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AlgebraQuestion and Answers: Page 70

Question Number 193542 Answers: 1 Comments: 2

f(x) = x^3 +3x^2 −1 1) calcul h(X) = f(a+X) −b 2) determine a and b such that h is odd

$$\mathrm{f}\left(\mathrm{x}\right)\:=\:\mathrm{x}^{\mathrm{3}} +\mathrm{3x}^{\mathrm{2}} −\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calcul}\:\:\:\:\mathrm{h}\left(\mathrm{X}\right)\:=\:\mathrm{f}\left(\mathrm{a}+\mathrm{X}\right)\:−\mathrm{b} \\ $$2) determine a and b such that h is odd

Question Number 193525 Answers: 2 Comments: 0

((27t73)/(11)) and R=0 then t=? how is explontry solution

$$\:\:\:\:\:\frac{\mathrm{27t73}}{\mathrm{11}} \\ $$$$\:\:\:\:\:\mathrm{and}\:\mathrm{R}=\mathrm{0}\:\:\:\mathrm{then}\:\mathrm{t}=? \\ $$$$\:\:\:\:\:\:\mathrm{how}\:\mathrm{is}\:\mathrm{explontry}\:\mathrm{solution} \\ $$

Question Number 193492 Answers: 1 Comments: 0

Question Number 193484 Answers: 0 Comments: 0

Nice problem: Find 8 distinctive numbers ∈N\{0} such that these are simultaniously true: (1) a+b+c+d = e+f+g+h (2) a^2 +b^2 +c^2 +d^2 = e^2 +f^2 +g^2 +h^2 (3) a^3 +b^3 +c^3 +d^3 = e^3 +f^3 +g^3 +h^3 [Find a method to generate such numbers]

$$\mathrm{Nice}\:\mathrm{problem}: \\ $$$$\mathrm{Find}\:\mathrm{8}\:\mathrm{distinctive}\:\mathrm{numbers}\:\in\mathbb{N}\backslash\left\{\mathrm{0}\right\}\:\mathrm{such} \\ $$$$\mathrm{that}\:\mathrm{these}\:\mathrm{are}\:\mathrm{simultaniously}\:\mathrm{true}: \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\:\:\:\:\:\:{a}+{b}+{c}+{d}\:=\:{e}+{f}+{g}+{h} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} \:=\:{e}^{\mathrm{2}} +{f}^{\mathrm{2}} +{g}^{\mathrm{2}} +{h}^{\mathrm{2}} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +{d}^{\mathrm{3}} \:=\:{e}^{\mathrm{3}} +{f}^{\mathrm{3}} +{g}^{\mathrm{3}} +{h}^{\mathrm{3}} \\ $$$$\left[\mathrm{Find}\:\mathrm{a}\:\mathrm{method}\:\mathrm{to}\:\mathrm{generate}\:\mathrm{such}\:\mathrm{numbers}\right] \\ $$

Question Number 193469 Answers: 0 Comments: 0

Question Number 193467 Answers: 3 Comments: 3

Proof : cot^(−1) ((((√(1+sint))+(√(1−sint)))/( (√(1+sint))−(√(1−sint)))))=(t/2)

$$\mathrm{Proof}\:: \\ $$$$\mathrm{cot}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{1}+\mathrm{sint}}+\sqrt{\mathrm{1}−\mathrm{sint}}}{\:\sqrt{\mathrm{1}+\mathrm{sint}}−\sqrt{\mathrm{1}−\mathrm{sint}}}\right)=\frac{\mathrm{t}}{\mathrm{2}}\: \\ $$

Question Number 193411 Answers: 2 Comments: 1

$$\:\:\:\:\: \\ $$$$ \\ $$

Question Number 193410 Answers: 2 Comments: 0

{ ((x=(√(3−(√(5+2(√3))))))),((y=(√(3+(√(5+2(√3))))))) :} x

$$\:\:\begin{cases}{\mathrm{x}=\sqrt{\mathrm{3}−\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\\{\mathrm{y}=\sqrt{\mathrm{3}+\sqrt{\mathrm{5}+\mathrm{2}\sqrt{\mathrm{3}}}}}\end{cases}\: \\ $$$$\:\:\:\:\underbrace{\boldsymbol{{x}}} \\ $$

Question Number 193371 Answers: 3 Comments: 0

Reduce to first order and solve , showing each step in detail. 1. y′′ +(y′)^3 siny=0 2. y′′=1+(y′)^2

$$\mathrm{Reduce}\:\mathrm{to}\:\mathrm{first}\:\mathrm{order}\:\mathrm{and}\:\mathrm{solve}\:, \\ $$$$\mathrm{showing}\:\mathrm{each}\:\mathrm{step}\:\mathrm{in}\:\mathrm{detail}. \\ $$$$\mathrm{1}.\:\mathrm{y}''\:+\left(\mathrm{y}'\right)^{\mathrm{3}} \mathrm{siny}=\mathrm{0} \\ $$$$\mathrm{2}.\:\mathrm{y}''=\mathrm{1}+\left(\mathrm{y}'\right)^{\mathrm{2}} \\ $$$$ \\ $$

