heureka

Thank you Melody,

thank you rosala

x=5.5 see the graph:

5x−8=3x+3 whats the x?

$$5x-8=3x+3 \quad | \quad-3x\\\\
\underbrace{5x-3x}_{2x}-8 = \underbrace{3x-3x}_{0}+3\\\\
2x-8=3\quad | \quad +8\\\\
2x\underbrace{-8+8}_{0} =3+8\\\\
2x=11\quad | \quad:2\\\\
{\not{2}x\over \not{2}} = {11 \over 2}\\\\
x={11 \over 2}\\\\
x=5.5$$

Hi Melody,

$$\\15848.9319246 * 10^{4.2} = 15848.9319246 * 10^{4+0.2}\\
= 15848.9319246 * 10^4*10^{0.2} \quad | \quad 10^{0.2} \approx 1.58489319246\\
= 15848.9319246 * 10^4*1.58489319246\\
\Rightarrow
=10^{0.2}*10^4* 10^4*10^{0.2}\\
=10^{8.4}$$

how to form the quadratic function into y=ax^2+bx+c  ?

$$\begin{array}{rcl}
y&=&
ax^2+bx+c\\\\
y&=&
a\left(
x^2+{b\over a}x\right)
+c\\\\
y&=&
a\left(
x^2+{b\over a}x \textcolor[rgb]{1,0,0}{
+ ({b\over 2a})^2 - ({b\over 2a})^2 }
\right)
+c\\\\
y&=&
a\left[(x+{b\over 2a})^2 - ({b\over 2a})^2
\right]
+c\\\\
y&=&a (x+{b\over 2a})^2 - {b^2\over 4a}+c\\\\
y&=&a (x+{b\over 2a})^2 + c - {b^2\over 4a}
\end{array}$$

15848.9319246 * 10^4.2 = ?

$$\boxed{15848.9319246 \approx 10^{4.2} }\\
15848.9319246 * 10^{4.2} \approx 10^{4.2} *10^{4.2}=10^{4.2+4.2}=10^{8.4}$$

15848.9319246 * 10^4.2 $$\approx$$ 10^8.4 = 251188643.151

what is 2 over 3 divided by 5   ?

$$\begin{array}{rcl}
\dfrac {
\frac {2}{3}
}
{
\textcolor[rgb]{1,0,0}{5}
}
=
\dfrac {
\frac {2}{3}
}
{
\textcolor[rgb]{1,0,0}{\frac {5}{1} }
}
=
\frac {2}{3}
\times
\textcolor[rgb]{1,0,0}{\frac {1}{5}}
=
\frac {2*\textcolor[rgb]{1,0,0}{1}}{3*\textcolor[rgb]{1,0,0}{5}}
=
\frac {2}{15}
\end{array}$$

Point G is located at (3, −1), and point H is located at (−2, 3). Find the x value for the point that is 2 over 3 the distance from point G to point H.

$$\\x=(1-\lambda)*3 + \lambda*(-2) \quad | \quad x_g=3 \quad and \quad x_h = -2 \\\\
\text{if } \lambda=\textcolor[rgb]{1,0,0}{0} \Rightarrow x = (1-\textcolor[rgb]{1,0,0}{0})*3+\textcolor[rgb]{1,0,0}{0}*(-2) = 3 = x_g \\
\text{if } \lambda=\textcolor[rgb]{1,0,0}{1} \Rightarrow x = (1-\textcolor[rgb]{1,0,0}{1})*3+\textcolor[rgb]{1,0,0}{1}*(-2) = -2 = x_h\\\\
\text{if } \lambda=\textcolor[rgb]{1,0,0}{{2\over3 }} \Rightarrow x = (1-\textcolor[rgb]{1,0,0}{{2\over3}} )*3+\textcolor[rgb]{1,0,0}{{2\over3}} *(-2)\\
\Rightarrow x={1\over3}*3 - {4\over3}\\
\Rightarrow x=-{1\over 3}$$

$$\begin{array}{rclclcl}
x &=& x^1 &=& x^{(1)} &=& x^1\\
x*x &=& x^1*x^1 &=& x^{(1+1)} &=& x^2\\
x*x*x &=& x^1*x^1*x^1 &=& x^{(1+1+1)} &=& x^3\\
x*x*x*x &=& x^1*x^1*x^1*x^1 &=& x^{(1+1+1+1)} &=& x^4\\
x*x*x*x*x &=& x^1*x^1*x^1*x^1*x^1 &=& x^{(1+1+1+1+1)} &=& x^5
\end{array}$$

$$\\\left(
{
(-7)\over(x-8)
}
\right)
=
\left(
{
(a)\over(8-x)
}
\right)
}\\\\
\left(
{
(7)\over(8-x)
}
\right)
=
\left(
{
(a)\over(8-x)
}
\right)
}\\\\
\boxed{7=a}$$