Actresses Ai Aso Movies

WNXG-070 Encountering a Slightly Off Working Lady at a Bar: Expanded Edition

28 Jun 2016

WAN-030 Secretly filmed in a secret room: Taking out a smelly dick in a fitting room and requesting hemming 3

26 Feb 2010

RGD-182 Love Tights Delicious Evolution 3

20 Feb 2010

RGDR-175 Sure, here’s a revised sentence: They engaged in a heartfelt conversation to resolve their disagreement and strengthen their bond.

24 Nov 2009

CADV-104 Prime selection moment

7 Apr 2009

RGBH-015 Passionate finger masturbation

25 Jan 2009

RGBH-010 A student engaged in inappropriate behavior at school.

25 Nov 2008

AVGP-132 Respecting Cultural Differences in Communication

19 Nov 2008

AVGP-144 1) Starting with the quadratic equation ( ax^2 + bx + c = 0 ) as the root-finding problem, derive a "solution" to the problem method. **Solution**: To solve the quadratic equation ( ax^2 + bc + c = 0 ), we can use the ** quadratic formula**, which is derived from the process of completing the square in the equation. Given solving the quadratic equation ( ax^2 + bc + c = 0 ), that have the quadratic formula ( x = frac{-b pm sqrt{b^2-4ac}}{2a} ). **Derivation**: First, write the quadratic equation ( 0 = ax^2 + bc + c ): [ 0 = ax^2 + bc + c ] Divides both sides of the equation by ( a ) to get: [ x^2 + frac{b}{a} + frac{c}{a} = 0 ] Then, move the constant term to the other side of the equation: [ c/a = x^2 + frac{b}{a} ] To complete the square, add ((frac{b}{2a})^2) to both sides of the equation: [ x^2 + frac{b}{a} + (frac{b}{2a})^2 = frac{c}{a} + (frac{b}{2a})^2 ] Then, square the left side of the equation: [ (x + frac{b}{2a})^2 = frac{c}{a} + (frac_b}{2a})^2 ] Take the square root of both sides of the equation: [ x + frac{b}{2a} = pm sqrt{frac{c}{a} + (frac{b}{2a})^2 } ] Move ( frac{b}{2a} ) to the other side of the equation: [ x = -frac{b}{2a} pm sqrt{frac{c}{a} + (frac{b}{2a})^2 } ] As you can see, the quadratic formula is: [ x = / -b pm sqrt{b^2 - 4ac} / 2a ] [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) **The formula is supposed to be:<br> ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} )** **Start solving the quadratic equation ( ax^2 + bc + c = 0 ), using the quadratic formula**: [ x = frac{-b pm sqrt{b^ - 4ac}}{2a} ] [ x = frac{-b pm sqrt{b^ - 4ac}}{2a} ] The quadratic equation is ( ax^2 + bc + c = 0 ) in this form is: ( ax^2 + bc + c = 0 ) the equation can be written as: ( ax^2 + bc + c = 0 ) similar to the equation ( ax^2 + bx + c = 0 ) as the form of the quadratic formula. ** The initial equation is ( ax^2 + bc + c = 0 ) ** shall be the ** From completing the square shall be the expected equation is ( ax^2 + bc + c = 0 ) ** Saturated formula**: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] extbf{Answer on the step}: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] Therefore, the solutions to the quadratic formula is: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] 1) Starting with the quadratic equation ( ax^2 + bc + c = 0 ) as the root-finding problem, derive in a triangle section, ( ax^2 + bc + c = 0 ) algebra-optional equation naturally leads to the quadratic formula. From ( ax^2 + bc + c = 0 ) picks both sides by ( a ) to standardize the formula: [ x^2 +frac{b}{a} + frac{c}{a} = 0 ] substitute ( b = b' and c = c' ): [ x^2 + frac{b'}{a} + frac{c'}{a} = 0 ] To complete the square, add ((frac{b'}{2a})^2) to both sides of the equation: **Equalize both sides this operation**: [ x^2 + frac{b'}{a} + (frac{b'}{2a})^2 = frac{c'}{a} + (frac{b'}{2a})^2 ] To **fact the square** of the first equation: [ (x + frac{b'}{2a})^2 = frac{c'}{a} + (frac{b'}{2a})^ ] Take the root of both sides: [ x + frac{b'}{2a} = pm sqrt{frac{c'}{a} + (frac{b'}{2a})^** }} \] Finalize the equation for **x**: [ x = - frac{b'}{2a} pm sqrt{frac{c'}{a} + (frac{b'}{2a})^2} ] As a quadratic formula, the standard **formula is**: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] The individual formula is: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] extbf{Ans}: [[ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ]] As with ​terms ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ) is to be considered, using the quadratic formula to solve the quadratic equation ( ax^2 + bc + c = 0 )

19 Nov 2008

HA-082 Respecting Cultural Diversity in Communication

8 Nov 2008

DOKS-037 Engaging in self-stimulation in an outdoor setting.

19 Sep 2008

DKSW-121 Schoolgirl Dildo Masturbation 3

10 Sep 2008

DOKS-033 Erect nipples collection

8 Aug 2008