gibsonj338

\(W=2\),  \(I=5\),  \(N=8\),  \(E=6\)

\(3\times862=2586\)

This took me several hours to figure this out as there were 1001 possibilities.  Thanks to Microsoft Excel, I was able to get to the answer.

To prove that \({2}^{0}=1\), look at this example

\(\frac{{2}^{5}}{{2}^{5}}\)

\(\frac{32}{{2}^{5}}\)

\(\frac{32}{32}\)

\(1\)

\(\frac{{2}^{5}}{{2}^{5}}\)

\({2}^{5-5}\)

\({2}^{0}\)

\(1\)

Actually anything with an exponent of 0 except for 0 is not 0 but is 1.

\({0}^{0}≠0\)

\({0}^{0}\) is \(Undefined\) because you cannot take \(0\) and raise it to the \(0\) power.  If you take any integer except for \(0\) and raise it to the power of \(0\), you get \(1\).

\({0}^{0}\)

\(Undefined\)

To find the diameter of the circle with only the area known, use the formula \(A=\pi\times{r}^{2}\) where \(A=Area\), \(\pi≈3.1415926535897932\), \(r=radius\).  We know that Area is equal to \({56.75in}^{2}\)and we know that \(\pi≈3.1415926535897932\), but we do not know what the radius is equal to.  In order to figure out what the radius is equal to, solve for r and then plug in what we know.

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

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

\(\frac{A}{\pi}=1\times{r}^{2}\)

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

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

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

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

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

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

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

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

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

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

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

\(r=\frac{\sqrt{56.75\pi}}{\pi}\)

\(r≈\frac{\sqrt{56.75\times3.1415926535897932}}{3.1415926535897932}\)

\(r≈\frac{\sqrt{178.2853830912207641}}{3.1415926535897932}\)

\(r≈\frac{13.3523549642458488717}{3.1415926535897932}\)

\(r≈4.2501865889546685678486\)

Now we know that the radius is approximately equal to \(4.2501865889546685678486\); however, we want to know what the diamanter of the circle is, not the radius.  In order to find the diamater, use the formula \(d=2r\) where \(d=diamater\), \(r=radius\).

\(d=2r\)

\(d≈2\times4.2501865889546685678486\)

\(d≈8.5003731779093371356972\)

The diameter of a circle wth an area of \({56.75in}^{2}\) is approximately equal to \(8.5003731779093371356972\).

If you mean \(({\frac{1}{3})}^{11}\)

\(({\frac{1}{3})}^{11}\)

\(\frac{{1}^{11}}{{3}^{11}}\)

\(\frac{1\times1\times1\times1\times1\times1\times1\times1\times1\times1\times1}{{}^{{3}^{11}}}\)

\(\frac{1\times1\times1\times1\times1\times1\times1\times1\times1\times1}{{}^{{3}^{11}}}\)

\(\frac{1\times1\times1\times1\times1\times1\times1\times1\times1}{{}^{{3}^{11}}}\)

\(\frac{1\times1\times1\times1\times1\times1\times1\times1}{{}^{{3}^{11}}}\)

\(\frac{1\times1\times1\times1\times1\times1\times1}{{}^{{3}^{11}}}\)

\(\frac{1\times1\times1\times1\times1\times1}{{}^{{3}^{11}}}\)

\(\frac{1\times1\times1\times1\times1}{{}^{{3}^{11}}}\)

\(\frac{1\times1\times1\times1}{{}^{{3}^{11}}}\)

\(\frac{1\times1\times1}{{}^{{3}^{11}}}\)

\(\frac{1\times1}{{}^{{3}^{11}}}\)

\(\frac{1}{{}^{{3}^{11}}}\)

\(\frac{1}{3\times3\times3\times3\times3\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{9\times3\times3\times3\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{27\times3\times3\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{81\times3\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{243\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{729\times3\times3\times3\times3\times3}\)

\(\frac{1}{2187\times3\times3\times3\times3}\)

\(\frac{1}{6561\times3\times3\times3}\)

\(\frac{1}{19683\times3\times3}\)

\(\frac{1}{59049\times3}\)

\(\frac{1}{177147}\)

\(0.0000056450292695\)

If you mean \(\frac{1}{{3}^{11}}\)

\(\frac{1}{3\times3\times3\times3\times3\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{9\times3\times3\times3\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{27\times3\times3\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{81\times3\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{243\times3\times3\times3\times3\times3\times3}\)

\(\frac{1}{729\times3\times3\times3\times3\times3}\)

\(\frac{1}{2187\times3\times3\times3\times3}\)

\(\frac{1}{6561\times3\times3\times3}\)

\(\frac{1}{19683\times3\times3}\)

\(\frac{1}{59049\times3}\)

\(\frac{1}{177147}\)

\(0.0000056450292695\)

\(\frac{1}{3}\times\frac{1}{9}\)

\(\frac{1}{27}\)

\(0.037037037037037...\)

\(4+\frac{2}{3}x=9+\frac{1}{2}x\)

\([4+\frac{2}{3}x]\times6=[9+\frac{1}{2}x]\times6\)

\(24+\frac{12}{3}x=[9+\frac{1}{2}x]\times6\)

\(24+4x=[9+\frac{1}{2}x]\times6\)

\(24+4x=54+\frac{6}{2}x\)

\(24+4x=54+3x\)

\(24+4x-3x=54+3x-3x\)

\(24+1x=54+3x-3x\)

\(24+x=54+3x-3x\)

\(24+x=54+0x\)

\(24+x=54+0\)

\(24+x=54\)

\(24+x-24=54-24\)

\(x-0=54-24\)

\(x=54-24\)

\(x=30\)

To find the x intercept, subsitute 0 for x and slove for y:

\(3x+9y=9\)

\(3(0)+9y=9\)

\(0+9y=9\)

\(9y=9\)

\(\frac{9y}{9}=\frac{9}{9}\)

\(1y=\frac{9}{9}\)

\(y=\frac{9}{9}\)

\(y=1\)

x intersept: (0, 1)

To find the y intercept, subsitute 0 for y and slove for x:

\(3x+9y=9\)

\(3x+9(0)=9\)

\(3x+0=9\)

\(3x=9\)

\(\frac{3x}{3}=\frac{9}{3}\)

\(1x=\frac{9}{3}\)

\(x=\frac{9}{3}\)

\(x=3\)

y intersept: (3,0)

You can also find the x and y intersepts by graphing the equation by subsiting numbers for x and getting aswers for y or subsititing numbers for y and getting answers for x and then plot the points and then connect the points.  Click on the following link to see the graph of this equation.