To be blunt, it looks like you just copy/pasted a wall of text from whatever you're working on. Ask an actual question, and format the wall of text so that we don't have to decipher it ourselves. Don't expect people to help solve a problem like that if it looks like you gave no thought in posting it
Both rooms share a common side whose dimension is unknown. Call it x.
Then, the area of both squares have x as common factor.
So, x is the greatest common factor of 104 and 130.
You should know how to calculate the greatest common factor of two integers.
Just find the prime factors and choose the common factors raised to the lowest exponent.
104 = (2^3) (13)
130 = (2) (5)(13)
=> the greatest common factor is 2 * 13 = 26, and that is the greatest possible integer length of the shared wall.
Answer: 26
Step-by-step explanation:
hope this helps, happy learning!!!
The two equations given in the question are
y = - 3x
And
y = 6x - 9
Let us now substitute the value of y from the first equation to the second equation, we get
y = 6x - 9
- 3x = 6x - 9
-3x - 6x = - 9
- 9x = - 9
Multiplying both sides of the equation by -1 we get
9x = 9
x = 9/9
= 1
Now let us put the value of x in the first equation, we get
y = - 3x
= - (3 * 1)
= - 3
When you bisect something, you cut it into two equally sized pieces. (from Latin: "bi" = two, "sect" = cut)
Bisecting an interval creates two smaller intervals each with half the length of the original interval. Some examples:
• bisecting [0, 2] gives the intervals [0, 1] and [1, 2]
• bisecting [-1, 1] gives the intervals [-1, 0] and [0, 1]
• bisecting an arbitrary interval ![[a,b]](https://tex.z-dn.net/?f=%5Ba%2Cb%5D)
gives the intervals ![\left[a,\frac{a+b}2\right]](https://tex.z-dn.net/?f=%5Cleft%5Ba%2C%5Cfrac%7Ba%2Bb%7D2%5Cright%5D)
and ![\left[\frac{a+b}2,b\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cfrac%7Ba%2Bb%7D2%2Cb%5Cright%5D)
