4(1-t)=5t^(2) Let t=x and t=y be the solutions to the given equation. What is the value of -xy? ◻

Rewrite in standard form: First, let's rewrite the equation in standard quadratic form. 4 - 4t = 5t^2 Move all terms to one side to set the equation to zero. 5t^2 + 4t - 4 = 0Solve quadratic equation: Now, we need to solve the quadratic equation for t. We can use the quadratic formula: t = [-b ± sqrt(b^2 - 4ac)] / (2a) Here, a = 5, b = 4, and c = -4.Plug values into formula: Plug the values into the quadratic formula. t = [-4 ± sqrt(4^2 - 4*5*(-4))] / (2*5) t = [-4 ± sqrt(16 + 80)] / 10 t = [-4 ± sqrt(96)] / 10Find solutions for t: Simplify the square root and the fraction. t = [-4 ± 4sqrt(6)] / 10 t = -2/5 ± 2sqrt(6)/5 So, the solutions are t = -2/5 + 2sqrt(6)/5 and t = -2/5 - 2sqrt(6)/5.Let x and y be: Let x = -2/5 + 2sqrt(6)/5 and y = -2/5 - 2sqrt(6)/5. To find -xy, we multiply x and y and then take the negative. -xy = -((-2/5 + 2sqrt(6)/5) * (-2/5 - 2sqrt(6)/5))Simplify multiplication: Use the difference of squares formula to simplify the multiplication. -xy = -(((-2/5)^2 - (2sqrt(6)/5)^2))Calculate squares: Calculate the squares and simplify. -xy = -((4/25 - 24/25)) -xy = -((-20/25))Simplify negative sign: Simplify the negative sign and the fraction. -xy = 20/25 -xy = 4/5

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4(1-t)=5t^(2)
Let t=x and t=y be the solutions to the given equation.
What is the value of -xy?

◻

$4(1-t)=5t^{2}$ $\newline$ Let $t=x$ and $t=y$ be the solutions to the given equation. $\newline$ What is the value of $-xy$ ? $\newline$ $\square$

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4(1-t)=5t^{2}

\newline

Let

t=x

and

t=y

be the solutions to the given equation.

\newline

What is the value of

-xy

\newline

\square

Rewrite in standard form: First, let's rewrite the equation in standard quadratic form. $\newline$ $4 - 4t = 5t^2$ $\newline$ Move all terms to one side to set the equation to zero. $\newline$ $5t^2 + 4t - 4 = 0$
Solve quadratic equation: Now, we need to solve the quadratic equation for $t$ . We can use the quadratic formula: $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 5$ , $b = 4$ , and $c = -4$ .
Plug values into formula: Plug the values into the quadratic formula. $\newline$ t = [ $-4 \pm \sqrt{4^2 - 4\cdot5\cdot(-4)}] / (2\cdot5)$ $\newline$ t = [ $-4 \pm \sqrt{16 + 80}] / 10$ $\newline$ t = [ $-4 \pm \sqrt{96}] / 10$
Find solutions for $t$ : Simplify the square root and the fraction. $\newline$ $t = \frac{-4 \pm 4\sqrt{6}}{10}$ $\newline$ $t = -\frac{2}{5} \pm \frac{2\sqrt{6}}{5}$ $\newline$ So, the solutions are $t = -\frac{2}{5} + \frac{2\sqrt{6}}{5}$ and $t = -\frac{2}{5} - \frac{2\sqrt{6}}{5}$ .
Let $x$ and $y$ be: Let $x = -\frac{2}{5} + \frac{2\sqrt{6}}{5}$ and $y = -\frac{2}{5} - \frac{2\sqrt{6}}{5}$ . To find $-xy$ , we multiply $x$ and $y$ and then take the negative. -xy = -((-\frac{$2$}{$5$} + \frac{$2$\sqrt{$6$}}{$5$}) * (-\frac{$2$}{$5$} - \frac{$2$\sqrt{$6$}}{$5$}))
Simplify multiplication: Use the difference of squares formula to simplify the multiplication.\(\newline $-xy = -(((-\frac{2}{5})^2 - (\frac{2\sqrt{6}}{5})^2))$
Calculate squares: Calculate the squares and simplify. $\newline$ $-xy = -\left(\frac{4}{25} - \frac{24}{25}\right)$ $\newline$ $-xy = -\left(-\frac{20}{25}\right)$
Simplify negative sign: Simplify the negative sign and the fraction. $\newline$ $-xy = \frac{20}{25}$ $\newline$ $-xy = \frac{4}{5}$