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Question Number 193210    Answers: 1   Comments: 0

Question Number 193205    Answers: 1   Comments: 1

Question Number 193204    Answers: 1   Comments: 0

Question Number 193199    Answers: 1   Comments: 0

a_1 , a_2 ,...,a_n are mutually distinct and is a am sequence . if a_( 1) +a_( 2) +...+a_n =A and a_1 ^( 2) + a_2 ^2 +..+ a_n ^( 2) = B find the am sequence.

$$ \\ $$$$\:\:\:{a}_{\mathrm{1}} \:,\:{a}_{\mathrm{2}} \:,...,{a}_{{n}} \:{are}\:\:{mutually}\:{distinct} \\ $$$$\:\:{and}\:{is}\:{a}\:\:\:\:{am}\:\:{sequence}\:. \\ $$$$\:\:\:{if}\:{a}_{\:\mathrm{1}} \:+{a}_{\:\mathrm{2}} \:+...+{a}_{{n}} \:={A} \\ $$$$\:\:\:{and}\:\: \\ $$$$\:\:\:{a}_{\mathrm{1}} ^{\:\mathrm{2}} \:+\:{a}_{\mathrm{2}} ^{\mathrm{2}} \:+..+\:{a}_{{n}} ^{\:\mathrm{2}} =\:{B} \\ $$$$\:\:\:\:{find}\:\:{the}\:\:{am}\:\:{sequence}. \\ $$

Question Number 193198    Answers: 1   Comments: 0

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 193192    Answers: 1   Comments: 0

solve and solution Ω=∫(√(sin^(−1) x))dx=?

$$\mathrm{solve}\:\mathrm{and}\:\mathrm{solution} \\ $$$$\Omega=\int\sqrt{\mathrm{sin}^{−\mathrm{1}} \mathrm{x}}\mathrm{dx}=? \\ $$

Question Number 193203    Answers: 1   Comments: 0

Question Number 193183    Answers: 1   Comments: 0

There exists a unique positive integer a for which The sum u = Σ_(n=1) ^(2023) ⌊((n^2 −na)/5)⌋ is an integer strictly between −1000 & 1000 find a+u.

$$ \\ $$$${There}\:{exists}\:{a}\:{unique}\:{positive}\:{integer}\:{a}\:{for} \\ $$$${which}\:{The}\:{sum}\:{u}\:\:=\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{2023}} {\sum}}\lfloor\frac{{n}^{\mathrm{2}} −{na}}{\mathrm{5}}\rfloor\:{is}\:{an}\:{integer} \\ $$$${strictly}\:{between}\:−\mathrm{1000}\:\&\:\mathrm{1000}\:{find}\:{a}+{u}. \\ $$

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 193180    Answers: 3   Comments: 0

If a + b = 1001 & HCF(a, b) = 13 then how many set of a & b.

$$\mathrm{If}\:{a}\:+\:{b}\:=\:\mathrm{1001}\:\&\:\mathrm{HCF}\left({a},\:{b}\right)\:=\:\mathrm{13}\: \\ $$$$\mathrm{then}\:\mathrm{how}\:\mathrm{many}\:\mathrm{set}\:\mathrm{of}\:{a}\:\&\:{b}. \\ $$

Question Number 193179    Answers: 0   Comments: 0

f(θ) = ((v^2 sinθcosθ+vcos (√(v^2 sin^2 θ+Hg )) )/g) f(θ)_(max) =?

$$\:\:{f}\left(\theta\right)\:=\:\frac{{v}^{\mathrm{2}} \mathrm{sin}\theta\mathrm{cos}\theta+{v}\mathrm{cos}\:\sqrt{{v}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \theta+{Hg}\:}\:}{{g}} \\ $$$$\:\:\:{f}\left(\theta\right)_{{max}} =? \\ $$

Question Number 193175    Answers: 3   Comments: 0

please solve for x if 2x^2 =2^x

$${please}\:{solve}\:{for}\:{x}\:{if} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} =\mathrm{2}^{{x}} \\ $$

