\mathrm{3}} −\frac{\mathrm{2}}{{c}}{m}^{\mathrm{2}} −\mathrm{8}{m}+\mathrm{10}{c}=\mathrm{0} \\ $$$$... \\ $$

Question Number 187124 Answers: 2 Comments: 0

(a/x)=(b/y)=(c/z)=(1/3) , a−2b+c=2 and −2y+z=1 x=? An altered form of q#187020 (In this case solveable)

$$\frac{{a}}{{x}}=\frac{{b}}{{y}}=\frac{{c}}{{z}}=\frac{\mathrm{1}}{\mathrm{3}}\:, \\ $$$${a}−\mathrm{2}{b}+{c}=\mathrm{2}\:\:{and}\:\:−\mathrm{2}{y}+{z}=\mathrm{1}\:\:\:\: \\ $$$${x}=? \\ $$$${An}\:{altered}\:{form}\:{of}\:\:{q}#\mathrm{187020} \\ $$$$\left({In}\:{this}\:{case}\:{solveable}\right) \\ $$

Question Number 187008 Answers: 2 Comments: 0

(a/3)=(b/4)=(c/5) 3a+c=42 b=? how is solution

$$\frac{{a}}{\mathrm{3}}=\frac{{b}}{\mathrm{4}}=\frac{{c}}{\mathrm{5}}\:\:\:\:\:\:\:\:\:\:\:\mathrm{3}{a}+{c}=\mathrm{42}\:\:\:\:\:\:\:\:{b}=? \\ $$$${how}\:{is}\:{solution} \\ $$

Question Number 186941 Answers: 0 Comments: 1

((sinx)/x^b )=((Σ_(n=0) ^∞ (((−1)^n )/((2n+1)!))x^(2n+1) )/x^b ) prove that

$$\frac{{sinx}}{{x}^{{b}} }=\frac{\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}{x}^{\mathrm{2}{n}+\mathrm{1}} }{{x}^{{b}} } \\ $$$${prove}\:{that} \\ $$

Question Number 186929 Answers: 3 Comments: 0

If { ((B+R+P=−1)),((B^2 +R^2 +P^2 =17)),((B^3 +R^3 +P^3 =11)) :} then B^5 +R^5 +P^5 =?

$$\:\:{If}\:\begin{cases}{{B}+{R}+{P}=−\mathrm{1}}\\{{B}^{\mathrm{2}} +{R}^{\mathrm{2}} +{P}^{\mathrm{2}} =\mathrm{17}}\\{{B}^{\mathrm{3}} +{R}^{\mathrm{3}} +{P}^{\mathrm{3}} =\mathrm{11}}\end{cases} \\ $$$$\:{then}\:{B}^{\mathrm{5}} +{R}^{\mathrm{5}} +{P}^{\mathrm{5}} \:=? \\ $$

Question Number 186867 Answers: 2 Comments: 0

Question Number 186864 Answers: 2 Comments: 0

if AB,BA and BB three tow digits natural numbers if (((AB+BA))/(BB))=4 then find the maximum volue of (A+B)=?

$${if}\:\:{AB},{BA}\:{and}\:{BB}\:\:{three}\:\:{tow}\:{digits}\:{natural} \\ $$$${numbers}\:{if}\:\:\frac{\left({AB}+{BA}\right)}{{BB}}=\mathrm{4} \\ $$$${then}\:{find}\:{the}\:{maximum}\:{volue}\:{of} \\ $$$$\left({A}+{B}\right)=? \\ $$

Question Number 186772 Answers: 0 Comments: 0

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Question Number 186701 Answers: 2 Comments: 2

Question Number 186690 Answers: 1 Comments: 0

Question Number 186682 Answers: 0 Comments: 0

Prove the following set identities 1) A∪(B∪C)=(A∪B)∪C 2) A∪∅=A 3) A∩(B∪C)=(A∩B)∪(A∩C)

$$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{set}\:\mathrm{identities} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{A}\cup\left(\mathrm{B}\cup\mathrm{C}\right)=\left(\mathrm{A}\cup\mathrm{B}\right)\cup\mathrm{C} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{A}\cup\varnothing=\mathrm{A} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{A}\cap\left(\mathrm{B}\cup\mathrm{C}\right)=\left(\mathrm{A}\cap\mathrm{B}\right)\cup\left(\mathrm{A}\cap\mathrm{C}\right) \\ $$

Question Number 186732 Answers: 2 Comments: 0

Question Number 186657 Answers: 1 Comments: 0

Question Number 186616 Answers: 1 Comments: 0

Question Number 186627 Answers: 2 Comments: 1

Question Number 186582 Answers: 0 Comments: 0

(((10x^3 −c)/(4x^4 −x+1))+x^2 )^2 =x(x^3 +1)

$$\left(\frac{\mathrm{10}{x}^{\mathrm{3}} −{c}}{\mathrm{4}{x}^{\mathrm{4}} −{x}+\mathrm{1}}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} ={x}\left({x}^{\mathrm{3}} +\mathrm{1}\right) \\ $$

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