heureka

Thank you, Melody

$$t = \frac{5-b}{2}$$

$$v = \frac{D}{m}$$

2^n*2^(n+2)*2   ?

$$2^n*2^{(n+2)}*2^1 = 2^{n+n+2+1}= 2^{2n+3}$$

The formula for an object's final velocity is i =

$$\small{\text{
$
i = f + gt
$
}}$$

find the area between the curves y=x^2 and y=1-x^2

I. limits above and below the integral sign:

The Cut of two functions :

$$\\ y_c = x_c^2 = 1-x_c^2 \\
2x_c^2=1 \\
x_c^2=\frac{1}{2} \\
x_c=\pm\sqrt{\frac{1}{2}}\\
x_c=\pm\frac{\sqrt{2}}{2}$$ 

II. The area between is:

 $$\small{\text{
$
\int\limits_{ x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }(1-x^2)-x^2 \ dx
= \int\limits_{x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} } 1-2x^2 \ dx
= \int\limits_{x=-\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} } dx - 2\int\limits_{x=-1}^{+1}x^2 \ dx
$
}}\\\\
\small{\text{
$
=[x]_{- \frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }-2[\frac{x^3}{3}]_{ -\frac{\sqrt{2}}{2} }^{ \frac{\sqrt{2}}{2} }
=[\frac{ \sqrt{2} }{2} -(-\frac{\sqrt{2}}{2} )]-2[\frac{ (\frac{\sqrt{2}}{2})^3}{3}-\frac{(-(\frac{\sqrt{2}}{2}) )^3}{3}]
=\sqrt{2} - 2[\frac{4*\sqrt{2}}{24}-\frac{-4*\sqrt{2}}{24}]
$
}}\\\\
\small{\text{
$
=\sqrt{2} -2[ \frac{2*4 \sqrt{2} }{24}]
=\sqrt{2} - \frac{2}{3} \sqrt{2}
=\frac{1}{3} \sqrt{2}
$
}} \\\\
\small{\text{The area is $ \frac{1}{3} \sqrt{2} $}}$$

Thank you CPhill

4.9t^2+19.8t-20=0? How do i find t ?

$$\small{\text{
Use the formula:
$
\boxed{
ax^2+bx+c=0 \qquad x_{1,2}=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}
}
$
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{19.8^2-4*4.9*(-20)} }{2*4.9}
$
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{ 392.04+392 } }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm\sqrt{ 784.04 } }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1,2}=\dfrac{-19.8\pm 28.001 }{9.8}
$
}}
}}\\\\
\small{\text{
$
t_{1}=\dfrac{-19.8 + 28.001 }{9.8} = 0.8368
$
}}
}}\\\\
\small{\text{
$
t_{2}=\dfrac{-19.8 - 28.001 }{9.8} = -4.8777
$
}}$$

2x+3y=1 and -3x+y=15 how do you get points (-4,3)

$$\begin{array}{rrcl|lrcl}
(1) & 2x+3y &=& 1 \quad & \quad (2)& -3x+y &=& 15 \quad | \quad * 3 \\
& & & \quad & \quad (2)& -9x+3y &=& 45 \\
\hline
\end{array}$\\\\\\$
\small{\text{
$
\begin{array}{lrcl}
(1)-(2): & 2x-(-9x) +3y -3y &=& 1-45 \\
& 11x &=& -44 \\
& \textcolor[rgb]{1,0,0}{x} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{-4} \\\\
\hline \\
(1) & 2(-4) + 3y &=& 1 \\
& -8 + 3y &=& 1 \\
& 3y &=& 9 \quad | \quad : 3\\
& \textcolor[rgb]{1,0,0}{ y} &\textcolor[rgb]{1,0,0}{=}& \textcolor[rgb]{1,0,0}{3}
\end{array}
$
}}$$