x=0
because you must
determine the undefined range
write all numerators above the common denomonator
multiply the parenthesis
collect the alike terms
remove parenthesis
eliminate the opposites collect the like terms
set the numerator equal to 0
divide both sides by 6
check if thesolution is the defined range
if so then you would find your answer X=0
Answer:
The values of variables x and m are 11 and 17
Step-by-step explanation:
The question has missing details as the diagram of the trapezoid isn't attached.
(See attachment).
Given that trapezoid CHLE is isosceles then the angles at the base area equal (4x)
And
The angles at the top are also equal
8m = 11x + 15
At this point, the four angles in the trapezoid are 8m, 11x + 15, 4x and 4x..
The sum of interior= 360
So,
11x + 15 + 8m + 4x + 4x = 360
Collect like terms
11x + 4x + 4x + 8m = 360 - 15
19x + 8m = 345
Substitute 11x + 15 for 8m
19x + 11x + 15 = 345
30x + 15 = 345
30x = 345 - 15
30x = 330
Divide through by 30
30x/30 = 330/30
x = 11
Recall that 8m = 11x + 15;
8m = 11(11) + 15
8m = 121 + 15
8m = 136
Divide through by 8
8m/8 = 136/8
m = 17
Hence, the values of variables x and m are 11 and 17
Set it up like a regular equation. I used:

As you can see, however, I inserted the "less than" symbol rather than the equal sign because values greater than a certain value can also equal "x". Solve it like a normal equation and you will arrive at:
<u>
</u>

This means that, as long as x is greater than 45, "6x" will be greater than "3x+135". Thus, if you bring 46 students, the Science Center will cost less.
She works for 40 hours, and then 5 hours.
So, for these first 40 hours, Tyra earns $21.40 per hour. We can find how much she earns in 40 hours:
21.40 x 40 = 856.
So, in the first 40 hours, she makes $856.
Now, for any number of hours Tayra works more that 40 hours, she will earn $32.10 each hour.
She works 5 hours over 40 hours, since it is 45 hours in total.
Hence, Tayra will earn $32.10 every hour for the 5 hours:
5 x 32.10 = 160.5.
Therefore, she will earn $160.50 for the 5 hours.
Now we have both of the earning amounts that make up how much she makes in the 45 hours. We can add them together
40 hours + 5 hours = 45 hours.
Substitute the amounts in the hours.
856 + 160.5 = 1016.5.
Thus, the answer is that Tyra will warn $1,016.50 in the total 45 hours.
Ergo, the correct answer is C.
Hope this helps, let me know if you need anything else! :)
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)