An object is launched from the ground. The object’s height, in feet, can be described by the quadratic function h(t) = 80t – 16t
2 answers:
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Hey there, hope I can help!
Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
Lets use the quadratic formula now
Multiply the numbers 2 * 1 = 2
![\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%20%5Csqrt%5Bn%5D%7Bab%7D%3D%5Csqrt%5Bn%5D%7Ba%7D%5Csqrt%5Bn%5D%7Bb%7D%20%5C%20%5Ctextgreater%20%5C%20%20%5Csqrt%7B47%7D%5Csqrt%7B2%5E2%7D%20)
![\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}](https://tex.z-dn.net/?f=%5Cmathrm%7BApply%5C%3Aradical%5C%3Arule%7D%3A%20%5Csqrt%5Bn%5D%7Ba%5En%7D%3Da%20%5C%20%5Ctextgreater%20%5C%20%20%5Csqrt%7B2%5E2%7D%3D2%20%5C%20%5Ctextgreater%20%5C%20%202%5Csqrt%7B47%7D%20%5C%20%5Ctextgreater%20%5C%20%20%5Cfrac%7B2%2B2%5Csqrt%7B47%7D%7D%7B2%7D)
Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.
Therefore our final solutions are
Hope this helps!
Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
Lets use the quadratic formula now
Multiply the numbers 2 * 1 = 2
Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.
Therefore our final solutions are
Hope this helps!
Answer:
Step-by-step explanation:
1. Based on the information given in the problem, you can draw the right triangle shown in the figure attached.
2. Keeping the informationn given in the problem, you can calculate the angle that the sum makes with the ground ()as you can see below:
Where:
3. Then:
