Answer: D) The base is e^(-1)
We use the rule that x^(-k) = 1/(x^k). That allows us to say e^(-1) = 1/(e^1) = 1/e
The 1/e is the base of the exponential (1/e)^x
In general, the exponential b^x has base b.
Answer:
-3g
Step-by-step explanation:
It would probably be 56.55
Answer: x=-1, y=2, z=-3
Step-by-step explanation:
[1] 2x + 3y - 4z = 16
[2] -x + 2y - z = 8
[3] 2x - y - 2z = 2
Solve by Substitution :
Solve equation [3] for the variable y
[3] y = 2x - 2z - 2
Plug this in for variable y in equation [1]
2x + 3•(2x-2z-2) - 4z = 16
8x - 10z = 22
Plug this in for variable y in equation [2]
-x + 2•(2x-2z-2) - z = 8
3x - 5z = 12
Solve equation [2] for the variable x
3x = 5z + 12
x = 5z/3 + 4
Plug this in for variable x in equation [1]
8•(5z/3+4) - 10z = 22
10z/3 = -10
10z = -30
Solve equation for the variable z
10z = - 30
z = - 3
By now we know this much :
x = 5z/3+4
y = 2x-2z-2
z = -3
Use the z value to solve for x
x = (5/3)(-3)+4 = -1
Use the x and z values to solve for y
y = 2(-1)-2(-3)-2 = 2
