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  • 1 answers

Divina S.S 9 months, 3 weeks ago

∠OAP = 90° ∠AOP = 180° - (90° +35° ) = 55°  THATS ALL
  • 3 answers

Umang Soni 9 months, 3 weeks ago

Yes that's I what confused here by the way thanks divina

Divina S.S 9 months, 3 weeks ago

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Sahil Kumar 9 months, 3 weeks ago

First you draw a quadrilateral and inside the circle and name the square A B C D sequencely after that let circle touch the quadrilateral at point P Q R S , P between AB , Q between b and c , R between c and d and s between a and d SO, AP = AS , BP = BQ , RC = CQ AND RD = DS (PROPERTY OF TANGENT )SARA EQUATION KO ADD KR DO AP + BP + RC + RD = AS + BQ + CQ + DS AB + CD = AD + BC Apne prove me galti se BC ke jagah Da likh diye h
  • 4 answers

Piyush Barnwal 9 months, 3 weeks ago

Sorry d=6

Piyush Barnwal 9 months, 3 weeks ago

d=4

Umair Hadi 9 months, 3 weeks ago

6

Ayush Kesharwani 9 months, 3 weeks ago

d=6
  • 1 answers

Devansh Kumar 9 months, 3 weeks ago

f(x)= x^2-px+q a=1 b=-p C=q a+b=-b/a =-(-p)/1 =p a×b=c/a =q/1 =q
  • 1 answers

Ashish Raj 9 months, 3 weeks ago

4.27
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  • 1 answers

Alisha Kushwah 9 months, 3 weeks ago

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  • 1 answers

Piyush Barnwal 9 months, 3 weeks ago

Add and subtract both the equations you get the value of cosec and cot as cot=cos /sin and cosec =1/sin put the value of sin
  • 1 answers

Piyush Barnwal 9 months, 3 weeks ago

542 is not a sq. Root no.
  • 5 answers

Sarthak Varshney 9 months, 3 weeks ago

12

Lavya Sharma 9 months, 3 weeks ago

X^2 -3 =x^2 =-3 X= underroot -3

Kanu Choudhary 9 months, 3 weeks ago

X²-3 X²-√3² (X+√3)(x-√3) X=-√3, X=√3

Pramith Fernandes A 9 months, 3 weeks ago

Sorry 100%

Pramith Fernandes A 9 months, 3 weeks ago

x=−√3 and x=+√3 this is the correct answer 100
  • 1 answers

Praharshitha Samapada Lakshmi 9 months, 3 weeks ago

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  • 0 answers
  • 1 answers

Kanishk K 9 months, 2 weeks ago

CosecA= 5/3
  • 2 answers

Samya Wadhwani 9 months, 3 weeks ago

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Swapnil Verma 9 months, 3 weeks ago

Given: Height of cylinder(H) = 10cm Diameter of base of cylinder = 8cm Therefore, Radius of the cylinder(R) = (8/2)cm = 4cm Height of conical holes, made on both the ends of cylinder(h) = 4cm Diameter of conical holes = 6cm Therefore, Radius of conical holes(r) = (6/2)cm = 3cm Slant height of conical holes(l) = (h² + r²)^1/2 = {(4)² + (3)²}^1/2 = (16 + 9)^1/2 = (25)^1/2 = 5cm To find: i) T.S.A. of cylinder. ii) T.S.A. of remaining solid. Solution: i) T.S.A. of cylinder = 2 × (22/7) × R × (R + H) = 2 × (22/7) × 4 × (4 + 10) = 352 cm² ii) T.S.A. of remaining solid = { (T.S.A. of cylinder) - 2(Base area of conical hole) } + { 2(C.S.A. of conical hole) } = [ { 352 } - 2{ (22/7) × r² } ] + [ 2{ (22/7) × r × l } ] = { 352 - (396/7) } + { (660/7) } = (2464 + 660 - 396)/7 = 390 cm² (approximately)
  • 3 answers

Devendra 99 9 months, 3 weeks ago

2(sin^2@ - (1-sin^2@))= -1 sin^@ -1 + sin^2@ = -1/2 2sin^@ = - 1/2 + 1 sin^2@ = 1/2 So, @ = 30° Now, tan^@ = tan^2 30° i.e1/√3 So ans is 1/√3

Mohammad Izaan 9 months, 4 weeks ago

1

Kush Shukla 9 months, 3 weeks ago

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  • 3 answers

Mohammad Izaan 9 months, 4 weeks ago

1

Bhaskar Pali 9 months, 4 weeks ago

Foreground is a good night

Aditya Raghuwanshi 9 months, 4 weeks ago

0
  • 5 answers

Vishal Dubay 9 months, 3 weeks ago

2x2x3x3x3

Tushar S 9 months, 4 weeks ago

2×2×3×3×3

Meena Verma 9 months, 4 weeks ago

2×2×3×3×3

Badal Kshatriya 9 months, 4 weeks ago

2²×3³

Chaitanya Dubey 9 months, 4 weeks ago

34
  • 0 answers
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  • 2 answers

Palak Siwach 9 months, 3 weeks ago

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Palak Siwach 9 months, 3 weeks ago

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  • 2 answers

Meena Verma 9 months, 4 weeks ago

2x^2+3x-6 a=2 b=3 c=-6 Using quadratic formula = -b +-√b^2-4ac/2a = -3+-√3^2-4×2×(-6) /2×2 = -3+-√9+48/4 = -3+-√57/4 Roots are -3+√57/4 and -3-√57/4

Palak Siwach 9 months, 4 weeks ago

–3±「57/4
  • 2 answers

Vishal Dubay 9 months, 2 weeks ago

In △BMC and △EMD ∠BMC=∠EMD [Vertically opposite] MC=DM [Given] ∠BCM=∠EDM [Alternate angles] ∴△BMC≅△EMD [By ASA] Hence, BC=DE [By CPCT] →(1) AE=AD+DE=BC+BC=2BC →(2) Now, △BLC∼△ELA (AA Similarly) BLEL=BCAE [By CPCT] BLEL=BC2BC BLEL=12 EL=2BL 

Chetan Jha 10 months ago

In △BMC and △EMD ∠BMC=∠EMD [Vertically opposite] MC=DM [Given] ∠BCM=∠EDM [Alternate angles] ∴△BMC≅△EMD [By ASA] Hence, BC=DE [By CPCT] →(1) AE=AD+DE=BC+BC=2BC →(2) Now, △BLC∼△ELA (AA Similarly) BLEL=BCAE [By CPCT] BLEL=BC2BC BLEL=12 EL=2BL 
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  • 5 answers

Not Fine Bro 🎭 9 months, 4 weeks ago

Your question is wrong

Giriraj Joshi 10 months ago

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Giriraj Joshi 10 months ago

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Not Fine Bro 🎭 10 months ago

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Giriraj Joshi 10 months ago

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