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Thanks, may you explain why the constant is 16 over 3 instead of 32 over 3 please
It is 2^4 × (8)^(-1/3) × (2/3)=16 × (1/2) × (2/3) = 16/3.
Please what if we have limit 0 to 1 and (2-x)^4…will I take my x=2t or x=t…..because x=2t doesn’t satisfy the limit
Good work
Trying hard to cater the answers in the simple manner. Thanks 😊
Nice 👍, but you were too fast, Please take your time to break it down next time, you skipped a lot of solvings and got me lost many times.
note: SELF-OBSERVATION: Q1. Ans: [(𝟏/𝟔) 𝑩(𝟏/𝟔,𝟏/𝟐)]; Q3. Ans: [𝟐𝟒𝟑/𝟕𝟕𝟎]
Thanks, may you explain why the constant is 16 over 3 instead of 32 over 3 please
It is 2^4 × (8)^(-1/3) × (2/3)=16 × (1/2) × (2/3) = 16/3.
Please what if we have limit 0 to 1 and (2-x)^4…will I take my x=2t or x=t…..because x=2t doesn’t satisfy the limit
Good work
Trying hard to cater the answers in the simple manner. Thanks 😊
Nice 👍, but you were too fast,
Please take your time to break it down next time, you skipped a lot of solvings and got me lost many times.
note: SELF-OBSERVATION: Q1. Ans: [(𝟏/𝟔) 𝑩(𝟏/𝟔,𝟏/𝟐)]; Q3. Ans: [𝟐𝟒𝟑/𝟕𝟕𝟎]