.help me thanks much

Mathematics

1 answer:

DochEvi [55]3 years ago

8 0

(16x^1/6y^-2)^3/2 / (x^-1/6y^6)^3/2

=(4^2)^3/2 (x^1/6)^3/2 (y^-2)^3/2 / (x^-1/6)^3/2(y^6)^3/2
= 4^3 x^1/4 y^-3 / x^-1/4 y ^9
= 64 x^(1/4 +1/4) y^(-3-9)
= 64 x^1/2 y^-12
= 64 x^1/2 / y^12

answer is D.64 x^1/2 / y^12

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$\bf \qquad \textit{initial velocity}\\\\
\begin{array}{llll}
\qquad \textit{in feet}\\\\
h(t) = -16t^2+v_ot+h_o
\end{array} 
\quad 
\begin{cases}
v_o=\textit{initial velocity of the object}\\
h_o=\textit{initial height of the object}\\
h=\textit{height of the object at "t" seconds}
\end{cases}\\\\
-------------------------------\\\\$

$\bf h(t)=7+29t-16t^2\qquad h(t)=19\implies 19=7+29t-16t^2
\\\\\\
16t^2-29t+12=0\impliedby \textit{now, let's use the quadratic formula}
\\\\\\
t=\cfrac{-(-29)\pm\sqrt{(-29)^2-4(16)(12)}}{2(16)}\implies t=\cfrac{29\pm\sqrt{841-768}}{32}
\\\\\\
t=\cfrac{29\pm\sqrt{73}}{32}\implies t\approx 
\begin{cases}
1.17\\
0.64
\end{cases}$

so hmm check the picture below, thus the ball hits 19 feet of height twice, once on the way up, and once on the way down, at about 0.64 seconds and at 1.17 seconds.