gibsonj338

What are the size of the tiles?

\(f(x)={x}^{2}+3x-4\)

\(f(x+2)-f(2)={(x+2)}^{2}+3\times(x+2)-4-{2}^{2}+3\times2-4\)

\(f(x+2)-f(2)={x}^{2}+2x+2x+4+3\times(x+2)-4-{2}^{2}+3\times2-4\)

\(f(x+2)-f(2)={x}^{2}+4x+4+3\times(x+2)-4-{2}^{2}+3\times2-4\)

\(f(x+2)-f(2)={x}^{2}+4x+4+3x+6-4-4+3\times2-4\)

\(f(x+2)-f(2)={x}^{2}+4x+4+3x+6-4-4+6-4\)

\(f(x+2)-f(2)={x}^{2}+7x+4+6-4-4+6-4\)

\(f(x+2)-f(2)={x}^{2}+7x+10-4-4+6-4\)

\(f(x+2)-f(2)={x}^{2}+7x+6-4+6-4\)

\(f(x+2)-f(2)={x}^{2}+7x+2+6-4\)

\(f(x+2)-f(2)={x}^{2}+7x+8-4\)

\(f(x+2)-f(2)={x}^{2}+7x+4\)

\(f(x)={2x}^{2}\)

\(g(x)=4-2x\)

\(g(f(0))=4-2\times2\times{0}^{2}\)

\(g(f(0))=4-2\times2\times0\)

\(g(f(0))=4-4\times0\)

\(g(f(0))=4-0\)

\(g(f(0))=4\)

What do you need help with in geometry?

In order to add or subtract fractions the numerators have to be the same bfore you can add or subtract. The actual answer is x=9/10 or x=0.9.

Look at my answer. 10/9 is not the correct answer.  The answer is 9/10 or 0.9.

\(\frac{-5}{2x}=\frac{-7}{4x}-\frac{5}{6}\)

\(-\frac{5}{2x}=-\frac{7}{4x}-\frac{5}{6}\)

\((-\frac{5}{2x})\times4x=(-\frac{7}{4x}-\frac{5}{6})\times4x\)

\(-\frac{20x}{2x}=(-\frac{7}{4x}-\frac{5}{6})\times4x\)

\(-\frac{20x}{2x}=-\frac{28x}{4x}-\frac{20x}{6}\)

\(-10=-\frac{28x}{4x}-\frac{20x}{6}\)

\(-10=-7-\frac{20x}{6}\)

\(-10\times6=(-7-\frac{20x}{6})\times6\)

\(-60=(-7-\frac{20x}{6})\times6\)

\(-60=-42-\frac{120x}{6}\)

\(-60=-42-20x\)

\(-60+42=-42-20x+42\)

\(-18=-42-20x+42\)

\(-18=0-20x\)

\(-18=-20x\)

\(\frac{-18}{-20}=\frac{-20x}{-20}\)

\(\frac{9}{10}=\frac{-20x}{-20}\)

\(\frac{9}{10}=\frac{1x}{1}\)  

\(\frac{9}{10}=\frac{x}{1}\)

\(\frac{9}{10}=x\)

\(0.9=x\)

\(x=0.9\)

\(0.27\)

\(\frac{27}{100}\)

\(5\frac{1}{2}\)

\(5+\frac{1}{2}\)

\(\frac{10}{2}+\frac{1}{2}\)

\(\frac{11}{2}\)

\(5.5\)

\(log(0.001)=x\)

\(x=log(0.001)\)

\({10}^{x}=0.001\)

\(x=-3\)

Replace \(x\) with \(-3\) and solve to see if the answer is correct:

\({10}^{-3}=0.001\)

\(\frac{1}{{10}^{3}}=0.001\)

\(\frac{1}{{1000}}=0.001\)

\(0.001=0.001\)