Question Number 193363 Answers: 1 Comments: 0

y = 4 × 10^(2x) Express x in terms of y, giving an exact simplified answer in terms of log base 10.

$${y}\:=\:\mathrm{4}\:×\:\mathrm{10}^{\mathrm{2}{x}} \\ $$$$\mathrm{Express}\:{x}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{y},\:\mathrm{giving}\:\mathrm{an}\:\mathrm{exact} \\ $$$$\mathrm{simplified}\:\mathrm{answer}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{log}\:\mathrm{base}\:\mathrm{10}. \\ $$

Question Number 193356 Answers: 2 Comments: 0

Question Number 193346 Answers: 2 Comments: 0

(√(4x−3))−(√(2x−5))=2 solve for x

$$\sqrt{\mathrm{4x}−\mathrm{3}}−\sqrt{\mathrm{2x}−\mathrm{5}}=\mathrm{2} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{x} \\ $$

Question Number 193339 Answers: 1 Comments: 0

Prove that a group G of prime order is cyclic.

$${Prove}\:{that}\:{a}\:{group}\:{G}\:{of}\:{prime}\:{order}\:{is}\:{cyclic}. \\ $$$$ \\ $$

Question Number 193314 Answers: 1 Comments: 0

a ,b, c > 0 & a^2 +b^2 +c^2 =3 prove that (((1+(3/(ab+bc+ca)) )^((a+b+c)^2 ) ))^(1/3) ≤(1+(a/b))(1+(b/c))(1+(c/a))

$$ \\ $$$$\boldsymbol{{a}}\:,\boldsymbol{{b}},\:\boldsymbol{{c}}\:\:>\:\mathrm{0}\:\&\:\:\boldsymbol{{a}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} +\boldsymbol{{c}}^{\mathrm{2}} =\mathrm{3}\:\boldsymbol{{prove}}\:\boldsymbol{{that}}\: \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{1}+\frac{\mathrm{3}}{\boldsymbol{{ab}}+\boldsymbol{{bc}}+\boldsymbol{{ca}}}\:\right)^{\left(\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}\right)^{\mathrm{2}} } \:}\leqslant\left(\mathrm{1}+\frac{\boldsymbol{{a}}}{\boldsymbol{{b}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{b}}}{\boldsymbol{{c}}}\right)\left(\mathrm{1}+\frac{\boldsymbol{{c}}}{\boldsymbol{{a}}}\right) \\ $$

Question Number 193296 Answers: 2 Comments: 1

If a^2 + b^2 + c^2 = 16, x^2 + y^2 + z^2 = 25 and ax + by + cz = 20 then what is the value of ((a + b + c)/(x + y + z)) ?

$$\mathrm{If}\:{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} \:+\:{c}^{\mathrm{2}} \:=\:\mathrm{16},\:{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\:{z}^{\mathrm{2}} \:=\:\mathrm{25} \\ $$$$\mathrm{and}\:{ax}\:+\:{by}\:+\:{cz}\:=\:\mathrm{20}\:\mathrm{then}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the} \\ $$$$\mathrm{value}\:\mathrm{of}\:\:\frac{{a}\:+\:{b}\:+\:{c}}{{x}\:+\:{y}\:+\:{z}}\:? \\ $$

Question Number 193284 Answers: 1 Comments: 0

6y −2xy = 4 8z − yz = 9 10x − 4xz = 8 find x+y +z = ?

$$\:\:\:\mathrm{6}{y}\:−\mathrm{2}{xy}\:=\:\mathrm{4} \\ $$$$\:\:\:\:\mathrm{8}{z}\:−\:{yz}\:=\:\mathrm{9} \\ $$$$\:\:\:\mathrm{10}{x}\:−\:\mathrm{4}{xz}\:=\:\mathrm{8}\: \\ $$$${find}\:{x}+{y}\:+{z}\:=\:? \\ $$

