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  • 1 answers

Khushi ? 6 years, 10 months ago

Correct ur ques plz
  • 1 answers

Sia ? 6 years, 4 months ago

Steps of construction:-

Steps of construction:-

  1. Draw AB = 4 cm
  2. Draw RA perpendicular to AB, and mark C on it such that AC = 3 cm.
  3. Join BC, ABC is the original triangle
  4. Draw any ray AX making acute angle {tex}\angle BAX{/tex} with AB.
  5. Mark five points using a compass , A1, A2 ,….., A5 on AX such that AA= A1A2 = A2A3 = A3A4= A4A5 
  6. Join BA5, and draw a line parallel to it passing from A3, intersecting AB at B'.
  7. Draw line parallels to CB through B' to intersect AC at C'.
  8. {tex}\triangle{/tex}AB'C' is the required triangle.
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Sia ? 6 years, 4 months ago

When two coins are tossed simultaneously
Total number of outcomes = {HH, HT, TH, TT}
Total number of outcomes = 4
Favourable outcomes = {HT, TH} = 2
Probability of getting exactly one head ={tex}\frac { 2 } { 4 } = \frac { 1 } { 2 }{/tex}

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Sia ? 6 years, 4 months ago

Clearly,{tex}\angle M P O = 90 ^ { \circ }{/tex}
Since {tex}B K \perp H M , A H \perp H M \text { and } O P \perp H M{/tex}. Therefore, AH||OP||BK .
Let AH = x, BK = y and OP = r. Further, let  MB = z.
In {tex}\Delta M K B{/tex} and {tex}\Delta M H A{/tex},we have
{tex}\angle M K B = \angle M H A{/tex}= 90°
and {tex}\angle B M K = \angle A M H{/tex} [Common]
So, by AA criterion of similarity, we have
{tex}\Delta M K B \sim \Delta M H A{/tex}
 {tex}\Rightarrow \quad \frac { B K } { A H } = \frac { M B } { M A }{/tex}
{tex}\Rightarrow \frac { y } { x } = \frac { z } { 2 r + z }{/tex}
{tex}\Rightarrow 2 r y + y z = z x{/tex}
{tex}\Rightarrow z = \frac { 2 r y } { x - y }{/tex}...(i)
In {tex}\Delta M K B{/tex} and {tex}\Delta M P O{/tex}, We have 
{tex}\angle M K B = \angle M P O = 90 ^ { \circ }{/tex}
{tex}\angle B M K = \angle O M P{/tex} [Common]
So, by AA criterion of similarly, we obtain
{tex}\Delta M K B \sim \Delta M P O{/tex}
{tex}\Rightarrow \quad \frac { B K } { O P } = \frac { B M } { O M }{/tex}
{tex}\Rightarrow \frac { y } { r } = \frac { z } { r + z }{/tex}
{tex}\Rightarrow r y + y z = r z{/tex}
{tex}\Rightarrow \quad z = \frac { ry } { r - y }{/tex}...(ii)
From (i) and (ii), we get
{tex}\frac { r y } { r - y } = \frac { 2 r y } { x - y }{/tex}
{tex}\Rightarrow 2 r - 2 y = x - y{/tex}
{tex}\Rightarrow 2 r = x + y{/tex}
Thus, AB = AH +BK.

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Sia ? 6 years, 4 months ago

2x2 + kx +3=0
We know that quadratic equation has two equal roots only when the value of discriminant is equal to zero.
Comparing equation 2x 2 +kx +3=0 with general quadratic equation ax2 + bx + c =0, we get a = 2, b = k and c = 3
Discriminant = b2 − 4ac = k2 - 4(2)(3) = k2 - 24
Putting discriminant equal to zero
k2 - 24 = 0 ⇒ k2 = 24
{tex}\Rightarrow k = \pm \sqrt { 24 } = \pm 2 \sqrt { 6 } \Rightarrow k = 2 \sqrt { 6 } , - 2 \sqrt { 6 }{/tex}

  • 2 answers

Vedika Sopra 6 years, 10 months ago

Area of sector - area of tirangle

Samyukta ... 6 years, 10 months ago

Theta/360×πr^2- r^2 × sin theta/2
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Mayank ... 6 years, 10 months ago

Bhai rs aggarwal me diya hua hai.....
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Mayank ... 6 years, 10 months ago

Haaaain ye kya hai...?? Mehak Square or circle ka volume kb se hone lga...???

Mehak Arya 6 years, 10 months ago

Volume of circle = volume of square
  • 1 answers

Archi Mehta 6 years, 10 months ago

Secθ+tanθ=p ----------------------(1) ∵, sec²θ-tan²θ=1 or, (secθ+tanθ)(secθ-tanθ)=1 or, secθ-tanθ=1/p ----------------(2) Adding (1) and (2) we get, 2secθ=p+1/p or, secθ=(p²+1)/2p ∴, cosθ=1/secθ=2p/(p²+1) ∴, sinθ=√(1-cos²θ) =√[1-{2p/(p²+1)}²] =√[1-4p²/(p²+1)²] =√[{(p²+1)²-4p²}/(p²+1)²] =√[(p⁴+2p²+1-4p²)/(p²+1)²] =√(p⁴-2p²+1)/(p²+1) =√(p²-1)²/(p²+1) =(p²-1)/(p²+1) ∴, cosecθ=1/sinθ=1/[(p²-1)/(p²+1)]=(p²+1)/(p²-1) I HOPE THIS HELPS!
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Mayank ... 6 years, 10 months ago

Complete the question please....
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Mayank ... 6 years, 10 months ago

Question clear nhi ho paa rha hai...
  • 1 answers

Vikram Singh 6 years, 10 months ago

Copied from Brainly... → To find : HCF( 18 , 48 ) → To create : Linear Combination ( 18 , 48 , gcd( 18 , 48 ))  Steps :  ◘ Using Euclid's Algorithm :    → 48 = 18 x 2 + 12   → 18 = 12 x 1 + 6 ► Since '6' divides the last system, '6' is the HCF  ► Taking things the Anti-Algorithmic way :-     → 6 = 18 - 12             = 18 - ( 48 - ( 18 x 2 ) )             = 18 ( 3 ) - 48     → 6 = 18 ( 3 ) + 48 ( -1 ) ♦ Hence, the appropriate Linear Combination is :     •→ 6 = 18 ( 3 ) + 48 ( -1 )

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