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Coordinate GeometryQuestion and Answers: Page 6

Question Number 188998 Answers: 2 Comments: 3

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what is the answer A ? B ? C?.

$w h a t i s t h e a n s w e r$ $A ? B ? C ? .$

Question Number 188177 Answers: 2 Comments: 2

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Quel est la condition necessaire et sufisante pour que les medianes des triangles ABC et CDE soient alignes avec le point c

$Q u e l e s t l a c o n d i t i o n n e c e s s a i r e$ $e t s u f i s a n t e p o u r q u e l e s m e d i a n e s$ $d e s t r i a n g l e s A B C e t C D E s o i e n t$ $a l i g n e s a v e c l e p o i n t c$

Question Number 187362 Answers: 2 Comments: 1

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Montrer que: 4r^2 =a^2 +b^2 +c^2 +d^2

$M o n t r e r q u e :$ $4 r^{2} = a^{2} + b^{2} + c^{2} + d^{2}$

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Question Number 185169 Answers: 1 Comments: 1

calcul de l′ angle x △ABD ((sin y)/(AD))=((sin44 )/(BD)) (1) △ADC ((sin x)/(AD))=((sin48 )/(DC)) (2) (1) BD sin y=ADsin 44 (2) DCsinx =ADsin 48 (((1))/((2)))⇔ ((BDsin y)/(DCsin x))=((sin 44)/(sin 48)) (3) △BDC ((sin 14)/(BD))=((sin 28)/(DC)) ((BD)/(DC))=((sin 14)/(sin 28)) (3)⇔((sin 14×)/(sin 28×))((sin y)/(sin x))=((sin 44)/(sin 48)) Relation entre x et y △ABC (∡A +∡B+ ∡C)=180 48+44+y+28+14+x=180 x+y=46 ⇒ y=46−x ((sin x)/(sin y))=((sin 48×sin 28)/(sin 44×sin 14)) ((sinx)/(sin(46−x)))=((2sin 48×cos 14)/(sin 44)) =((sin x)/(sin 46cos x−cos 46sin x)) =((sin x)/(cos x(sin 46−tan x×cos 46))) =((tan x)/(sin 46−cos 46×tan x)) sin 44×tan x=2sin 48×cos 14(sin 46−cos 46tan x) (sin 44+2sin 48cos 14×cos 46)tan x= 2sin 48cos 14×sin 46 tan x=((2sin 48×cos 14×sin 46)/(sin 44+2sin 48×cos 14×cos 46)) sin 44= sin 48= sin 46 cos 14= cos 46= tan x=0,61150462 soit: X=31,446^°

$c a l c u l d e l^{'} a n g l e x$ $△ A B D \frac{\sin y}{AD} = \frac{\sin 44}{BD} (1)$ $△ A D C \frac{\sin x}{AD} = \frac{\sin 48}{DC} (2)$ $(1) BD \sin y = ADsin 44$ $(2) DCsinx = ADsin 48$ $\frac{(1)}{(2)} \Leftrightarrow \frac{BDsin y}{DCsin x} = \frac{\sin 44}{\sin 48} (3)$ $△ B D C \frac{\sin 14}{BD} = \frac{\sin 28}{DC}$ $\frac{BD}{DC} = \frac{\sin 14}{\sin 28}$ $(3) \Leftrightarrow \frac{\sin 14 \times}{\sin 28 \times} \frac{\sin y}{\sin x} = \frac{\sin 44}{\sin 48}$ $R e l a t i o n e n t r e x e t y$ $△ A B C (∡ A + ∡ B + ∡ C) = 180$ $48 + 44 + y + 28 + 14 + x = 180$ $x + y = 46 \Rightarrow y = 46 - x$ $\frac{\sin x}{\sin y} = \frac{\sin 48 \times \sin 28}{\sin 44 \times \sin 14}$ $\frac{sinx}{\sin (46 - x)} = \frac{2 \sin 48 \times \cos 14}{\sin 44}$ $= \frac{\sin x}{\sin 46 \cos x - \cos 46 \sin x}$ $= \frac{\sin x}{\cos x (\sin 46 - \tan x \times \cos 46)}$ $= \frac{\tan x}{\sin 46 - \cos 46 \times \tan x}$ $\sin 44 \times \tan x = 2 \sin 48 \times \cos 14 (\sin 46 - \cos 46 \tan x)$ $(\sin 44 + 2 \sin 48 \cos 14 \times \cos 46) \tan x =$ $2 \sin 48 \cos 14 \times \sin 46$ $\tan x = \frac{2 \sin 48 \times \cos 14 \times \sin 46}{\sin 44 + 2 \sin 48 \times \cos 14 \times \cos 46}$ $\sin 44 = \sin 48 = \sin 46$ $\cos 14 = \cos 46 =$ $\tan x = 0, 61150462$ $s o i t : X = 31, 446^{°}$

Question Number 184769 Answers: 0 Comments: 2

Question Number 184349 Answers: 1 Comments: 0

I_n = ∫_0 ^( 1) (1−x^2 )^n dx Relate I_n and I_(n−1) Find I_n in terms of n hence deduce that Σ_(k=0) ^n (((−1)^k ((n),(k) ))/(2k+1))=((2^(2n) (n!)^2 )/((2n+1)!))

$I_{n} = \int_{0}^{1} {(1 - x^{2})}^{n} d x$ $R e l a t e I_{n} a n d I_{n - 1}$ $F i n d I_{n} i n t e r m s o f n$ $h e n c e d e d u c e t h a t \underset{k = 0}{\sum^{n}} \frac{{(- 1)}^{k} (\begin{matrix} n \\ k \end{matrix})}{2 k + 1} = \frac{2^{2 n} {(n!)}^{2}}{(2 n + 1)!}$

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Question Number 183631 Answers: 1 Comments: 1

determiner r? AB=6 AE=5

$d e t e r m i n e r r ?$ $A B = 6 A E = 5$

Question Number 183426 Answers: 2 Comments: 3

surface de la partie bleu du graphe?

$s u r f a c e d e l a p a r t i e b l e u$ $d u g r a p h e ?$

Question Number 183324 Answers: 2 Comments: 1

Find the minimum distance between C_f , C_g ; C_f : y^2 = 4ax , C_g : x^2 + y^2 −24ay+ 128a^2 = 0

$F i n d t h e m i n i m u m d i s t a n c e b e t w e e n C_{f}, C_{g}$ $; C_{f} : y^{2} = 4 a x, C_{g} : x^{2} + y^{2} - 24 a y + 128 a^{2} = 0$

Question Number 183307 Answers: 1 Comments: 1

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Question Number 182772 Answers: 1 Comments: 0

χ { ((x= t+1)),((y= 2t−3)),((z= −t +2)) :} Δ { ((x= 3t +2)),((y= −t−1 )),((z= t+1)) :} { ((Are these two lines located)),((in the same plane or not?)),((where t∈ R)) :}

$χ {\begin{cases} x = t + 1 \\ y = 2 t - 3 \\ z = - t + 2 \end{cases} Δ {\begin{cases} x = 3 t + 2 \\ y = - t - 1 \\ z = t + 1 \end{cases} {\begin{cases} A r e t h e s e t w o l i n e s l o c a t e d \\ i n t h e s a m e p l a n e o r n o t ? \\ w h e r e t \in R \end{cases}$

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Area of Circle ?

$A r e a o f C i r c l e ?$

Pg 1 Pg 2 Pg 3 Pg 4 Pg 5 Pg 6 Pg 7 Pg 8 Pg 9 Pg 10

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