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  • 1 answers

Parul Negi 3 years, 2 months ago

Take part in a feast.
  • 4 answers

Vani Singh 3 years, 2 months ago

They formed metal oxide

Siddharth Singh 3 years, 3 months ago

They Form Metal Oxides by reacting with O² present in the air . For Eg :- 2Mg+0²=2Mgo

H I M A N S H U ... 3 years, 3 months ago

The metal convert into metal oxide due to the presence of oxygen present in the atmosphere

Vansh Singh 3 years, 3 months ago

Metallic Oxides are formed as metals react with oxygen present in air.
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Preeti Dabral 3 years, 1 month ago

{tex}\mathop {\lim }\limits_{x \to {2^ - }} f(x) = \mathop {\lim }\limits_{x \to {2^ - }} 2x + 1{/tex}
{tex}\mathop {\lim }\limits_{h \to 0} \left[ {2\left( {2 - h} \right) + 1} \right]{/tex} = 5
{tex}= \mathop {\lim }\limits_{x \to {2^ + }} f(x) = \mathop {\lim }\limits_{x \to {2^ + }} \left( {3x -1} \right){/tex}
{tex} = \mathop {\lim }\limits_{h \to 0} 3\left( {2 + h} \right) - 1{/tex} = 5
In given that question f(x) is continuous at x=2,therefore
{tex} \mathop {\lim }\limits_{x \to {2^ - }} f(x) = f(2) = \mathop {\lim }\limits_{x \to {2^ + }} f(x){/tex}
5 = k 
{tex}\Rightarrow k = 5{/tex}

  • 2 answers

Ritu Raj 3 years, 3 months ago

No of chairs = 85.Cost of each chair = ₹ 180 ,Cost of 85 chairs = ₹85×180.Number of tables =₹ 25,Cost of each table = ₹ 140,Cost of 25 tables = ₹25 x 140,Total cost of all chairs and tables = ₹85 x 180 + 25 x 140₹) = ₹ 15300 + 3500 = ₹ 18800 (Money given for advance = ₹ 2500 ∴ Money pay to the dealer = ₹ 18800 – ₹ 2500 = ₹ 16300.
(Number of chairs = 85) (Cost of one chair = ₹ 180) (Cost of 85 chairs = ₹ (85 x 180) (Number of tables = 25) (Cost of one table = ₹ 140) (Cost of 25 tables = ₹ (25 x 140) (Total cost of all chairs and tables = ₹(85 x 180 + 25 x 140) = ₹ (15300 + 3500) = ₹ 18800 (Money given in advance = ₹ 2500) /∴ Balance money to be paid to the dealer = ₹ 18800 – ₹ 2500 = ₹ 16300.
  • 3 answers

Shreyansh Koserwal 3 years, 3 months ago

Inamgaon
Inamgaon is situated on the river Ghod. Ghod River is located in Pune District, Maharashtra, western India. It is a tributary of the Bhima River

Navya Awari 3 years, 3 months ago

H
  • 1 answers

Anshul Pratap Singh 2 years, 8 months ago

See ncert book 📚
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Abhinav Sharma 3 years, 3 months ago

-14a4b + 2a3b3
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Q.Z Khan 3 years, 2 months ago

Demand deposit basically means that if we have deposit money in banks or any formal sector we can withdraw our money according to our wish aur our demand ...
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Anuj Kumar 3 years, 3 months ago

<font face ="Times New Roman">Incomplete question
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Ritu Raj 3 years, 3 months ago

Burning of wastes Leaves which fall from tree
  • 2 answers

Ravinsh Kumar 3 years, 2 months ago

King Tut

Deepika Pandey 3 years, 3 months ago

He was a tenegar when hi died
  • 2 answers

Parul Negi 3 years, 2 months ago

A man with a conscience.

Putuli Kumari P K 3 years, 2 months ago

From where has this story been taken?
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Shivani Kanhaiya 3 years, 1 month ago

She is right now for

Priya Singh 3 years, 1 month ago

Bangal ki Nawab aur East India company ke bich Vivad company dwara banae Gaye Kille aur Vyapar mein Lagan Na Dena company ki rajnitik mamalon mein Aana ke dalna Aadi the
  • 3 answers

Sasmita Prusty 3 years, 1 month ago

The father of nesan is mahatma Gandhi

Nirupama Behera 3 years, 2 months ago

Which of the father of nesan

Lithika Shree 3 years, 2 months ago

A substance which makes the leaves green
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Preeti Dabral 3 years ago

Akbar, in full Abū al-Fatḥ Jalāl al-Dīn Muḥammad Akbar, (born October 15?, 1542, Umarkot [now in Sindh province, Pakistan]—died c. October 25, 1605, Agra, India), the greatest of the Mughal emperors of India. He reigned from 1556 to 1605 and extended Mughal power over most of the Indian subcontinent.

  • 3 answers
2 is the correct answer

Student Student 3 years, 3 months ago

2

Shreya Gupta 3 years, 3 months ago

SinA=√2 - cosA Perpendicular =√2 - cosA Hypotenuse = 1 H^2=P^2+B^2 1^2=(√2 - cosA)^2+ B^2 1=√2^2+CosA^2 - 2×√2×CosA + B^2 1=2+cos^2A - 2√2cosA + B^2 -1=cos^2A - 2√2cosA+ B^2 B^2=-1- cos^2A + 2√2cosAB B=√(-1- cos^2A + 2√2cosAB) tanA + CotA= ? TanA=P/B= √2 - cosA/√(-1- cos^2A + 2√2cosAB) CotA=B/P √(-1- cos^2A + 2√2cosAB)/√2 - cosA tanA + CotA= ? √2 - cosA/√(-1- cos^2A + 2√2cosAB) + √(-1- cos^2A + 2√2cosAB)/√2 - cosA
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