Question

a quadratic equation is shown:
$4n^2 + 9n + 11 = 0$
what are the exact solutions to this equation?

$\bigcirc\\ n = \frac{-9 \pm i\sqrt{95}}{8}$
$\bigcirc\\ n = \frac{-9 \pm \sqrt{257}}{8}$
$\bigcirc\\ n = \frac{-9 \pm \sqrt{226}}{8}$
$\bigcirc\\ n = -9 \pm \sqrt{19}$

question 26
2.5 pts
a soccer ball is kicked from an initial height of 0 meters with an upward velocity of 32 m/s. the ball’s height, $h$ (in meters), after $t$ seconds is given by the following equation:
$h = 32t - 16t^2$
find the values of $t$ when the ball’s height is 10 meters. round to the nearest hundredth.
select all that apply.

$square$ 1.61 seconds
$square$ 0.77 seconds
$square$ 0.39 seconds
$square$ 1.37 seconds

Explanation:

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First Problem

Step1: Identify quadratic coefficients

For $4n^2 + 9n + 11 = 0$, $a=4$, $b=9$, $c=11$

Step2: Calculate discriminant

$\Delta = b^2 - 4ac = 9^2 - 4(4)(11) = 81 - 176 = -95$

Step3: Apply quadratic formula

$n = \frac{-b \pm \sqrt{\Delta}}{2a} = \frac{-9 \pm i\sqrt{95}}{8}$

Step1: Set height equation to 10

$10 = 32t - 16t^2$

Step2: Rearrange to standard quadratic

$16t^2 - 32t + 10 = 0$

Step3: Simplify the equation

Divide by 2: $8t^2 - 16t + 5 = 0$

Step4: Identify coefficients

$a=8$, $b=-16$, $c=5$

Step5: Calculate discriminant

$\Delta = (-16)^2 - 4(8)(5) = 256 - 160 = 96$

Step6: Apply quadratic formula

$t = \frac{16 \pm \sqrt{96}}{16} = \frac{16 \pm 4\sqrt{6}}{16} = \frac{4 \pm \sqrt{6}}{4}$

Step7: Compute decimal values

$\frac{4 + \sqrt{6}}{4} \approx \frac{4 + 2.449}{4} \approx 1.61$, $\frac{4 - \sqrt{6}}{4} \approx \frac{4 - 2.449}{4} \approx 0.39$

Answer:

$n = \frac{-9 \pm i\sqrt{95}}{8}$

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Second Problem