gibsonj338

\(0.51=1-({x}^{2}-{300000000}^{2})\)

\(0.51=1-({x}^{2}-90000000000000000)\)

\(51=100-100\times({x}^{2}-90000000000000000)\)

\(51=100-{100x}^{2}+ 9000000000000000000\)

\(51=-{100x}^{2}+ 9000000000000000100\)

\(0={-100x}^{2}+ 9000000000000000049\)

\({-100x}^{2}+ 9000000000000000049=0\)

\({-100x}^{2}=-9000000000000000049\)

\({x}^{2}= 90000000000000000.49\)

\(x=± 300000000.0000000008166667\)

\(\sqrt{2x-5}=\sqrt{3x-12}\)

\(2x-5=3x-12\)

\(2x=3x-7\)

\(-x=-7\)

\(x=7\)

No one really knows.

\({({5x}^3{})}^{2}\)

\({5}^{2}{x}^{3\times2}\)

\({5}^{2}{x}^{6}\)

\(25{x}^{6}\)

radius = 9 cm into three equal sectors.

a) formua for arc length is \(s=r\Theta \) where \(s\) = arc length, \(r\) = radius, and \(\Theta \) = angle in radians

In order to find the arc length in the problem, more inforormation is needed.  You need to say what the angle is either in radians.

b) formula for area of a sector is \(A=\frac{1}{2}{r}^{2}\Theta \) where \(A\) = area of the sector of a circle, \(r\) = radius, and \(\Theta \) = angle in radians

In oreder to find the area of a section in the problem, more inforormation is needed.  You need to say what the angle is in radians.

\({log}_{10}(y)=2.0\)

\({log}_{10}(y)=2\)

\(y={10}^{2}\)

\(y=100\)

\(3:50\)

or

\(\frac{3}{50}\times100\)

\(\frac{300}{50}\)

6%

You have a \(3\) in \(50\) chance or 6% chance of contracting the virus.

Likewise

\(50:50-3:50\)

\(47:50\)

or

\(\frac{50}{50}-\frac{3}{50}\times100\)

\(\frac{47}{50}\times100\)

\(\frac{4700}{50}\)

94%

You have a \(47\)in \(50\) chance or 94% chance of not contracting the virus.

You mean 84.615384615%?

\(x-(-2\frac{4}{5})=3\frac{7}{10}\)

\(x+2\frac{4}{5}=3\frac{7}{10}\)

\(x+\frac{14}{5}=\frac{37}{10}\)

\(x=\frac{37}{10}-\frac{14}{5}\)

\(x=\frac{37}{10}-\frac{28}{10}\)

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

\(x=0.9\)

\(\frac{1}{4}\times(60+16s)=15+4s\)

\(\frac{60}{4}+\frac{16}{4}s=15+4s\)

\(15+4s=15+4s\)

\(4s=0+4s\)

\(0=0\)

\(s\) = all real numbers