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  • 2 answers

Samya Wadhwani 1 year, 8 months ago

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Swapnil Verma 1 year, 8 months 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 1 year, 8 months 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 1 year, 8 months ago

1

Kush Shukla 1 year, 8 months ago

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  • 3 answers

Mohammad Izaan 1 year, 8 months ago

1

Bhaskar Pali 1 year, 8 months ago

Foreground is a good night

Aditya Raghuwanshi 1 year, 8 months ago

0
  • 5 answers

Vishal Dubay 1 year, 8 months ago

2x2x3x3x3

Tushar R 1 year, 8 months ago

2×2×3×3×3

Meena Verma 1 year, 8 months ago

2×2×3×3×3

Badal Kshatriya 1 year, 8 months ago

2²×3³

Chaitanya Dubey 1 year, 8 months ago

34
  • 0 answers
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Meena Verma 1 year, 8 months 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 1 year, 9 months ago

–3±「57/4
  • 1 answers

Kaushik Verma 1 year, 9 months ago

substract 7 in each number then find LCM
  • 2 answers

Vishal Dubay 1 year, 8 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 

Chetan Jha 1 year, 9 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 
  • 0 answers
  • 1 answers

Bhavesh Singh 1 year, 9 months ago

Show that 3-2√2 is irrational number when it is given that √2 is an irrational number.
  • 2 answers

Tara Bai 1 year, 9 months ago

Thank u...but can u show the steps to solve dis

Lovejeet Kaur 1 year, 9 months ago

Hence, the usual speed of the train is 40 km/h. Q. A train covers a distance of 480 km at a uniform speed. If speed had been 8km /hr less then it would have taken 3 hours more to cover the same distance.
  • 3 answers

Palak Siwach 1 year, 8 months ago

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Shiran Khan 1 year, 8 months ago

Your question is wrong because AC is already given

Vaishnavi Chaudhary 1 year, 9 months ago

Your answer is wrong cuz AC is already given in question
  • 1 answers

Deepti Pandey 1 year, 9 months ago

To prove that this is equal to sin²A.cos²A
  • 2 answers

Divyansh Kuntal 1 year, 9 months ago

40/2 (2×6 +(40-1) ×6) =4920

Swanand Gajkeshwar 1 year, 9 months ago

4920
  • 1 answers

Shubha Bm 1 year, 8 months ago

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Few rules to keep homework help section safe, clean and informative.
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  • 0 answers
  • 0 answers
  • 2 answers

Swanand Gajkeshwar 1 year, 9 months ago

Let∅=30° sin30°×cos30° =√3/4

Suhaani .K 1 year, 9 months ago

Can a tangent be drawn from the radius of a circle to outside the circle if not then what would be call that line that is passing through radius to outside of the circle

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