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  • 2 answers

Chirag Sharma 2 years ago

180

Uttam Verma 2 years ago

Match box is a cuboid having its length(l), breadth(b) and height(h) as 4 cm, 2.5 cm and 1.5 cm respectively. Volume of 1 match box = l × b × h = ( 4 × 2.5 × 1.5 ) c m 3 = 15 c m 3 Hence, Volume of 12 such matchboxes = ( 15 × 12 ) c m 3 = 180 c m 3
  • 3 answers

Arshdeep Kaur 1 year, 10 months ago

P(x)=5x+8/11=0≈5x=11/8≈x=11/8×5≈x=11/40

Uttam Verma 2 years ago

p(0)=5(0)+8/11 p(0)=8/11

R K 2 years ago

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  • 2 answers

Uttam Verma 2 years ago

Inner radius of hemispherical bowl = 5 cm Thickness of the bowl = 0.25 cm ∴ Outer radius (r) of hemispherical bowl = (5 + 0.25) cm = 5.25 cm Outer CSA of hemispherical bowl = 2r2 Therefore, the outer curved surface area of the bowl is 173.25 cm2.

Uttam Verma 2 years ago

r = 5 c m , thickness of steel sheet = 0.25 c m ⇒ R = 5 c m + 0.25 c m = 5.25 c m outer curved surface area of the bowl = 2 π R 2 = 2 × 22 7 × 525 100 × 525 100 c m 2 = 173.25 c m 2
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  • 3 answers

Manshi Nayak 2 years ago

60x+20 is the right 👍

Muskan Jaiswal 2 years ago

By long division

Muskan Jaiswal 2 years ago

66
  • 1 answers

Mudit Jhanwar 2 years ago

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

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  • 5 answers

Piru Singh 2 years ago

Two circles of ready 5 cm and 3 cm in the sector 2 and the distance between centres is 4 cm find the length of the common car

Shourya Sharma 2 years ago

4
4
What a question 😂

Dona Mathew 2 years ago

4
  • 3 answers
Bhag idhrse

Krushna Solanke 2 years ago

a= 2 3+ 5​​∴ a 1​= 3+ 5​2​On rationaliging the denominator , we get , a 1​= (3+ 5​)(3− 5​) 2(3− 5​)​= 9−5 6−2 5​​= 4 6−2 5​​= 2 3− 5​​Also, (a+ a 1​) 2 =a 2 + a 2 1​+2 Substituting the values of a  = a 1​, We get , `( 2 3+ 5​​+ 2 3− 5​​) 2 =(a 2 + a 2 1​+2) ∴a 2 + a 2 1​+2=( 2 3+ 5​+3− 5​​) 2 =(3) 2 =9 ∴a 2 + a 2 1​=9−2=7.
a= 2 3+ 5 ​ ​ ∴ a 1 ​ = 3+ 5 ​ 2 ​ On rationaliging the denominator , we get , a 1 ​ = (3+ 5 ​ )(3− 5 ​ ) 2(3− 5 ​ ) ​ = 9−5 6−2 5 ​ ​ = 4 6−2 5 ​ ​ = 2 3− 5 ​ ​ Also, (a+ a 1 ​ ) 2 =a 2 + a 2 1 ​ +2 Substituting the values of a  = a 1 ​ , We get , `( 2 3+ 5 ​ ​ + 2 3− 5 ​ ​ ) 2 =(a 2 + a 2 1 ​ +2) ∴a 2 + a 2 1 ​ +2=( 2 3+ 5 ​ +3− 5 ​ ​ ) 2 =(3) 2 =9 ∴a 2 + a 2 1 ​ =9−2=7.
  • 5 answers

Samruddhi Parik 2 years ago

The answer 20 is not the area

Ajeet Yadav 2 years ago

Thanks 🙏
Half yearly exam paper òuestion and answers

Khushi Khushi 2 years ago

20

Ayush Raj 2 years ago

20
  • 2 answers

Ambika Sharma 2 years, 1 month ago

1.732

Garvit Varshney 2 years, 1 month ago

1.73205080576...
  • 5 answers

Account Deleted 2 years ago

eight

Ambika Sharma 2 years, 1 month ago

8

Garvit Varshney 2 years, 1 month ago

08

Shivam Kumar 2 years, 1 month ago

8

Dhaval Agrawal 2 years, 1 month ago

8
  • 3 answers

Naman Bansal 1 year, 11 months ago

2

Ambika Sharma 2 years, 1 month ago

11

Pawar Mrunal 2 years, 1 month ago

6*5=30 hence 28-30= -2

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