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p(x)=7x+14p(x)=3x(2)+7x+4

1

2

8xy -(-2xy +3xy^2)8xy +2xy - 3xy^210xy-3xy^2

the points that is equidistant from the vertices of the triangle is known as circumcenter

This point is known as the circumcenter of the circle which is equidistant from all the vertices of triangle ABC.

Cost of 8 book is ₹72 So, cost of 1 book is ₹ 72/8 = ₹9so cost of 20 books is ₹ 9 × 20 = ₹180

thank you sir

i do not got equation 10

TQ

it is has no real roots

thnQ u so much

Given: In parallelogram ABCD, P & Q any twopoints lying on the sides DC and AD.

To show:

ar (APB) = ar (BQC).Proof:

Here, ΔAPB and ||gm ABCD stands on the samebase AB and lie between same parallel AB and DC.

Therefore,

ar(ΔAPB) = 1/2 ar(||gm ABCD) — (i)

Similarly,

Parallelogram ABCD and ∆BQC stand on the samebase BC and lie between the same parallel BC and AD.

ar(ΔBQC) = 1/2 ar(||gm ABCD) — (ii)

From eq (i) and (ii),

we have

ar(ΔAPB)= ar(ΔBQC)

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Given: ABCD is a parallelogram

So, AB||CD & AD|| BC

To show:

(i) ar (APB) + ar (PCD) = ar (ABCD)

(ii) ar (APD) + ar (PBC) = ar (APB) + ar (PCD)

Proof:

(i)

Through the point P ,draw GH parallel to AB.

In a parallelogram,

AB || GH (by construction) — (i)

Thus,

AD || BC ⇒ AG || BH — (ii)

From equations (i) and (ii),

ABHG is a parallelogram.

Now,

In ΔAPB and parallelogram ABHG are lying on thesame base AB and between the same parallel lines AB and GH.

∴ar(ΔAPB) = 1/2 ar(ABHG) — (iii)

also,

In ΔPCD and parallelogram CDGH are lying on thesame base CD and between the same parallel lines CD and GH.

∴ar(ΔPCD) = 1/2 ar(CDGH) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPB)+ ar(ΔPCD) = 1/2 {ar(ABHG)+ar(CDGH)}

ar(APB) +ar(PCD) = 1/2 ar(ABCD)

(ii)

A line EF is drawn parallelto AD passing through P.

In a parallelogram,

AD || EF (by construction) — (i)

Thus,

AB || CD ⇒ AE || DF — (ii)

From equations (i) and (ii),

AEDF is a parallelogram.

Now,

In ΔAPD and parallelogram AEFD are lying on thesame base AD and between the same parallel lines AD and EF.

∴ar(ΔAPD) = 1/2 ar(AEFD) — (iii)

also,

In ΔPBC and parallelogram BCFE are lying on thesame base BC and between the same parallel lines BC and EF.

∴ar(ΔPBC) = 1/2 ar(BCFE) — (iv)

Adding equations (iii) and (iv),

ar(ΔAPD)+ ar(ΔPBC) = 1/2 {ar(AEFD)+ar(BCFE)}

ar(ΔAPD)+ ar(ΔPBC) =1/2ar(ABCD)

ar(APD) +ar(PBC) = ar(APB) + ar(PCD) ( From parti)

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LHS=cosA/(1-tanA)+sinA/(1-cotA)=cosA/(1 - sinA/cos A) + sin A/(1 - cos A/sin A)=cos²A/ (cos A - sin A) + sin²A / (sin A - cos A)=cos²A/ (cos A - sin A) - sin²A / (cos A - sin A)=(cos ² A - sin ² A) / (cos A - sin A)=(cos A - sin A)(cos A + sin A) / (cos A - sin A)=cos A + sin A i.e RHS

please check the questions once

it's wrong

da Ace

Consider cosec theta - sin theta = a³⇒ !/sin theta - sin theta = a³⇒ 1 - sin² theta/sin theta = a³cos² theta/ sin theta = a³→ (1)⇒ (cos² theta/sin theta)²/³ = (a³)²/³⇒ cos⁴/³ theta/sin²/³ theta = a²→ (2)Now consider, sec theta - cos theta = b³⇒ 1/cos theta - cos theta = b³⇒ 1 - cos²theta/cos theta = b³⇒ sin² theta/cos theta = b³→ (3)⇒ (sin² theta/cos theta)²/³ = (b³)²/³⇒ sin⁴/³ theta/cos²/³ theta = b²→ (4)Multiply (2) and (4), we get(cos⁴/³ theta/sin²/³ theta)× (sin⁴/³ theta/cos²/³ theta) = a²b²→ (5)a² + b² =(cos⁴/³ theta/sin²/³ theta) + (sin⁴/³ theta/cos²/³ theta)(cos² theta + sin² theta)/(sin²/³ theta cos²/³ theta)= 1/sin²/³ thetacos²/³ thetaConsider, a²b²(a²+b²) =(sin²/³ theta cos²/³ theta)× 1/sin²/³ theta cos²/³ theta= 1 Hence proved.

pg no. 1

pg no. 2

hand writing me bhejo

3x - 5 = 7x + 5+12=> -5-5-12 = 7x -3x=> 4x = - 22=> X = -5.5

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9×3=27 this is your answer

To solve: x^2+4x+3=0Solutionsum=4product=3numbers are 3 and 1Therefore, x^2+3x+1x+3=0x(x+3) (x+3)(x+1)(x+3)x=-1, -3

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tan49° / cot41°= tan(90°-41°)/cot41°. (as tan (90°-a)=cota)=cot41°/cot 41°=1

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L.H.S. = (cos4x + cos2x) +cos3x / (sin4x + sin2x) + sin3x

= 2cos[(4x+2x)/2].cos[(4x-2x)/2] + cos3x / 2sin[(4x+2x)/2].cos[(4x-2x)/2] + sin3x

= 2cos3x.cosx + cos3x / 2sin3x.cosx + sin3x

= cos3x(2cosx + 1) / sin3x(2cosx + 1) // Taking common

=cos3x(2cosx + 1) / sin3x(2cosx + 1)

= cos3x / sin3x

= cot3x

Hence,L.H.S = R.H.S.