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lina2011 [118]
2 years ago
11

Drag the ordered pair that are 5.5 units apart.

Mathematics
1 answer:
Alexus [3.1K]2 years ago
6 0

Answer:

  (-1.5, -3.5) and (-1.5, 2)

Step-by-step explanation:

The x-coordinates in increasing order are ...

  -3, -1.5, 5

The distances between pairs of these coordinates are ...

  -1.5 -(-3) = 1.5

  5 -(-3) = 8

  5 -(-1.5) = 6.5

Clearly, any pair with x = 5 is too far away from any pair with a different x-value.

<u>∆x = 1.5</u>

If the distance between pairs is 5.5, then the distance formula tells us the y-coordinate differences for pairs with x-coordinates of -3 and -1.5 must be ...

  d = √((x2 -x1)^2 +(y2 -y1)^2)

  5.5 = √(1.5^2 +(y2 -y1)^2) . . . substitute known values

  30.25 = 2.25 +(y2 -y1)^2 . . . . square both sides

  √28 = (y2 -y1) . . . . . . . . . . . . . . subtract 2.25 and take the square root

This value is irrational. Clearly none of the y-coordinates is irrational, so there are no point pairs with x-coordinates -3 and -1.5 that are 5.5 units apart.

__

<u>∆x = 0</u>

If the x-coordinates are the same, then the y-coordinates must differ by 5.5 in order for the points to be 5.5 units apart. The coordinates in order are ...

  for x = -1.5, y = -3.5, 2, 2.5 . . . . . differences of 5.5, 6, 0.25

  for x = 5, y = -3.5, 1.5 . . . . . . . difference of 5

The only pair we can find that is 5.5 units apart is ...

  (-1.5, -3.5) and (-1.5, 2).

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Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.
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d) \int a dt = v(t) = [L][T]^{-1}=LT^{-1}

e) \frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

v = \frac{ds}{dt}

a= \frac{dv}{dt}

Part a

If we do the dimensional analysis for v we got:

v = \frac{[L]}{[T]} = LT^{-1}

Part b

For the acceleration we can use the result obtained from part a and we got:

a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}

Part c

From definition if we do the integral of the velocity respect to t we got the position:

\int v dt = s(t)

And the dimensional analysis for the position is:

\int v dt = s(t) = [L]=L

Part d

The integral for the acceleration respect to the time is the velocity:

\int a dt = v(t)

And the dimensional analysis for the position is:

\int a dt = v(t) = [L][T]^{-1}=LT^{-1}

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}

7 0
3 years ago
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