Question Number 193238 Answers: 1 Comments: 0

s=a+b+c+d+..... number terms :n {a;b;c;d.....}>0 then E=s/(s−a)+s/(s−b)+s/s−c)+.... a) E>=n^2 b)E>=n^2 /(n−1) c) E>=n/(n+1) d) E>=n^2 /(n+1) e) E>=n^2 −1

$${s}={a}+{b}+{c}+{d}+..... \\ $$$${number}\:{terms}\::{n} \\ $$$$\left\{{a};{b};{c};{d}.....\right\}>\mathrm{0} \\ $$$$\left.{then}\:{E}={s}/\left({s}−{a}\right)+{s}/\left({s}−{b}\right)+{s}/{s}−{c}\right)+.... \\ $$$$\left.{a}\left.\right)\:{E}>={n}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:{b}\right){E}>={n}^{\mathrm{2}} /\left({n}−\mathrm{1}\right) \\ $$$$\left.{c}\left.\right)\:{E}>={n}/\left({n}+\mathrm{1}\right)\:\:\:\:\:\:{d}\right)\:{E}>={n}^{\mathrm{2}} /\left({n}+\mathrm{1}\right) \\ $$$$\left.{e}\right)\:{E}>={n}^{\mathrm{2}} −\mathrm{1} \\ $$$$ \\ $$$$ \\ $$

Question Number 193221 Answers: 1 Comments: 0

find the cube root of 9ab^2 + (b^2 +24a^2 )(√(b^2 −3a^2 ))

$$ \\ $$$$\boldsymbol{{find}}\:\boldsymbol{{the}}\:\boldsymbol{{cube}}\:\boldsymbol{{root}}\:\boldsymbol{{of}} \\ $$$$\mathrm{9}\boldsymbol{{ab}}^{\mathrm{2}} \:+\:\left(\boldsymbol{{b}}^{\mathrm{2}} +\mathrm{24}\boldsymbol{{a}}^{\mathrm{2}} \right)\sqrt{\boldsymbol{{b}}^{\mathrm{2}} −\mathrm{3}\boldsymbol{{a}}^{\mathrm{2}} } \\ $$

Question Number 193213 Answers: 3 Comments: 0

((a^m /a^(−n) ))^(m−n)

$$\left(\frac{{a}^{{m}} }{{a}^{−{n}} }\right)^{{m}−{n}} \\ $$

Question Number 193197 Answers: 0 Comments: 0

(−(3/4))^(666) mod1000=?

$$\left(−\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{666}} {mod}\mathrm{1000}=? \\ $$

Question Number 193195 Answers: 0 Comments: 0

(−3)^(666) mod1000=?

$$\left(−\mathrm{3}\right)^{\mathrm{666}} {mod}\mathrm{1000}=? \\ $$

Question Number 193182 Answers: 1 Comments: 0

(x^2 /(2^2 −1^2 ))+(y^2 /(2^2 −3^2 ))+(z^2 /(2^2 −5^2 ))+(w^2 /(2^2 −7^2 ))=1 (x^2 /(4^2 −1^2 ))+(y^2 /(4^2 −3^2 ))+(z^2 /(4^2 −5^2 ))+(w^2 /(4^2 −7^2 ))=1 (x^2 /(6^2 −1^2 ))+(y^2 /(6^2 −3^2 ))+(z^2 /(6^2 −5^2 ))+(w^2 /(6^2 −7^2 ))=1 (x^2 /(8^2 −1^2 ))+(y^2 /(8^2 −3^2 ))+(z^2 /(8^2 −5^2 ))+(w^2 /(8^2 −7^2 ))=1 find (x^2 +y^2 +z^2 +w^2 ).

$$ \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }+\frac{{w}^{\mathrm{2}} }{\mathrm{2}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }+\frac{{w}^{\mathrm{2}} }{\mathrm{4}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }+\frac{{w}^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{8}^{\mathrm{2}} −\mathrm{1}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{\mathrm{8}^{\mathrm{2}} −\mathrm{3}^{\mathrm{2}} }+\frac{{z}^{\mathrm{2}} }{\mathrm{8}^{\mathrm{2}} −\mathrm{5}^{\mathrm{2}} }+\frac{{w}^{\mathrm{2}} }{\mathrm{8}^{\mathrm{2}} −\mathrm{7}^{\mathrm{2}} }=\mathrm{1}\: \\ $$$$ \\ $$$${find}\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} +{w}^{\mathrm{2}} \right). \\ $$

Question Number 193087 Answers: 1 Comments: 0

Question Number 193130 Answers: 3 Comments: 1

Solve for x x^2 −c=(√(c−x))

$$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}^{\mathrm{2}} −{c}=\sqrt{{c}−{x}} \\ $$

Question Number 193049 Answers: 0 Comments: 0

Question Number 193045 Answers: 1 Comments: 0

what is the HCF of 8k+1 and 9k ? where k ∈ Z^+

$${what}\:{is}\:{the}\:{HCF}\:{of}\: \\ $$$$\mathrm{8}{k}+\mathrm{1}\:{and}\:\mathrm{9}{k}\:?\:{where}\:{k}\:\in\:\mathbb{Z}^{+} \\ $$

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