Answer:
3/25
Step-by-step explanation:
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Ok so let us label our equations first
1- 3x+2y+z=7
2- 5x+5y+4z=3
3- 3x+2y+3z=1
subtracting equation 3 from equation 1
3x+2y+3z=1
(-). 3x+2y+z= 7
----------------------
2z=-6----->z=-3
since we already chose equation 1 and 3 we must involve equation 2
substituting the value if z in the second equation
5x+5y-12=3
5x+5y=15
choosing either the first or the third
3x+2y-3=7
3x+2y=10
solving the system
5x+5y=15
3x+2y=10
multipling the first equation by 2 and the second equation by 5
10x+10y=30
15x+10y=50
subtracting the two equations
-5x=-20--->x=4 substituting for the value of x
40+10y=30--->10y=-10y--->y=-1
so our soultion is x=4,y=-1,z==-3 or (4,-1,-3)
Answer:
0
Step-by-step explanation:
Assuming the six sided dice are numbered 1 through six
The max each could roll is 6
6+6 = 12
They cannot reach a sum of 13
Answer:
I don't really know if this is a trick question or not but, is 6363 minutes the answer?
Step-by-step explanation:
We have that
<span>[6x+5]=1+2*(3x+2)
[6x+5]=1+2*3x+2 --------> is not correct ------->1+ 2*(3x+2)=1+2*3x+2*2
then
</span>[6x+5]=1+6x+4---------------> [6x+5]=[6x+5]
<span>this equation is an identity, all real numbers are solutions.</span>
