Plz help..............refresh the page before u answer

Mathematics

1 answer:

pshichka [43]3 years ago

6 0

Like any other time you divide one fraction by the other...

(a/b)/(c/d)=(a/b)(d/c)=(ad)/(bc)

So in this case you actually have:

3x^3/(4y) * 9/x multiply

27x^3/(4xy) one x cancels out leaving you with

27x^2/(4y)

Second answer down :P

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H=7 + 29t-16t^2 find all values of t for which the balls height is 19ft

NikAS [45]

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

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

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.