Answer:
-3/4
Step-by-step explanation:
-5/4 + 2/4 = -3/4
Hope this helps :)
Answer:
Total cost = <em>$</em> 4.55
Step-by-step explanation:
2(8a + 5b) -(2a + 3b)
<em>Part A: Simplify the expression shown above</em>
Exapanding the brackets;
16a + 10b - 2a - 3b
Collecting like terms;
16a - 2a + 10b - 3b
14a + 7b
<em>Part B: If a represents apples that cost $0.25 each and b represents bananas that cost SO. 15 each, what is the total cost based on the expression above?</em>
a = 0.25
b = 0.15
14a + 7b
Inserting the values into the equation;
14 (0.25) + 7(0.15)
Total cost = <em>$</em> 4.55
The answer would be 30 since for 180 pounds and 60 feet you would need to divide 180 by 60 to get the unit rate for the pressure and then multiply by 10 which would get you 30. Correct me if I’m wrong. Hope it helps :)
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)
this answer is 9% it is bigger than 0.5