Answer:
Times 30 with every time
Step-by-step explanation:
Answer:
Most black horses will fade to a brownish color if the horse is exposed to sunlight regularly
Step-by-step explanation:
So hmm check the picture below
![\bf \qquad \textit{initial velocity}\\\\ h = -16t^2+v_ot+h_o \qquad \text{in feet}\\ \\ \begin{cases} v_o=\textit{initial velocity of the object}\to &64\\ h_o=\textit{initial height of the object}\to &12\\ h=\textit{height of the object at "t" seconds} \end{cases}\\\\ -----------------------------\\\\](https://tex.z-dn.net/?f=%5Cbf%20%5Cqquad%20%5Ctextit%7Binitial%20velocity%7D%5C%5C%5C%5C%0Ah%20%3D%20-16t%5E2%2Bv_ot%2Bh_o%20%5Cqquad%20%5Ctext%7Bin%20feet%7D%5C%5C%0A%5C%5C%20%0A%5Cbegin%7Bcases%7D%0Av_o%3D%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%5Cto%20%2664%5C%5C%0Ah_o%3D%5Ctextit%7Binitial%20height%20of%20the%20object%7D%5Cto%20%2612%5C%5C%0Ah%3D%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%20%5Cend%7Bcases%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C)
![\bf \textit{vertex of a parabola}\\ \quad \\ \begin{array}{lccclll} h(t)=&-16t^2&+64t&+12\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvertex%20of%20a%20parabola%7D%5C%5C%20%5Cquad%20%5C%5C%0A%0A%5Cbegin%7Barray%7D%7Blccclll%7D%0Ah%28t%29%3D%26-16t%5E2%26%2B64t%26%2B12%5C%5C%0A%26%5Cuparrow%20%26%5Cuparrow%20%26%5Cuparrow%20%5C%5C%0A%26a%26b%26c%0A%5Cend%7Barray%7D%5Cqquad%20%0A%5Cleft%28-%5Ccfrac%7B%7B%7B%20b%7D%7D%7D%7B2%7B%7B%20a%7D%7D%7D%5Cquad%20%2C%5Cquad%20%20%7B%7B%20c%7D%7D-%5Ccfrac%7B%7B%7B%20b%7D%7D%5E2%7D%7B4%7B%7B%20a%7D%7D%7D%5Cright%29)
part 1)
it takes
![\bf -\cfrac{{{ b}}}{2{{ a}}}\quad seconds](https://tex.z-dn.net/?f=%5Cbf%20-%5Ccfrac%7B%7B%7B%20b%7D%7D%7D%7B2%7B%7B%20a%7D%7D%7D%5Cquad%20seconds)
part 2)
![\bf \textit{now, doubling }v_o\\\\ \begin{cases} v_o=\textit{initial velocity of the object}\to &128\\ h_o=\textit{initial height of the object}\to &12\\ h=\textit{height of the object at "t" seconds}\end{cases}\\\\ -----------------------------\\\\ \textit{vertex of a parabola}\\ \quad \\ \begin{array}{lccclll} h(t)=&-16t^2&+128t&+12\\ &\uparrow &\uparrow &\uparrow \\ &a&b&c \end{array}\qquad \left(-\cfrac{{{ b}}}{2{{ a}}}\quad ,\quad {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\right)](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bnow%2C%20doubling%20%7Dv_o%5C%5C%5C%5C%0A%5Cbegin%7Bcases%7D%0Av_o%3D%5Ctextit%7Binitial%20velocity%20of%20the%20object%7D%5Cto%20%26128%5C%5C%0Ah_o%3D%5Ctextit%7Binitial%20height%20of%20the%20object%7D%5Cto%20%2612%5C%5C%0Ah%3D%5Ctextit%7Bheight%20of%20the%20object%20at%20%22t%22%20seconds%7D%5Cend%7Bcases%7D%5C%5C%5C%5C%0A-----------------------------%5C%5C%5C%5C%0A%5Ctextit%7Bvertex%20of%20a%20parabola%7D%5C%5C%20%5Cquad%20%5C%5C%0A%0A%5Cbegin%7Barray%7D%7Blccclll%7D%0Ah%28t%29%3D%26-16t%5E2%26%2B128t%26%2B12%5C%5C%0A%26%5Cuparrow%20%26%5Cuparrow%20%26%5Cuparrow%20%5C%5C%0A%26a%26b%26c%0A%5Cend%7Barray%7D%5Cqquad%20%0A%5Cleft%28-%5Ccfrac%7B%7B%7B%20b%7D%7D%7D%7B2%7B%7B%20a%7D%7D%7D%5Cquad%20%2C%5Cquad%20%20%7B%7B%20c%7D%7D-%5Ccfrac%7B%7B%7B%20b%7D%7D%5E2%7D%7B4%7B%7B%20a%7D%7D%7D%5Cright%29)
it will reach the maximum height at
![\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}\quad feet](https://tex.z-dn.net/?f=%5Cbf%20%7B%7B%20c%7D%7D-%5Ccfrac%7B%7B%7B%20b%7D%7D%5E2%7D%7B4%7B%7B%20a%7D%7D%7D%5Cquad%20feet)
how much higher than before is that? well, what was the y-coordinate for when the vₒ was 64? what did you get for
![\bf {{ c}}-\cfrac{{{ b}}^2}{4{{ a}}}](https://tex.z-dn.net/?f=%5Cbf%20%7B%7B%20c%7D%7D-%5Ccfrac%7B%7B%7B%20b%7D%7D%5E2%7D%7B4%7B%7B%20a%7D%7D%7D)
?
subtract that from this height when vₒ is 128 or doubled, to get their difference, that's how much higher it became
Hello! Using the law of sines, it would be 4.0212 for side a, or just 4 since you wanted it rounded to the nearest tenth.