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avatar+1317 

Help Me, please.

 

Solve for g

 

h=\(\sqrt{\frac{g^2}{d}-5t}-1\)

ManuelBautista2019  Mar 4, 2018

Best Answer 

 #1
avatar+7266 
+2

\(h=\sqrt{\frac{g^2}{d}-5t}-1\)

                                          Add  1  to both sides of the equation.

\(h+1=\sqrt{\frac{g^2}{d}-5t}\)

                                          Square both sides of the equation.

\((h+1)^2=\frac{g^2}{d}-5t\)

                                          Add  5t  to both sides.

\((h+1)^2+5t=\frac{g^2}{d}\)

                                                Multiply both sides by  d .

\(d[(h+1)^2+5t]=g^2\)

                                                Take the ± square root of both sides.

\(±\sqrt{d[(h+1)^2+5t]}=g \\~\\ g=±\sqrt{d[(h+1)^2+5t]}\)

hectictar  Mar 4, 2018
 #1
avatar+7266 
+2
Best Answer

\(h=\sqrt{\frac{g^2}{d}-5t}-1\)

                                          Add  1  to both sides of the equation.

\(h+1=\sqrt{\frac{g^2}{d}-5t}\)

                                          Square both sides of the equation.

\((h+1)^2=\frac{g^2}{d}-5t\)

                                          Add  5t  to both sides.

\((h+1)^2+5t=\frac{g^2}{d}\)

                                                Multiply both sides by  d .

\(d[(h+1)^2+5t]=g^2\)

                                                Take the ± square root of both sides.

\(±\sqrt{d[(h+1)^2+5t]}=g \\~\\ g=±\sqrt{d[(h+1)^2+5t]}\)

hectictar  Mar 4, 2018
 #2
avatar+1317 
+2

Thank you!!

ManuelBautista2019  Mar 4, 2018

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