Mathmasteriw
Junior Member
- Joined
- Oct 22, 2020
- Messages
- 85
so this what i get??
Any good?
Any good?
Yes! Since \(\displaystyle t^0=1\) then it can be simplified to \(\displaystyle 6t^2 +4\)
The answer is right, but it's wrong to say [MATH]3(2t)^{2-1}[/MATH]; the 2 shouldn't be inside the parentheses. What you mean is [MATH]3(2t^{3-1})[/MATH] or [MATH]3\cdot2(t^{3-1})[/MATH]so this what i get??
Any good?
Yes you're correct Dr P. I missed that.The answer is right, but it's wrong to say [MATH]3(2t)^{2-1}[/MATH]; the 2 shouldn't be inside the parentheses. What you mean is [MATH]3(2t^{3-1})[/MATH] or [MATH]3\cdot2(t^{3-1})[/MATH]
Don't forget the original problem:Great stuff guys thanks to you all for your help once again!!
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Second line from bottom. (an equation)How dose this look now?
Thanks all for your help
Seems to be dv/duSecond line from bottom. (an equation)
What did you write on the left-hand-side of that equation?
Ahh now I see, should be dv/dt ?Second line from bottom. (an equation)
What did you write on the left-hand-side of that equation?
Not quite. Look carefully, and think before writing. What are you differentiating there? Is it v? or ...Ahh now I see, should be dv/dt ?