Answer:
x = -1, y = 2 and z = 1
Step-by-step explanation:
The given system of equations are :
2x - y + 3z= -1 ....(1)
x + 2y - 4z = -1 ......(2)
y – 2z = 0 .....(3)
Equation (3) can be written as :
y = 2z
Use y = 2z in equation (2)
x + 2(2z) - 4z = -1
x + 4z - 4z = -1
x = -1
Put the value of x in equation (1) :
-2 -y +3z = -1
-y+3z = 1 ....(4)
Adding equation (3) and (4)
y-2z+(-y+3z)=1
z = 1
Now put z = 1 in equation (4)
-y+3=1
-y = -2
y = 2
Hence, the values of x,y and z are -1, 2 and 1 respectively.
There is a division symbol, division is subtracted. Are you sure you have the problem stated clearly?
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f(x) being even means
f(x) = f(-x)
So the zeros come in positive and negative pairs. If there are an odd number of intercepts like there are here, it's because one of them is x=0 which is its own negation.
Given zero x=6 we know x=-6 is also a zero.
So we know three zeros, and know the other two zeros are a positive and negative pair.
The only choice with (-6,0) and (0,0) is A.
Choice A
Answer:
3.5 min
Step-by-step explanation:
