Question

c. Calculate the oxidation number of
underlined atoms.

g. Na2HPO4​

Answer 1

Answer:

ᴋʜᴀ ᴛʜᴀ ʀᴇ ɪᴛᴛᴇ ᴅᴇʀ sᴇ.??

Here the number of machines and the number of days are in inverse proportion.

63/54 = x/42

63×42 = 54x

x = 63×42/54

x= 49

Hence 49 machines would be required.

Answer 2

Answer:

ᴡᴀɪᴛ sɪʀғ ᴍᴀɪ ᴋʀᴛɪ ʜᴜ.. ʜᴜᴍᴇsʜᴀ ᴛᴜ ᴋᴀʙʜɪ ɴʜɪ..

ᴛᴇᴋᴏ ᴍɪss ʙʜɪ ᴍᴀɪ ʜɪ ᴋʀᴛɪ ʜᴜ ᴛᴜ ɴᴏɪ ᴋʀᴛᴀ ᴍᴇᴋᴏ ʙɪʟᴋᴜʟ ʙʜɪ

sɪʀғ ᴇᴋ ʜɪ ᴍᴇʜɴᴀᴛ ᴋʀᴇ ᴀᴜʀ ᴅᴜsʀᴇ ᴋᴇ ᴘᴀᴀs ᴛɪᴍᴇ ʜɪ ɴʜɪ ʙᴀᴀᴛ ᴛᴀᴋ ᴋʀɴᴇ ᴋᴀ..

(￣へ ￣ 凸 (◞ ‸ ◟ㆀ)

⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)

Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)

Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s2)

Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )