We have a three unknown, 4 equation homogeneous system. These always have at least (0,0,0) as a solution. Let's write the equations, one column at a time.
1a + 0b + 0c = 0
-1a + 1b +0c = 0
0a - 1b + c = 0
0a + 0b + -1 c = 0
We could do row reduction but these are easy enough not to bother.
Equation 1 says
a = 0
Equation 4 says
c = 0
Substituting in the two remaining,
-1(0) + 1b + 0c = 0
b = 0
0(0) - 1b + 0 = 0
b = 0
The only 3-tuple satisfying the vector equation is (a,b,c)=(0,0,0)
Answer:
y = (6/11)x + 13/11
Step-by-step explanation:
y = mx + c
m = (-1-5)/(-4-7) = -6/-11 = 6/11
y = (6/11)x + c
When x = 7, y = 5
5 = (6/11)(7) + c
5 = 42/11 + c
c = 5 - 42/11
c = 13/11
y = (6/11)x + 13/11
11y = 6x + 13
The highest number that both 9 and 27 can be multiplied into is 9. 9 can go into 9 one time, and 9 can go into 27 three times.
9•1=9
9•3=27
(9, 18, 27 )=times
I hope I helped.
Answer:
18 is the answer
Step-by-step explanation:
We know this because The factors of 36 are: 1, 2, 3, 4, 6, 9, 12, 18, 36
The factors of 54 are: 1, 2, 3, 6, 9, 18, 27, 54
Answer: -n-4
Step-by-step explanation:
Distribute the - to n and 4
