The equation $y = -6t^2 + 43t$ describes the height (in feet) of a projectile $t$ seconds after it is launched from the surface of Mars at 43 feet per second. In how many seconds will the projectile first reach 77 feet in height? Express your answer as a decimal rounded to the nearest tenth.
No one wants to solve it, steve
Bruh, steve you in first grade or something? Because you are nowhere near the correct answer
(-3t + 11)(2t - 7) = roots t = 11/3, 7/2
Ya lol and he thinks hes a genius but hes in kindergarten
To find the time at which the height of the projectile is 77 feet, we need to solve the equation $-6t^2 + 43t = 77$ for $t$.
Let's rearrange the equation to get it in the form $-6t^2 + 43t - 77 = 0$. To solve this quadratic equation, we can use the quadratic formula:
$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},$$
where $a$, $b$, and $c$ are the coefficients of the quadratic equation $at^2 + bt + c = 0$. In this case, $a = -6$, $b = 43$, and $c = -77$.
Plugging these values into the quadratic formula gives:
$$t = \frac{-43 \pm \sqrt{43^2 - 4(-6)(-77)}}{2(-6)}.$$
Simplifying further:
$$t = \frac{-43 \pm \sqrt{1849 - 1848}}{-12}.$$
$$t = \frac{-43 \pm \sqrt{1}}{-12}.$$
As $\sqrt{1} = 1$, we have two possible solutions:
$$t_1 = \frac{-43 + 1}{-12} = \frac{-42}{-12} = 3.5,$$
$$t_2 = \frac{-43 - 1}{-12} = \frac{-44}{-12} = 3.7.$$
Rounding both solutions to the nearest tenth, we find that the projectile will first reach a height of 77 feet after approximately 3.5 seconds.
huh? Just solve
-6t^2 + 43t = 77