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Arada [10]
4 years ago
15

An object moves along the x-axis, its position at each time t> 0 given by \"x(t) = 81/2*t^2-3/4*t^4\" Determine the time inte

rval(s), if any, during which the object slows down.
Mathematics
1 answer:
Tamiku [17]4 years ago
3 0
You can say an object slow down when the acceleration is negative which can be observed on a negative slope during the graph of acceleration, from a position equation lets derive 2 times to attain the acceleration equation

<span>x(t) = 81/2*t^2-3/4*t^4
V(t) = 81X - 3t^3
A(t) = 81 - 9t^2

</span>81 - 9t^2 < 0<span>
t^2 > 9
-3 > t > 3


</span>
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Leni [432]

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the 3rd one

Step-by-step explanation:


6 0
3 years ago
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Solve for x. Find all solutions from 0 to 2 π . ​Show all work as you solve each problem - you may need more space to show your
gtnhenbr [62]

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x=189

Step-by-step explanation:

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3 0
2 years ago
Determine whether y2=3x+1 is a function.
bazaltina [42]

Answer: it is not a function

Step-by-step explanation:

Given :

y^{2} = 3x + 1

The first thing is to make y the subject of the formula, that means we will take the square root of both sides

y = \sqrt{3x + 1}

since y =  \sqrt{3x + 1} , the the value of y could be positive or negative.

For example , \sqrt{4} = ± 2 , since -2 x -2 will also give + 4

For y =  \sqrt{3x + 1} to be a function , there must be a unique value for y for each value of x , but this is not the case as y could be positive or negative for a single value of x , so it is not a function.

4 0
3 years ago
Consider the hyperbola represented by the equation: (picture)
forsale [732]

Answer:

- The center of hyperbola is (-2 , 3)

- The left vertex, if the hyperbola opens horizontally, or the

  bottom vertex, if it opens vertically, is (-6 , 3)

- The other vertex is (2 , 3)

Step-by-step explanation:

* Lets study the equation of the hyperbola

- The standard form of the equation of a hyperbola with  

  center (h , k) and transverse axis parallel to the x-axis is

  (x - h)²/a² - (y - k)²/b² = 1, Where

- The length of the transverse axis is 2a

- The coordinates of the vertices are (h ± a , k)

- The length of the conjugate axis is 2b

- The coordinates of the co-vertices are (h , k ± b)

- The coordinates of the foci are (h ± c , k), where c² = a² + b²

* Now lets solve the problem

∵ (x + 2)²/4² - (y - 3)²/5² = 1

∵  (x - h)²/a² - (y - k)²/b² = 1

∴ a = 4 , b = 5

∴ h = -2 , k = 3

∵ The center of the hyperbola is (h , k)

∴ The center of hyperbola is (-2 , 3)

∵ The coordinates of the vertices are (h + a , k)  , (h - a , k)

∴ The coordinates of the vertices are (-2 + 4 , 3)  , (-2 - 4 , 3)

∴ The coordinates of the vertices are (2 , 3)  , (-6 , 3)

∴ The left vertex, if the hyperbola opens horizontally, or the

  bottom vertex, if it opens vertically, is (-6 , 3)

∴ The other vertex is (2 , 3)

6 0
4 years ago
2 1/8x+3/4x+1/6-7/12x+5 2/3
Georgia [21]
55
—— x + 35/6
24

this is the final answer
7 0
3 years ago
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