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

Studdy Solution

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

Studdy Solution

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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Question Find the value of ttt that satisfies the equation V(t)=4t3−27t2V(t)=4t^3-27t^2V(t)=4t3−27t2.

Studdy Solution

STEP 1

Assumptions1. The equation is V(t)=4t3−27tV(t)=4 t^{3}-27 t^{}V(t)=4t3−27t . We need to find the value of ttt for which V(t)=0V(t)=0V(t)=0

STEP 2

To find the value of ttt, we need to set the equation equal to zero and solve for ttt.4t−27t2=04 t^{}-27 t^{2} =04t−27t2=0

STEP 3

We can factor out a t2t^{2}t2 from each term to simplify the equation.t2(t−27)=0t^{2}(t -27) =0t2(t−27)=0

STEP 4

Now we can set each factor equal to zero and solve for ttt.t2=0t^{2} =0t2=04t−27=04t -27 =04t−27=0

STEP 5

olving the first equation t2=0t^{2} =0t2=0 gives us one solution.t=0t =0t=0

STEP 6

olving the second equation 4t−27=04t -27 =04t−27=0 gives us the second solution.4t=274t =274t=27t=274=6.75t = \frac{27}{4} =6.75t=427​=6.75So, the values of ttt for which V(t)=0V(t)=0V(t)=0 are t=0t=0t=0 and t=6.75t=6.75t=6.75.

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

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

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

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

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

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

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 $t=6.75$ .