Let the three numbers be x, y, and z.
If the sum of the three numbers is 3, then x+y+z=3
If subtracting the second number from the sum of the first and third numbers gives 9, then x+z-y=9
If subtracting the third number from the sum of the first and second numbers gives -5, then x+y-z=-5
This forms the system of equations:
[1] x+y+z=3
[2] x-y+z=9
[3] x+y-z=-5
First, to find y, let's take do [1]-[2]:
x+y+z=3
-x+y-z=-9
2y=-6
y=-3
Then, to find z, let's do [1]-[3]:
x+y+z=3
-x+-y+z=5
2z=8
z=4
Now that you have y and z, plug them into [1] to find x:
x+y+z=3
x-3+4=3
x=2
So the three numbers are 2,-3, and 4.
X is obviously 1
1x2=2
Entao...(So...)
Please mark as brainliest
Answer:$0.60
Step-by-step explanation:
Divide younr cost by the amount of soda.
Answer:
<em>The last choice is correct</em>
<em />
<em />
Step-by-step explanation:
<u>Least Common Multiple (LCM)</u>
To find the LCM we can follow this procedure:
List the prime factors of each monomial.
Multiply each factor the greatest number of times it occurs in either factor.
We have two monomials:
![12a^4b^2c^5](https://tex.z-dn.net/?f=12a%5E4b%5E2c%5E5)
![40a^3b^7c^1](https://tex.z-dn.net/?f=40a%5E3b%5E7c%5E1)
The prime factors of the first monomial are:
![2^2,3,a^4,b^2,c^5](https://tex.z-dn.net/?f=2%5E2%2C3%2Ca%5E4%2Cb%5E2%2Cc%5E5)
The prime factors of the second monomial are:
![2^3,5,a^3b^7c^1](https://tex.z-dn.net/?f=2%5E3%2C5%2Ca%5E3b%5E7c%5E1)
LCM = Multiply ![2^3*3*5*a^4*b^7*c^5](https://tex.z-dn.net/?f=2%5E3%2A3%2A5%2Aa%5E4%2Ab%5E7%2Ac%5E5)
These are all the factors the greatest number of times they occur.
Operating:
![LCM=8*15*a^4*b^7*c^5](https://tex.z-dn.net/?f=LCM%3D8%2A15%2Aa%5E4%2Ab%5E7%2Ac%5E5)
![\boxed{LCM=120a^4b^7c^5}](https://tex.z-dn.net/?f=%5Cboxed%7BLCM%3D120a%5E4b%5E7c%5E5%7D)
The last choice is correct