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AlgebraQuestion and Answers: Page 69

Question Number 192661 Answers: 1 Comments: 2

Question Number 192630 Answers: 0 Comments: 0

z=xy−5x+2y. find (dz/dx) and (dz/dy) at(2,4)

$${z}={xy}−\mathrm{5}{x}+\mathrm{2}{y}.\:{find}\:\frac{{dz}}{{dx}}\:{and}\:\frac{{dz}}{{dy}}\:{at}\left(\mathrm{2},\mathrm{4}\right) \\ $$

Question Number 192629 Answers: 0 Comments: 0

Z=f(x_(1,) x_(2,) x_3 )=x_1 x_2 +x_1 ^5 −x_2 ^2 x_3 find f_1 ,f_(11) ,and f_(21)

$${Z}={f}\left({x}_{\mathrm{1},} {x}_{\mathrm{2},} {x}_{\mathrm{3}} \right)={x}_{\mathrm{1}} {x}_{\mathrm{2}} +{x}_{\mathrm{1}} ^{\mathrm{5}} −{x}_{\mathrm{2}} ^{\mathrm{2}} {x}_{\mathrm{3}} \:{find}\:{f}_{\mathrm{1}} ,{f}_{\mathrm{11}} ,{and}\:{f}_{\mathrm{21}} \\ $$

Question Number 192625 Answers: 3 Comments: 1

lim_(x→0) ((sin^4 (πcos(x)))/(1−cos(1−cos(1−cos(x)))))

$${lim}_{{x}\rightarrow\mathrm{0}} \frac{{sin}^{\mathrm{4}} \left(\pi{cos}\left({x}\right)\right)}{\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left(\mathrm{1}−{cos}\left({x}\right)\right)\right)} \\ $$

Question Number 192596 Answers: 2 Comments: 0

Question Number 192594 Answers: 0 Comments: 0

Question Number 192542 Answers: 2 Comments: 0

Question Number 192508 Answers: 1 Comments: 1

Question Number 192481 Answers: 1 Comments: 0

Question Number 192477 Answers: 2 Comments: 0

Find: ((((25)/(42)) − (5/(16)) + ((10)/9) − (2/3))/((3/8) + (4/5) − (5/7) − (4/3))) = ?

$$\mathrm{Find}:\:\:\:\:\:\frac{\frac{\mathrm{25}}{\mathrm{42}}\:−\:\frac{\mathrm{5}}{\mathrm{16}}\:+\:\frac{\mathrm{10}}{\mathrm{9}}\:−\:\frac{\mathrm{2}}{\mathrm{3}}}{\frac{\mathrm{3}}{\mathrm{8}}\:+\:\frac{\mathrm{4}}{\mathrm{5}}\:−\:\frac{\mathrm{5}}{\mathrm{7}}\:−\:\frac{\mathrm{4}}{\mathrm{3}}}\:=\:? \\ $$

Question Number 192463 Answers: 3 Comments: 0

Question Number 192440 Answers: 1 Comments: 0

Given { ((A=(((p^2 +q^2 +r^2 )^2 )/((pq)^2 +(pr)^2 +(qr)^2 )))),((B=((q^2 −pr)/(p^2 +q^2 +r^2 )) )) :} If p+q+r=0 then A^2 −4B=?

$$\:\:\:\:\mathrm{Given}\:\begin{cases}{\mathrm{A}=\frac{\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{pq}\right)^{\mathrm{2}} +\left(\mathrm{pr}\right)^{\mathrm{2}} +\left(\mathrm{qr}\right)^{\mathrm{2}} }}\\{\mathrm{B}=\frac{\mathrm{q}^{\mathrm{2}} −\mathrm{pr}}{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }\:}\end{cases}\:\:\:\:\:\: \\ $$$$\:\mathrm{If}\:\mathrm{p}+\mathrm{q}+\mathrm{r}=\mathrm{0}\:\mathrm{then}\:\mathrm{A}^{\mathrm{2}} −\mathrm{4B}=? \\ $$$$ \\ $$

Question Number 192437 Answers: 1 Comments: 0

(1/a) + (1/b) + (1/c) = (1/(a + b + c)) . Prove that (1/a^5 ) + (1/b^5 ) + (1/c^5 ) = (1/(a^5 + b^5 + c^5 )) = (1/((a + b + c)^5 ))

$$\frac{\mathrm{1}}{{a}}\:+\:\frac{\mathrm{1}}{{b}}\:+\:\frac{\mathrm{1}}{{c}}\:=\:\frac{\mathrm{1}}{{a}\:+\:{b}\:+\:{c}}\:.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\frac{\mathrm{1}}{{a}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{b}^{\mathrm{5}} }\:+\:\frac{\mathrm{1}}{{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{{a}^{\mathrm{5}} \:+\:{b}^{\mathrm{5}} \:+\:{c}^{\mathrm{5}} }\:=\:\frac{\mathrm{1}}{\left({a}\:+\:{b}\:+\:{c}\right)^{\mathrm{5}} } \\ $$

Question Number 192425 Answers: 1 Comments: 0

Question if “k” is odd & A=1^k +2^k +...+n^(k ) & B=1+2+...+n prove that : B ∣ A

$${Question} \\ $$$${if}\:\:``{k}''\:{is}\:{odd}\:\:\&\:{A}=\mathrm{1}^{{k}} +\mathrm{2}^{{k}} +...+{n}^{{k}\:\:} \:\&\:\:{B}=\mathrm{1}+\mathrm{2}+...+{n} \\ $$$${prove}\:{that}\:\::\:\:{B}\:\mid\:{A}\: \\ $$

