Question #225915

The value of 1/D²+3D (2 + 6x) =


A) x


B) x² C) 0 D) 1


1
Expert's answer
2021-08-19T15:23:02-0400

D2+3D(2+6x)=0SubtractD2frombothsidesD2+3D(2+6x)D2=0D2Simplify3D(2+6x)=D2Dividebothsidesby3D;D03D(2+6x)3D=D23D;D02+6x=D3;D0Subtract2frombothsides2+6x2=D32;D0Simplify6x=D32;D0Dividebothsidesby6;D06x6=D3626;D0x=D618;D0D^2+3D\left(2+6x\right)=0\\ \mathrm{Subtract\:}D^2\mathrm{\:from\:both\:sides}\\ D^2+3D\left(2+6x\right)-D^2=0-D^2\\ \mathrm{Simplify}\\ 3D\left(2+6x\right)=-D^2\\ \mathrm{Divide\:both\:sides\:by\:}3D;\quad \:D\ne \:0\\ \frac{3D\left(2+6x\right)}{3D}=\frac{-D^2}{3D};\quad \:D\ne \:0\\ 2+6x=-\frac{D}{3};\quad \:D\ne \:0\\ \mathrm{Subtract\:}2\mathrm{\:from\:both\:sides}\\ 2+6x-2=-\frac{D}{3}-2;\quad \:D\ne \:0\\ \mathrm{Simplify}\\ 6x=-\frac{D}{3}-2;\quad \:D\ne \:0\\ \mathrm{Divide\:both\:sides\:by\:}6;\quad \:D\ne \:0\\ \frac{6x}{6}=-\frac{\frac{D}{3}}{6}-\frac{2}{6};\quad \:D\ne \:0\\ x=\frac{-D-6}{18};\quad \:D\ne \:0


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