solving for v_i

[speed4baseball]

hello there viewers

i have found a new problem

v^2_f=v^2_i+2ad *solve for v_i

*_ means subscript

so i tried dividing both sides by v and the result was v_i = 2ad/f
and i dont think thats right
experts?

 

[galactus]

\(\displaystyle v^{2}_{f}=v^{2}_{i}+2ad\)

Subtract 2ad:

\(\displaystyle v^{2}_{f}-2ad=v^{2}_{i}\)

Take square root:

\(\displaystyle \pm\sqrt{v^{2}_{f}-2ad}=v_{i}\)

That's it.

 

[soroban]

Hello, speed4baseball!

Your problem is impossible to read . . . ever heard of parentheses and spaces?

v^2_f=v^2_i+2ad *solve for v_i

*_ means subscript

. . v^2_i+2ad=v^2_f

. . v^2_i=v^2_f-2ad

. . v=sqrt v^2_f-2ad

Got it?

 

[speed4baseball]

thnx for the reminder galactus of square root properties
but i have another question for you
how do u put v_1 or exponents with out the ^ ?
i would like to know so i wont confuse people
thnx for your help

@soroban
im sorry i am new to this
forum i try to put symbols like_ or ^ to express what i am trying to say and thank u as well

 

[galactus]

In the upper right corner of my last post, click on 'quote'. This will show you the code I used to make it display in LaTeX format, and show up nice and 'purdy'

 

[speed4baseball]

yea!! i got it thnx

 

[Deleted member 4993]

Or you could simply write (v_1)^2

or use the <sub> and <sup> buttons above to write v[sub:1sib18s8]1[/sub:1sib18s8][sup:1sib18s8]2[/sup:1sib18s8]

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