G(a/2)= 2(a/2)-3
=2a/2-3
then just subtract 2a/2 by 3/1 by finding the common denominator and same for the next one.
365717262cm2
I think they will be able with the one that
The mean GPA is greater than 2.75.
<h3>
What is GPA?</h3>
- A grade point average is a figure that represents the average of the final grades obtained in courses over time.
- A student's grade point average, sometimes known as a GPA, is computed by adding all accumulated final grades and dividing the total by the number of grades granted.
To find the GPA:
- GPA is calculated and could be utilized by the college to tell choices, which include the subsequent educational progression.
- Divide the total range of grade factors earned with the aid of the total number of letter-graded units undertaken.
- Mean GPA = total/50
- = 2.75.
Therefore, the mean GPA is greater than 2.75.
Know more about GPA here:
brainly.com/question/17010398
#SPJ4
Answer:
a) ![v = \frac{[L]}{[T]} = LT^{-1}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B%5BL%5D%7D%7B%5BT%5D%7D%20%3D%20LT%5E%7B-1%7D)
b) ![a = \frac{[L}{T}^{-1}]}{{T}}= L T^{-1} T^{-1}= L T^{-2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5BL%7D%7BT%7D%5E%7B-1%7D%5D%7D%7B%7BT%7D%7D%3D%20L%20T%5E%7B-1%7D%20T%5E%7B-1%7D%3D%20L%20T%5E%7B-2%7D)
c) ![\int v dt = s(t) = [L]=L](https://tex.z-dn.net/?f=%20%5Cint%20v%20dt%20%3D%20s%28t%29%20%3D%20%5BL%5D%3DL)
d) ![\int a dt = v(t) = [L][T]^{-1}=LT^{-1}](https://tex.z-dn.net/?f=%20%5Cint%20a%20dt%20%3D%20v%28t%29%20%3D%20%5BL%5D%5BT%5D%5E%7B-1%7D%3DLT%5E%7B-1%7D)
e) ![\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bda%7D%7Bdt%7D%3D%20%5Cfrac%7B%5BL%5D%5BT%5D%5E%7B-2%7D%7D%7BT%7D%20%3D%20%5BL%5D%5BT%5D%5E%7B-2%7D%20%5BT%5D%5E%7B-1%7D%20%3D%20LT%5E%7B-3%7D)
Step-by-step explanation:
Let define some notation:
[L]= represent longitude , [T] =represent time
And we have defined:
s(t) a position function


Part a
If we do the dimensional analysis for v we got:
![v = \frac{[L]}{[T]} = LT^{-1}](https://tex.z-dn.net/?f=%20v%20%3D%20%5Cfrac%7B%5BL%5D%7D%7B%5BT%5D%7D%20%3D%20LT%5E%7B-1%7D)
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}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7B%5BL%7D%7BT%7D%5E%7B-1%7D%5D%7D%7B%7BT%7D%7D%3D%20L%20T%5E%7B-1%7D%20T%5E%7B-1%7D%3D%20L%20T%5E%7B-2%7D)
Part c
From definition if we do the integral of the velocity respect to t we got the position:

And the dimensional analysis for the position is:
![\int v dt = s(t) = [L]=L](https://tex.z-dn.net/?f=%20%5Cint%20v%20dt%20%3D%20s%28t%29%20%3D%20%5BL%5D%3DL)
Part d
The integral for the acceleration respect to the time is the velocity:

And the dimensional analysis for the position is:
![\int a dt = v(t) = [L][T]^{-1}=LT^{-1}](https://tex.z-dn.net/?f=%20%5Cint%20a%20dt%20%3D%20v%28t%29%20%3D%20%5BL%5D%5BT%5D%5E%7B-1%7D%3DLT%5E%7B-1%7D)
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}](https://tex.z-dn.net/?f=%20%5Cfrac%7Bda%7D%7Bdt%7D%3D%20%5Cfrac%7B%5BL%5D%5BT%5D%5E%7B-2%7D%7D%7BT%7D%20%3D%20%5BL%5D%5BT%5D%5E%7B-2%7D%20%5BT%5D%5E%7B-1%7D%20%3D%20LT%5E%7B-3%7D)