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Question Number 224637 Answers: 0 Comments: 0

Question Number 224635 Answers: 1 Comments: 0

Prove that: sin(54°) = (((√5) + 1)/4)

$$\mathrm{Prove}\:\mathrm{that}:\:\:\:\mathrm{sin}\left(\mathrm{54}°\right)\:=\:\:\frac{\sqrt{\mathrm{5}}\:+\:\mathrm{1}}{\mathrm{4}} \\ $$

Question Number 224634 Answers: 1 Comments: 0

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

Question Number 224633 Answers: 0 Comments: 0

x+y=2,4 x=2 y=4

$${x}+{y}=\mathrm{2},\mathrm{4} \\ $$$${x}=\mathrm{2} \\ $$$${y}=\mathrm{4} \\ $$

Question Number 224629 Answers: 4 Comments: 0

Question Number 224632 Answers: 0 Comments: 0

Question Number 224623 Answers: 2 Comments: 2

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

Question Number 224615 Answers: 0 Comments: 1

The vectors OP, OQ and OR represented by a,b and c respectively: where a=10i+j, b=−2i+7j, c=a+3b, and O is the origin. OR and PR intersects at M where OM=kOR and PM=lPQ and k, l are constants. Find: (i) The equation of the lines of PQ and OR (ii) The coordinayes of the point M (iii)The values of the constants k and l

$${The}\:{vectors}\:{OP},\:{OQ}\:{and}\:{OR}\:{represented} \\ $$$${by}\:{a},{b}\:{and}\:{c}\:{respectively}:\:{where}\:{a}=\mathrm{10}{i}+{j}, \\ $$$${b}=−\mathrm{2}{i}+\mathrm{7}{j},\:{c}={a}+\mathrm{3}{b},\:{and}\:{O}\:{is}\:{the}\:{origin}. \\ $$$${OR}\:{and}\:{PR}\:{intersects}\:{at}\:{M}\:{where} \\ $$$${OM}={kOR}\:{and}\:{PM}={lPQ}\:{and}\:{k},\:{l}\:{are} \\ $$$${constants}.\:{Find}: \\ $$$$\left({i}\right)\:{The}\:{equation}\:{of}\:{the}\:{lines}\:{of}\:{PQ}\:{and} \\ $$$${OR} \\ $$$$\left({ii}\right)\:{The}\:{coordinayes}\:{of}\:{the}\:{point}\:{M} \\ $$$$\left({iii}\right){The}\:{values}\:{of}\:{the}\:{constants}\:{k}\:{and}\:{l} \\ $$$$ \\ $$

Question Number 224614 Answers: 1 Comments: 0

A binary operation ∗ is defined on a set real numbers, R by x∗y = 2x + 2y −((xy)/3) . find: (i) The inverse of x under the operation ∗ (ii) Truth set when m∗7=−2∗m

$${A}\:{binary}\:{operation}\:\ast\:{is}\:{defined}\:{on}\:{a}\:{set} \\ $$$${real}\:{numbers},\:{R}\:{by} \\ $$$${x}\ast{y}\:=\:\mathrm{2}{x}\:+\:\mathrm{2}{y}\:−\frac{{xy}}{\mathrm{3}}\:. \\ $$$${find}: \\ $$$$\left({i}\right)\:{The}\:{inverse}\:{of}\:{x}\:{under}\:{the}\:{operation}\:\ast \\ $$$$\left({ii}\right)\:{Truth}\:{set}\:{when}\:{m}\ast\mathrm{7}=−\mathrm{2}\ast{m} \\ $$

Question Number 224608 Answers: 1 Comments: 1

Question Number 224606 Answers: 1 Comments: 0

α β = 9 (( (α+β))/( (α−β))) =?

$$\:\:\: \alpha\: \beta\:=\:\mathrm{9}\:\: \\ $$$$\:\:\:\:\frac{ \left(\alpha+\beta\right)}{ \left(\alpha−\beta\right)}\:=?\: \\ $$

Question Number 224600 Answers: 0 Comments: 5

i am back guys

$$\mathrm{i}\:\mathrm{am}\:\mathrm{back}\:\mathrm{guys} \\ $$

Question Number 224594 Answers: 0 Comments: 1

let I be the incenter of a non−isosceles ΔABC and let the incircle be tanget to the sides point D,E,F. the line AI intersects (ABC) at A and S. the line SD intersects (ABC) at S and T. let IT ∩ EF =M, (BIC) ∩ (DEF)=K,L. prove that KDLM is a kite.

$$ \\ $$$$\:\:\:\:\mathrm{let}\:{I}\:\mathrm{be}\:\mathrm{the}\:\mathrm{incenter}\:\mathrm{of}\:\mathrm{a}\:\mathrm{non}−\mathrm{isosceles}\:\:\Delta{ABC}\:\:\:\:\:\: \\ $$$$\:\:\:\:\mathrm{and}\:\mathrm{let}\:\mathrm{the}\:\mathrm{incircle}\:\mathrm{be}\:\mathrm{tanget}\:\mathrm{to}\:\mathrm{the}\:\mathrm{sides}\:\mathrm{point}\:{D},{E},{F}.\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{line}\:{AI}\:\mathrm{intersects}\: \left({ABC}\right)\:\mathrm{at}\:{A}\:\mathrm{and}\:{S}. \\ $$$$\:\:\:\:\:\mathrm{the}\:\mathrm{line}\:{SD}\:\mathrm{intersects}\: \left({ABC}\right)\:\mathrm{at}\:{S}\:\mathrm{and}\:{T}.\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\mathrm{let}\:{IT}\:\cap\:{EF}\:={M},\: \left({BIC}\right)\:\cap\: \left({DEF}\right)={K},{L}. \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{that}}\:{KDLM}\:\mathrm{is}\:\mathrm{a}\:\mathrm{kite}. \\ $$$$ \\ $$$$ \\ $$

Question Number 224592 Answers: 0 Comments: 3