Question Number 192409 Answers: 1 Comments: 0

Question Number 192399 Answers: 1 Comments: 1

Algebra (1 ) G, is a group and o(G ) = p^( 2) . prove that G is an abelian group. hint: ( p is prime number ) −−−−−−−−−−−−−

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{Algebra}\:\left(\mathrm{1}\:\right) \\ $$$$\:\:{G},\:{is}\:{a}\:{group}\:\:{and}\:\:\:{o}\left({G}\:\right)\:=\:{p}^{\:\mathrm{2}} \:. \\ $$$$\:\:\:{prove}\:{that}\:{G}\:{is}\:{an}\:{abelian}\:{group}. \\ $$$$\:\:\:{hint}:\:\:\left(\:{p}\:{is}\:{prime}\:{number}\:\:\right) \\ $$$$\:\:\:\:\:−−−−−−−−−−−−− \\ $$

Question Number 192387 Answers: 1 Comments: 0

Simplify (√(2(1+(√(4+(((2017^4 −1)/(2017^2 )))^2 ))))) is ....

$$\:\mathrm{Simplify}\: \\ $$$$\:\sqrt{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{4}+\left(\frac{\mathrm{2017}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2017}^{\mathrm{2}} }\right)^{\mathrm{2}} }\right)}\: \\ $$$$\:\mathrm{is}\:....\: \\ $$

Question Number 192374 Answers: 3 Comments: 1

why “ 200!<100^(200) ” ?

$${why}\:\:\:``\:\mathrm{200}!<\mathrm{100}^{\mathrm{200}} \:''\:? \\ $$

Question Number 192370 Answers: 2 Comments: 0

Solve the equation x^4 − 2x^3 + 4x^2 + 6x − 21 = 0, Given that the sum of two of its roots is zero

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}\:\:\:\mathrm{x}^{\mathrm{4}} \:\:−\:\:\mathrm{2x}^{\mathrm{3}} \:\:+\:\:\mathrm{4x}^{\mathrm{2}} \:\:+\:\:\mathrm{6x}\:\:\:−\:\:\mathrm{21}\:\:\:=\:\:\:\mathrm{0},\:\: \\ $$$$\mathrm{Given}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{of}\:\mathrm{its}\:\mathrm{roots}\:\mathrm{is}\:\mathrm{zero} \\ $$

Question Number 192363 Answers: 0 Comments: 0

Solve for x x_i −x+(2cx−cb)(y_i +cx^2 −cbx)=0 the following is true for this equaition i)c>0 ii)b>0 iii)there is only one real solution

$$\mathrm{Solve}\:\mathrm{for}\:{x} \\ $$$${x}_{{i}} −{x}+\left(\mathrm{2}{cx}−{cb}\right)\left({y}_{{i}} +{cx}^{\mathrm{2}} −{cbx}\right)=\mathrm{0} \\ $$$$\mathrm{the}\:\mathrm{following}\:\mathrm{is}\:\mathrm{true}\:\mathrm{for}\:\mathrm{this}\:\mathrm{equaition} \\ $$$$\left.{i}\right)\mathrm{c}>\mathrm{0} \\ $$$$\left.{ii}\right)\mathrm{b}>\mathrm{0} \\ $$$$\left.{iii}\right)\mathrm{there}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{solution} \\ $$

Question Number 192330 Answers: 0 Comments: 0

Question Number 192350 Answers: 2 Comments: 0

Question Number 192186 Answers: 3 Comments: 0

If α, β and γ are the roots of x^3 + px + q = 0, find Σα^4 .

$$\mathrm{If}\:\:\alpha,\:\beta\:\:\mathrm{and}\:\gamma\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\:\:\mathrm{x}^{\mathrm{3}} \:\:+\:\:\mathrm{px}\:\:+\:\:\mathrm{q}\:\:=\:\:\mathrm{0},\:\:\:\:\mathrm{find}\:\:\:\Sigma\alpha^{\mathrm{4}} . \\ $$

Question Number 192171 Answers: 1 Comments: 2

2^x^x^x =2^(√2) x=?

$$\mathrm{2}^{{x}^{{x}^{{x}} } } =\mathrm{2}^{\sqrt{\mathrm{2}}} \\ $$$${x}=? \\ $$

Question Number 192173 Answers: 1 Comments: 0

prove that (x^2 +a^2 )^4 = (x^4 −6x^2 a^2 +a^4 )^2 +(4x^3 a−4xa^3 )^2

$${prove}\:{that} \\ $$$$\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)^{\mathrm{4}} \:=\:\left({x}^{\mathrm{4}} −\mathrm{6}{x}^{\mathrm{2}} {a}^{\mathrm{2}} +{a}^{\mathrm{4}} \right)^{\mathrm{2}} +\left(\mathrm{4}{x}^{\mathrm{3}} {a}−\mathrm{4}{xa}^{\mathrm{3}} \right)^{\mathrm{2}} \\ $$

Question Number 192149 Answers: 5 Comments: 0

Find: (1/2) + (3/2^3 ) + (5/2^5 ) + (7/2^7 ) + ...

$$\mathrm{Find}: \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:+\:\frac{\mathrm{3}}{\mathrm{2}^{\mathrm{3}} }\:+\:\frac{\mathrm{5}}{\mathrm{2}^{\mathrm{5}} }\:+\:\frac{\mathrm{7}}{\mathrm{2}^{\mathrm{7}} }\:+\:... \\ $$

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