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!!
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 ?