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Solved on Nov 07, 2023

Find the value of $t$ that satisfies the equation $V(t)=4t^3-27t^2$ .

STEP 1

Assumptions1. The equation is $V(t)=4 t^{3}-27 t^{}$ . We need to find the value of $t$ for which $V(t)=0$

STEP 2

To find the value of $t$ , we need to set the equation equal to zero and solve for $t$ .
$4 t^{}-27 t^{2} =0$

STEP 3

We can factor out a $t^{2}$ from each term to simplify the equation.
$t^{2}(t -27) =0$

STEP 4

Now we can set each factor equal to zero and solve for $t$ .
$t^{2} =0$ $4t -27 =0$

STEP 5

olving the first equation $t^{2} =0$ gives us one solution.
$t =0$

STEP 6

olving the second equation $4t -27 =0$ gives us the second solution.
$4t =27$ $t = \frac{27}{4} =6.75$ So, the values of $t$ for which $V(t)=0$ are $t=0$ and $t=6.75$ .

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