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r-ruslan [8.4K]

3 years ago

10

The position of an object at time t is given by s(t) = -8 - 9t. Find the instantaneous velocity at t = 1 by finding the derivati

ve. USING THE LIMIT FORMULA TO FIND THE DERIVATIVE

2 answers:

IgorLugansk [536]3 years ago

7 0

Answer:

Instantaneous velocity at t = 1 by finding the derivative is -9 units.

Step-by-step explanation:

The position of an object at time t is given by s(t) = -8 - 9t

Displacement, s(t) = -8 - 9t.

We need to find the instantaneous velocity at t = 1 by finding the derivative.

We have

s'(t) = 0 - 9 = -9

s'(1) = -9

So Instantaneous velocity at t = 1 by finding the derivative is -9 units.

finlep [7]3 years ago

3 0

Ughhh...the limit formula...this should be thrown in the trash the minute you start doing derivatives :P

(-8-9(t+d)-(-8-9t))/(t+d-t)

(-8-9t-9d+8+9t)/d

-9d/d (which is -9 for any value of d as well as when d approaches zero)

-9

So the instantaneous velocity is regardless of t or delta t

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Step-by-step explanation:

Required

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$D = (5,-8)$ -- Fourth quadrant

The above quadrants satisfy the condition:

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<u>HOW TO KNOW THE PERIMETER IS 42</u>

To do this, we simply calculate the distance between the edges and add them up

<u>Distance is calculated as:</u>

<u></u> $D = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2$ <u></u>

<u></u>

<u>For AB</u>

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<u>For BC</u>

$B = (-5,3)$

$C = (-5,-8)$

$D_2 = \sqrt{(-5 - (-5))^2 + (3 - (-8))^2}= \sqrt{(0)^2 + (11)^2} = \sqrt{121} = 11$

<u>For CD</u>

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$D_3 = \sqrt{(-5 -5)^2 + (-8 - (-8))^2}= \sqrt{(-10)^2 + (0)^2} = \sqrt{100} = 10$

<u>For DA</u>

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See attachment for rectangle

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