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avatar+470 

y=mx+b (solve for b)

 

A=h(b+c) (solve for b)

 

A=4r^2 (solve for r^2)

 

7x-y=14 (solve for x)

 

R= (E/i) (solve for i)

 

A= (r/2L) (solve for L)

 Oct 19, 2017

Best Answer 

 #1
avatar+8842 
+2
Solve for  b .          
y   =   mx + b Subtract  mx  from both sides of the equation.
y - mx   =   b  
   
Solve for  b .  
A   =   h(b + c) Divide both sides of the equation by  h  .
A / h   =   b + c Subtract  c  from both sides of the equation.
A / h  -  c   =   b  
   
Solve for r2 .  
A   =   4r2  Divide both sides of the equation by  4 .
A / 4   =   r2  
   
Solve for  x .  
7x - y  =  14 First add  y  to both sides. See if you can figure the rest out. smiley
   
Solve for  i .  
R  =  (E / i) Multiply both sides of the equation by  i .
i * R  =  E Divide both sides of the equation by  R.
i  =  E / R  
   
Solve for  L . Using this as the equation  A = \(\frac{r}2\)L ,
A  =  \(\frac{r}2\)L multiply both sides by  \(\frac2r\) .
\(\frac2r\) * A  = \(\frac{2}{r}\) * \(\frac{r}{2}\) * L  
\(\frac{2A}{r}\)  =  L If you meant for the equation to be  A = \(\frac{r}{2L}\) , then it is different!
 Oct 20, 2017
 #1
avatar+8842 
+2
Best Answer
Solve for  b .          
y   =   mx + b Subtract  mx  from both sides of the equation.
y - mx   =   b  
   
Solve for  b .  
A   =   h(b + c) Divide both sides of the equation by  h  .
A / h   =   b + c Subtract  c  from both sides of the equation.
A / h  -  c   =   b  
   
Solve for r2 .  
A   =   4r2  Divide both sides of the equation by  4 .
A / 4   =   r2  
   
Solve for  x .  
7x - y  =  14 First add  y  to both sides. See if you can figure the rest out. smiley
   
Solve for  i .  
R  =  (E / i) Multiply both sides of the equation by  i .
i * R  =  E Divide both sides of the equation by  R.
i  =  E / R  
   
Solve for  L . Using this as the equation  A = \(\frac{r}2\)L ,
A  =  \(\frac{r}2\)L multiply both sides by  \(\frac2r\) .
\(\frac2r\) * A  = \(\frac{2}{r}\) * \(\frac{r}{2}\) * L  
\(\frac{2A}{r}\)  =  L If you meant for the equation to be  A = \(\frac{r}{2L}\) , then it is different!
hectictar Oct 20, 2017

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