<em>Your question has confusing details. So, I will rephrase as:</em>
<em>How many 2' x 4' would be needed for Bottom Plates and Double Top Plates on a Building 25' x 50' ?
</em>
<em></em>
Answer:
12.5 plates
Step-by-step explanation:
Given
Building dimension:
![Length = 25ft](https://tex.z-dn.net/?f=Length%20%3D%2025ft)
![Width = 50ft](https://tex.z-dn.net/?f=Width%20%3D%2050ft)
Plates
![Length = 2ft](https://tex.z-dn.net/?f=Length%20%3D%202ft)
![Width= 4ft](https://tex.z-dn.net/?f=Width%3D%204ft)
To do this, we have to divide the dimension of the building by the dimension of the plates
So, we have:
![n = \frac{Length\ of\ Building}{Length\ of\ Plates}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BLength%5C%20of%5C%20Building%7D%7BLength%5C%20of%5C%20Plates%7D)
and
![n = \frac{Width\ of\ Building}{Width\ of\ Plates}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BWidth%5C%20of%5C%20Building%7D%7BWidth%5C%20of%5C%20Plates%7D)
For the length, we have:
![n = \frac{25}{2}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B25%7D%7B2%7D)
![n = 12.5](https://tex.z-dn.net/?f=n%20%3D%2012.5)
For the width, we have:
![n = \frac{50}{4}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B50%7D%7B4%7D)
![n = 12.5](https://tex.z-dn.net/?f=n%20%3D%2012.5)
So, 12.5 of the plates are needed
Substitute for p and q.
pq/3 = 3x4/3 = 12/3
q/2 = 4/2
You can't add fractions when the denominators are different. The common denominator of 2 and 3 would be 6. So multiply 12/3 by 2 and 4/2 by 3.
12/3 x 2 = 24/6
4/2 x 3 = 12/6
24/6 - 12/6 = 12/6
Simplify 12/6 to get 2.
A. 2
X-axis = A (-7,-3) B (-7,-10) C (-1,-3) first they changed to the x-axis which is this. Just so you know on the x-axis the x-axis stay and the y-axis changed which is this.
When they reflect again on the y-axis the verter of A' is
A' (7, 3) = answer
On the y-axis the y- axis doesnt change but the x-axis changed.
Hope this help!