+0  
 
+2
288
9
avatar+152 

\(a=\frac{v}{t}\)                            solve for t

 

\(A=\pi r^2\)                        solve for r

 

\(P=I^2R\)                       solve for I

xRainexSiderisx  Mar 21, 2017
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9+0 Answers

 #1
avatar+10657 
+3

a = v/t     Multiply both sides of the equation by 't

at = v      Divide both sides by 'a'

t = v/a

 

A=pi r^2     Divide both sides by pi

A/pi = r^2    Sqrt both sides

sqrt(A/pi) = r                                 Do you see how it is done?   Can you do the third one?

ElectricPavlov  Mar 21, 2017
 #3
avatar+5552 
+2

Hey you beat me to it...the feature that shows other users who are composing doesn't seem to be working anymore

hectictar  Mar 21, 2017
edited by hectictar  Mar 21, 2017
 #4
avatar+152 
+2

Yes I can! Thanks!

xRainexSiderisx  Mar 21, 2017
 #6
avatar+10657 
+2

I think you are correct....can honestly say I have not found ONE SINGLE thing in the 'update' that was a good change.

ElectricPavlov  Mar 21, 2017
 #7
avatar+152 
+1

Thats so true honestly

xRainexSiderisx  Mar 21, 2017
 #8
avatar+5552 
+4

Lol how about the hearts for points EP?

You gotta at least like them a little x)

hectictar  Mar 21, 2017
 #9
avatar+10657 
+3

Nope.  Posted earlier that I thought the hearts were cheesy and the symbol should be something math-related like # or the pi symbol !   Just not feelin' the heart thingie ....

ElectricPavlov  Mar 21, 2017
 #2
avatar+5552 
+4

All "solve for" means is get the variable by itself on one side of the equal sign.

\(a=\frac{v}{t}\)

 

Multiply both sides of the equation by t.

\(ta=v\)

 

Divide both sides of the equation by a.

\(t=\frac{v}{a}\)

Hey! time = velocity / acceleration

 

\(A = \pi r^2\)

 

Divide both sides of the equation by pi.

\(\frac{A}{\pi} = r^2\)

 

Take the + square root of both sides.

\(\sqrt{\frac{A}{\pi}} = r\)

radius = the square root of (area / pi)

 

The third one is exactly the same as the second one just with different letters.

hectictar  Mar 21, 2017
 #5
avatar+152 
+2

Thank you so much!

xRainexSiderisx  Mar 21, 2017

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