AnswerMachine

no problemo

We know that $\triangle BDF$ and $\triangle ECF$ have the same area, and adding the area of quadrilateral $ABFE$ to each of them shows us that$$\text{(area of $\triangle ADE$)}=\text{(area of $\triangle ABC$)}.$$Using the area formula for triangles, we can turn the last equation into $$\dfrac{1}{2}\cdot AD\cdot AE = \dfrac{1}{2}\cdot AB\cdot AC.$$Substituting in the given lengths, we get $$\dfrac{1}{2}\cdot 5\cdot 4 = \dfrac{1}{2}\cdot 3\cdot AC,$$so $AC=\dfrac{20}{3}$. We compute the length we are interested in as $EC=AC-AE=\dfrac{20}{3}-4=\boxed{\dfrac{8}{3}}$.

this problem is everywhere and most of them are wrong so feel free to check out some websites on google

https://math.stackexchange.com/questions/1387302/jackpot-probablity

https://web2.0calc.com/questions/please-help_76234

https://brainly.com/question/15438769

https://istudy-helper.com/mathematics/question19931329

https://www.freemathhelp.com/forum/threads/probability.124026/

https://web2.0calc.com/questions/stuck-on-a-problem

https://web2.0calc.com/questions/insert-title-here_1

You are right guest. Here is a more detailed ex

1.  We complete the square twice. \begin{align*} x^2 + 6x + y^2 + 6y &= 18 \\ x^2 + 6x + 9+ y^2 + 6y +9 &= 18  +9+9\\ (x +3)^2 + (y + 3)^2 &= 36 \end{align*} This is in the form $(x - h)^2 + (y - k)^2 = r^2$, so the center is $\boxed{(-3, -3)}$.

2.  We are given $x^2 - 10x + y^2 - 10y + 25 = 0$. Factoring, this is $x^2 - 10x + (y - 5)^2 = 0$. We use completing the square and add $25$ to both sides of the equation, giving us $x^2 - 10x +25  + (y - 5)^2 = 25$. Factoring again, this is $(x - 5)^2 + (y-5)^2 = 25$. This equation represents a circle of radius $5$, centered at $(5, 5)$. Thus, this circle will touch the $y$-axis at $\boxed{(0, 5)}$.

1.  First, subtract $4x$ and $6y$ from both sides to set up completing the square: $$x^2 -4x + y^2 -6y \leq 13.$$We add $((-4)/2)^2 = 4$ to complete the square in $x$ and $((-6)/2)^2 = 9$ to complete the square in $y$: $$x^2 -4x + 4 + y^2 -6y + 9 \leq 13+4+9,$$so $$(x-2)^2 + (y-3)^2 \leq 26.$$The graph of this inequality is a solid circle with center $(2,3)$ and radius $\sqrt{26}$, so the desired area is $\pi\left(\sqrt{26}\right)^2=\boxed{26\pi}$.

2.  First, we factor out the constants of the squared terms to get $9(x^2-2x)+9(y^2+4y)=-44.$ To complete the square, we need to add $\left(\dfrac{2}{2}\right)^2=1$ after the $-2x$ and $\left(\dfrac{4}{2}\right)^2=4$ after the $4y,$ giving$$9(x-1)^2+9(y+2)^2=-44+9+36=1.$$Dividing the equation by $9$ gives$$(x-1)^2+(y+2)^2=\frac{1}{9},$$so the radius is $\sqrt{\frac{1}{9}}=\boxed{\frac{1}{3}}.$

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