If $x$, $y$, and $z$ are positive integers such that $6xyz+30xy+21xz+2yz+105x+10y+7z=812$, find $x+y+z$.
1 answer:
Answer:
10
Step-by-step explanation:
When we simplify we get
Then we continue to factor to get:
We then see that we can factor into
we then do the prime factorization of 847, which i think is, . we have to find the numbers that multiply to 847 and then plug them into z+5, 3x=1,2y+7.
It has to be a positive, non-negative integer, right?
We also see that 3x+1=11 so we see that x=10/3 (which wont work).
So 3x+1=7, so x=2.
So 11 has to be in another term. It has to be in 2y+7=11 so y=2
for the last term we get z+5=11 so z=6
2+2+6=10
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Answer:
The answer is in the picture. I hope it is helpful.
Answer:
( -1 , 26)
Step-by-step explanation:
So since you need to find the other endpoint, you would follow these steps:
1.)
2.) 8 = 9 + x ( you just multiplied the 2 to the 4 to get 8)
3.) -1 = x (just solve it like a regular equation, so just subtract 9 on both sides to get rid of it and that leaves you with -1 = x)
You took the x values of both points and put them in the equation.
And its the same for y
1.)
2.) 16 = -10 + y
3.) 26 = y (you added the 10 on both sides because the 10 was negative and that took the 10 out and so it left you with 26 = y)