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AlgebraQuestion and Answers: Page 62

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