Question Number 193171    Answers: 0   Comments: 0

Question Number 193170    Answers: 1   Comments: 0

solve : 7^x =−2

$$\mathrm{solve}\::\:\mathrm{7}^{\mathrm{x}} =−\mathrm{2} \\ $$

Question Number 193162    Answers: 1   Comments: 2

Question Number 193161    Answers: 1   Comments: 0

Question Number 193153    Answers: 0   Comments: 0

Question Number 193149    Answers: 1   Comments: 0

Question Number 193139    Answers: 3   Comments: 0

Solve the inequalities: ∣x−(1/2)∣>1 Thank you

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{inequalities}: \\ $$$$\mid\mathrm{x}−\frac{\mathrm{1}}{\mathrm{2}}\mid>\mathrm{1} \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$

Question Number 193138    Answers: 2   Comments: 0

Show that for all a,b∈R i) ab≤(1/2)(a^2 +b^2 ) ii) (((a+b)/2))^2 ≤(a^2 +b^2 ) iii) (√(ab))≤(1/2)(a+b), for a,b≥0 such that a and b have square roots. Help

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{a},\mathrm{b}\in\mathbb{R} \\ $$$$\left.\mathrm{i}\right)\:\mathrm{ab}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{ii}\right)\:\left(\frac{\mathrm{a}+\mathrm{b}}{\mathrm{2}}\right)^{\mathrm{2}} \leqslant\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{iii}\right)\:\sqrt{\mathrm{ab}}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{a}+\mathrm{b}\right),\:\mathrm{for}\:\mathrm{a},\mathrm{b}\geqslant\mathrm{0}\:\mathrm{such}\:\mathrm{that} \\ $$$$\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:\mathrm{have}\:\mathrm{square}\:\mathrm{roots}. \\ $$$$ \\ $$$$\mathrm{Help} \\ $$

Question Number 193137    Answers: 2   Comments: 0

1) Prove that: ∣a+b+c∣≥∣a∣−∣b∣−∣c∣ 2) Find all x∈R that satify the follow− ing inequalities i) ∣x^2 −4∣<5 ii) ∣x∣+∣x+2∣<5 Help!

$$\left.\mathrm{1}\right)\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mid\mathrm{a}+\mathrm{b}+\mathrm{c}\mid\geqslant\mid\mathrm{a}\mid−\mid\mathrm{b}\mid−\mid\mathrm{c}\mid \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:\mathrm{Find}\:\mathrm{all}\:\mathrm{x}\in\mathbb{R}\:\mathrm{that}\:\mathrm{satify}\:\mathrm{the}\:\mathrm{follow}− \\ $$$$\mathrm{ing}\:\mathrm{inequalities}\: \\ $$$$\left.\mathrm{i}\right)\:\mid\mathrm{x}^{\mathrm{2}} −\mathrm{4}\mid<\mathrm{5} \\ $$$$\left.\mathrm{ii}\right)\:\mid\mathrm{x}\mid+\mid\mathrm{x}+\mathrm{2}\mid<\mathrm{5} \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 193117    Answers: 3   Comments: 0

lim_(x→0) (cosx)^(1/x)

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{cosx}\right)^{\frac{\mathrm{1}}{\mathrm{x}}} \\ $$

Question Number 193116    Answers: 2   Comments: 0

Show that for all a,b,c,d ∈ R with a,b,c,d ≥ 0 1) (√(ab))(√(cd)) ≤ (1/4)(a^2 +b^2 +c^2 +d^2 ) 2) (abcd)^(1/4) ≤ (1/4)(a+b+c+d) Help!

$$\mathrm{Show}\:\mathrm{that}\:\mathrm{for}\:\mathrm{all}\:\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\in\:\mathbb{R}\:\mathrm{with} \\ $$$$\mathrm{a},\mathrm{b},\mathrm{c},\mathrm{d}\:\geqslant\:\mathrm{0}\: \\ $$$$\left.\mathrm{1}\right)\:\sqrt{\mathrm{ab}}\sqrt{\mathrm{cd}}\:\leqslant\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{d}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{2}\right)\:\left(\mathrm{abcd}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \:\leqslant\:\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}\right) \\ $$$$ \\ $$$$\mathrm{Help}! \\ $$

Question Number 193111    Answers: 2   Comments: 0

⇃

$$\:\:\:\:\:\:\downharpoonleft\underline{\:} \\ $$

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