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.

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

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

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