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Fudgin [204]
3 years ago
8

What is common characteristics of 2 miles, 72,000 inches, 3,000 feet

Mathematics
1 answer:
evablogger [386]3 years ago
6 0

They're all distances or lengths.

They're all in imperial units.


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solniwko [45]
The answer is 13/25 13 is odd so it can't be evenly divided by anything so it is in its simplest form
6 0
3 years ago
Help meeeeeeeeeeeeeeeeee pplllzzzzzzzzz asap
Yakvenalex [24]

Answer: 16

Step-by-step explanation:

8+6+2

6 0
3 years ago
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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.
finlep [7]

Answer:

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

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

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

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
What is the difference of 13 and 5 times a number that equals 29
Anit [1.1K]
(13 - 5n) = 29
-5n = 29 - 13
-5n = 16
n = -16/5 or - 3 1/5
3 0
3 years ago
Read 2 more answers
There are 9 marbles representing 3 different colors. Write a problem where 2 marbles are selected at random without replcement a
Nata [24]
There are 9 marbles in the bag. We pick 2 without replacement and get a probability of 1/6.

Each draw of a marble has a probability associated with it. Multiplying these gives 1/6 so let us assume the probabilities are (1/3) and (1/2).

In order for the first draw to have a probability of 1/3 we need to draw a color that has (1/3)(9)=3 marbles. So let's say there are 3 red marbles. The P(a red marble is drawn) = 1/3.

Now that a marble has been drawn there are 8 marbles left. In order for the second draw to have a probability of 1/2 we must draw a color that has (1/2)(8) = 4 marbles. So let's say there are 4 blue marbles out of the 8.

Since there are 9 marbles to start and we have 3 red marbles and 4 blue marbles, the remaining 2 marbles must be a different color. Let us say they are green.

The problem is: There are 3 red marbles, 4 blue marbles and 2 green marbles in a jar. A marble is picked at random, it's color is noted and the marble is not replaced. A second marble is drawn at random and its color noted. What is the probability that the first marble is red and the second blue?
8 0
3 years ago
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