Solved He Quadratic Equation Y 16t2 64t 2 ï Represents The Chegg There are 2 steps to solve this one. to determine when the object hits the ground, we need to find the time t when the height y is zero . not the question you’re looking for? post any question and get expert help quickly. The quadratic equation y = − 16 t 2 4 t 2 represents a moving object's trajectory where y is the object's height in feet above the ground after t seconds. at what time will the object hit the ground? a. 2 seconds b. 0.25 seconds.
Solved The Quadratic Equation Y 1612 56t 2 Represents Chegg To solve the equation d = − 16 t 2 4 t, we rearranged it into the standard quadratic form and applied the quadratic formula. the correct solution for t corresponds to option a: t = 8 12 ± 41 − 4 d . thus, the answer is option a. Enter the equation you want to solve using the quadratic formula. the quadratic formula calculator finds solutions to quadratic equations with real coefficients. for equations with real solutions, you can use the graphing tool to visualize the solutions. quadratic formula: x = − b ± b 2 − 4 a c 2 a. step 2: click the blue arrow to submit. The quadratic equation $y= 16 t^{2} 4 t 2$ represents a moving object's trajectory where $y$ is the object's height in feet above the ground after $t$ seconds. at what time will the object hit the ground? a.) 1 second b.) 0.25 seconds c.) 2 seconds d.) 0.5 seconds. The quadratic equation y= 16t2 4t 2 represents a moving object's trajectory where y is the object's height in feet above the ground after t seconds.at what time will the object hit the ground? your solution’s ready to go! our expert help has broken down your problem into an easy to learn solution you can count on.
Solved The Quadratic Equation Y 16t2 64t 2 ï Represents The Chegg The quadratic equation $y= 16 t^{2} 4 t 2$ represents a moving object's trajectory where $y$ is the object's height in feet above the ground after $t$ seconds. at what time will the object hit the ground? a.) 1 second b.) 0.25 seconds c.) 2 seconds d.) 0.5 seconds. The quadratic equation y= 16t2 4t 2 represents a moving object's trajectory where y is the object's height in feet above the ground after t seconds.at what time will the object hit the ground? your solution’s ready to go! our expert help has broken down your problem into an easy to learn solution you can count on. X^2: x^{\msquare} \log {\msquare} \sqrt{\square} \nthroot[\msquare]{\square} \le \ge \frac{\msquare}{\msquare} \cdot \div: x^{\circ} \pi \left(\square\right)^{'} \frac{d}{dx} \frac{\partial}{\partial x} \int \int {\msquare}^{\msquare} \lim \sum \infty \theta (f\:\circ\:g) f(x). All equations of the form ax^{2} bx c=0 can be solved using the quadratic formula: \frac{ b±\sqrt{b^{2} 4ac}}{2a}. the quadratic formula gives two solutions, one when ± is addition and one when it is subtraction. The quadratic equation y = −16t2 4t 2 represents a moving object's trajectory, where y is the object's height in feet above the ground after t seconds. at what time will the object hit the ground? since y is the object's height, it will be on the ground when y = 0. so let's do that: 0 = −16t2 4t 2. The quadratic equation $y= 16 t^{2} 4 t 2$ represents a moving object's trajectory where $y$ is the object's height in feet above the ground after $t$ seconds. at what time will the object hit the ground? a.) 1 second b.) 2 seconds c.) 0.25 seconds d.) 0.5 seconds.