gibsonj338

\(\sqrt7\)

Exact Value

\(\sqrt7\)

Approximate Value

\(2.6457513110645906\)

\((q-5)(q-6)\)

\(F.O.I.L.\)

F - First Times First \((q\times q)\)

O - Outside Times Outside \((q\times[-6])\)

I - Inside Times Inside \(([-5]\times q)\)

L - Last Times Last \(([-5]\times[-6])\)

When you do the \(F.O.I.L.\) method for this expression, you get :

\({q}^{2}-6q-5q+30\)

Now combine like terms

\({q}^{2}-11q+30\)

I asume you mean \({x}^{2}+9x-10\), so I will go with that.

\({x}^{2}+9x-10\)

Subsitute \(x\) for 13.

\({13}^{2}+9\times13-10\)

\(169+9\times13-10\)

\(169+117-10\)

\(286-10\)

\(276\)

\(872+x=981\)

\(872+x-872=981-872\)

\(0+x=981-872\)

\(0+x=109\)

\(x=109\)

\(\frac{2}{{3}^{2}}+{2.5}^{2}\)

\(\frac{2}{9}+{2.5}^{2}\)

\(\frac{2}{9}+6.25\)

\(\frac{2}{9}+6.25\times\frac{100}{100}\)

\(\frac{2}{9}+\frac{625}{100}\)

\(\frac{2}{9}+\frac{625}{100} ÷\frac{25}{25}\)

\(\frac{2}{9}+\frac{25}{4}\)

\(\frac{2}{9}\times\frac{4}{4}+\frac{25}{4}\times\frac{9}{9}\)

\(\frac{8}{36}+\frac{25}{4}\times\frac{9}{9}\)

\(\frac{8}{36}+\frac{225}{36}\)

\(\frac{233}{36}\)

\(6 \frac{17}{233}\)

\(\frac{1}{5n}+6=-4\)

\(\frac{1}{5n}+\frac{6}{1}=-4\)

\(\frac{1}{5n}+\frac{6\times5n}{1\times5n}=-4\)

\(\frac{1}{5n}+\frac{30n}{5n}=-4\)

\(\frac{1+30n}{5n}=-4\)

\(\frac{1+30n}{5n}\times5n=-4\times5n\)

\(1+30n=-4\times5n\)

\(1+30n=-20n\)

\(1+30n+20n=-20n+20n\)

\(1+50n=-20n+20n\)

\(1+50n=0n\)

\(1+50n=0\)

\(1+50n-1=0-1\)

\(50n=0-1\)

\(50n=-1\)

\(\frac{50n}{50}=\frac{-1}{50}\)

\(1n=\frac{-1}{50}\)

\(1n=-0.02\)

\(n=-0.02\)

\(f(x)=5+\frac{1}{x+3}\)

\(f(x)=\frac{5}{1}+\frac{1}{x+3}\)

\(f(x)=\frac{5\times(x+3)}{1\times(x+3)}+\frac{1}{x+3}\)

\(f(x)=\frac{5x+15}{x+3}+\frac{1}{x+3}\)

\(f(x)=\frac{5x+15+1}{x+3}\)

\(f(x)=\frac{5x+16}{x+3}\)

\(0.1x+0.05y=30\)

\(x+y=500\)

Solve for x in the second equation

\(x+y=500\)

\(x=500-y\)

Subsitute x in th first equation and solve for y

\(0.1\times(500-y)+0.05y=30\)

\(50-0.1y+0.05y=30\)

\(50-0.05y=30\)

\(-0.05y=-20\)

\(y=400\)

Subsitute y in the second equation

\(x+400=500\)

\(x=100\)

There would be \(400\) nickels and \(100\) dimes totaling \(500\) coins with a face value of \($30.\)

\(4\times(x-3)=2\)

\(4x-12=2\)

\(4x=14\)

\(x=3.5\)

\({z}^{2}-\frac{4}{z}-\frac{3}{z}+\frac{2}{{z}^{2}}+z-12\)

\({z}^{2}-\frac{7}{z}+\frac{2}{{z}^{2}}+z-12\)

\({z}^{2}+z-\frac{7}{z}+\frac{2}{{z}^{2}}-12\)

The only thing that can be done is to simplify the expression.  The only restrictions would be if you subsitute \(z\) for a number and then solve.  \(z\) cannot be \(0\) because if you divide \(7\) by \(0\) or \(2\) by \({0}^{2}\) (or \(0\)), you would get \(undefined\) as an answer.