4 years ago

Can someone please help me with this one problem

1 answer:

5 0

The answer is: a) y=.33× 1.11^x

You might be interested in

lesya692 [45]

Answer:

B. {1, -4}

Step-by-step explanation:

Just rotate the screen and look at the position.

4 0

3 years ago

Zepler [3.9K]

Step-by-step explanation:

The particle is at rest when the derivative of x(t), which is its velocity, is zero, i.e.,

$\dfrac{d}{dt}[x(t)] = \dot{x}(t) = 0$

Since $x(t) = 2t^3 - 21t^2 + 72t - 53,$ its derivative is

$\dot{x}(t) = 6t^2 - 42t + 72$

Equating this to zero, we get

$6t^2 - 42t + 72 = 0 \Rightarrow t^2 - 7t + 12 = 0$

This is a familiar quadratic equation whose roots are

$t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$\;\;\;= \dfrac{7 \pm \sqrt{(7)^2 - 4(1)(12)}}{2}$

$\;\;\;= \dfrac{7 \pm 1}{2}$

$\;\;\;= 3\;\text{and}\;4$

This means that the particle will be at rest at t = 3 and t= 4.

8 0

3 years ago

sergiy2304 [10]

Answer = 12x3−7x2+9x−2

6 0

4 years ago

docker41 [41]

The answer is D because the answers are put into vertex form and all you have to do is look at your (h,k) values

4 0

3 years ago

padilas [110]

The answer is C. -1+-6=-7, not 7.

4 0

4 years ago