The position of an object at any time t is given by
s(t)=3t4−40t3+126t2−9
.
1 Determine the velocity of the object at any time t.
2 Does the object ever stop changing?
3 when is the object moving to the right and when is the object moving to the left​

Answers

Answer 1

Answer:

Step-by-step explanation:

I will ASSUME that your equation is

s(t) = 3t⁴ - 40t³ + 126t² - 9

1) velocity is the derivative of position

v(t) = s'(t) = 12t³ - 120t² + 252t

2) changing what? position? velocity? acceleration? gender? color?

position No   the plot is continuous

velocity Yes  both magnitude and direction. Zeros at t = 0, 3, and 7 s

acceleration Yes, both magnitude and direction. Zeros at approx. t = 1.306 and 5.361 s

gender? who knows

color? probably.

3) velocity is positive (moving to the right on a standard number line) for

0 < t < 3 and t > 7

Moves to the left for

t < 0 and 3 < t < 7

Answer 2

1. the velocity of the object at any time "t" is v(t) = 12t³ - 120t² + 252t.

2. the object stops changing at t = 6 and t = 3.5.

3. the object is moving to the right in the interval 0 < t < 3.5 and moving to the left in the interval t > 6.

To determine the velocity of the object at any time "t," we need to find the derivative of the position function "s(t)" with respect to "t."

1. Velocity of the object at any time "t":

The velocity is the rate of change of position with respect to time, which is given by the derivative of the position function, s(t).

s(t) = 3t⁴ - 40t³ + 126t² - 9

To find the velocity function, we take the derivative of s(t) with respect to t:

v(t) = d/dt (3t⁴ - 40t³ + 126t²- 9)

v(t) = 12t³ - 120t² + 252t

So, the velocity of the object at any time "t" is v(t) = 12t³ - 120t² + 252t.

2. Does the object ever stop changing?

The object stops changing when its velocity becomes zero. To find when the velocity is zero, we need to solve the equation v(t) = 0.

12t³ - 120t² + 252t = 0

Now, we can factor out common terms:

t(12t² - 120t + 252) = 0

Now, set each factor equal to zero and solve for "t":

1st factor: t = 0

2nd factor: 12t² - 120t + 252 = 0

To solve the quadratic equation, we can either factor it further or use the quadratic formula. Factoring the quadratic equation gives us:

(t - 6)(12t - 42) = 0

Setting each factor to zero:

1. t - 6 = 0

t = 6

2. 12t - 42 = 0

12t = 42

t = 3.5

So, the object stops changing at t = 6 and t = 3.5.

3. When is the object moving to the right and when is the object moving to the left?

The object is moving to the right when its velocity (v(t)) is positive, and it is moving to the left when its velocity (v(t)) is negative.

Let's analyze the velocity function

v(t) = 12t³ - 120t² + 252t

To find the intervals where the object is moving to the right or left, we need to determine when v(t) is positive or negative.

1. Find critical points by setting v(t) = 0 and solving for "t":

12t³ - 120t² + 252t = 0

As we already found earlier, the critical points are t = 6 and t = 3.5.

2. Test each interval created by these critical points using test points to determine the sign of v(t) in those intervals.

Pick a value of "t" in each of these intervals:

- When t < 3.5 (e.g., t = 0),

- When 3.5 < t < 6 (e.g., t = 4),

- When t > 6 (e.g., t = 7).

Now, substitute these test values into the velocity function v(t):

- v(0) = 12(0)³ - 120(0)² + 252(0) = 0

- v(4) = 12(4)³ - 120(4)² + 252(4) = 192

- v(7) = 12(7)³ - 120(7)² + 252(7) = -588

Based on the test points, we can conclude:

- The object is moving to the right when 0 < t < 3.5 (positive velocity).

- The object is moving to the left when t > 6 (negative velocity).

Therefore, the object is moving to the right in the interval 0 < t < 3.5 and moving to the left in the interval t > 6.

Learn more about velocity here

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Solve
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B. 19

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A.35 because 2 minuses equal a positive so you would cancel out and add them together

The correct alternative that, matches the correct value of this expression is the letter A. That is, the answer will be 35.

Step-by-step explanation:

To find the value of this expression, let's eliminate the parentheses, and add the numbers, where the signs are equal.

_ When the signs are the same: just add them up.

_ When the signs are different: just subtract.

Resolution:

[tex]\large \sf =27-(-8)[/tex]

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can someone pls help me with dis ill give brainliest​

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Step-by-step explanation:

Convert a number to and from scientific notation, e notation, engineering notation, standard form, and real numbers. Enter a number or a decimal number or scientific notation and the calculator converts to scientific notation, e notation, engineering notation, standard form and word form formats.

To enter a number in scientific notation use a carat ^ to indicate the powers of 10. You can also enter numbers in e notation. Examples: 3.45 x 10^5 or 3.45e5.

Order of magnitude will also be identified for the calculated standard form. The order of magnitude when written in standard form, is the nth power of 10. For example, 3.4 x 10^5 has an order of magnitude of 5 since 10 is raised to the 5th power.

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Yes it is written in scientific notation.

Step-by-step explanation:

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Step-by-step explanation:

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Answers

Answer:

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Step-by-step explanation:

Answer:

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