PLEASE HELP ON A TEST
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<u>ANSWER:</u> The correct answer is .
<u>Explanation</u>
The composition of a function of and its inverse will produce the independent variable
.
For instance, let .
The inverse of this function is .
If we compose these two functions, we will obtain;
This means we have to substitute the whole inverse function in to the function itself.
The other way round will also produce the same result.
Thus;
This does not only apply to the given example it applies to all functions and their inverse.
Rearrange the equation,
0x + y - ½z = 0 ==> Equation 1
x + 2y + 5z = 2 ==> Equation 2
3x + 6y - 3z = 9 ==> Equation 3
Put the 3 equation in matrix form,
Find the determinant, D, of the matrix,
By Sarrus' rule,
D = [(0 × 2 × -3) + (1 × 5 × 3) + (-½ × 1 × 6)] - [(-½ × 2 × 3) + (5 × 6 × 0) + (-3 × 1 × 1)]
D = [0 + 15 + (-3)] - [(-3) + 0 + (-3)]
D = 12 + 6
D = 18
By cramer's rule,
x = Determinant of x ÷ D
Let determinant of x = Dx,
To get this determinant, substitude 0, 2 and 9 with the unknown you're finding, in this case, x.
By Sarrus' rule,
Dx = [(0 × 2 × -3) + (1 × 5 × 9) + (-½ × 2 × 6)] - [(-½ × 2 × 9) + (0 × 5 × 6) + (-3 × 1 × 2)]
Dx = [0 + 45 + (-6)] - [-9 + 0 + (-6)]
Dx = 39 + 15
Dx = 54
x = Dx ÷ D
x = 54 ÷ 18
x = 3
By cramer's rule,
y = Determinant of y ÷ D
Let determinant of y = Dy
By Sarrus' rule,
Dy = [(0 × 2 × -3) + (0 × 5 × 3) + (-½ × 1 × 9)] - [(-½ × 2 × 3) + (5 × 9 × 0) + (-3 × 0 × 1)
Dy = [0 + 0 - 4½] - [-3 + 0 + 0]
Dy = -4½ + 3
Dy = -1½
y = Dy ÷ D
y = -1½ ÷ 18
y ≈ 0.0833 (3 significant figure)
Solving z using equation rule,
y = ½z
0.0833 = ½z
z = 0.0833 ÷ ½
z ≈ 0.167 (3 significant figure)
∴ x = 3, y = 0.0833 and z = 0.167.
0x + y - ½z = 0 ==> Equation 1
x + 2y + 5z = 2 ==> Equation 2
3x + 6y - 3z = 9 ==> Equation 3
Put the 3 equation in matrix form,
Find the determinant, D, of the matrix,
By Sarrus' rule,
D = [(0 × 2 × -3) + (1 × 5 × 3) + (-½ × 1 × 6)] - [(-½ × 2 × 3) + (5 × 6 × 0) + (-3 × 1 × 1)]
D = [0 + 15 + (-3)] - [(-3) + 0 + (-3)]
D = 12 + 6
D = 18
By cramer's rule,
x = Determinant of x ÷ D
Let determinant of x = Dx,
To get this determinant, substitude 0, 2 and 9 with the unknown you're finding, in this case, x.
By Sarrus' rule,
Dx = [(0 × 2 × -3) + (1 × 5 × 9) + (-½ × 2 × 6)] - [(-½ × 2 × 9) + (0 × 5 × 6) + (-3 × 1 × 2)]
Dx = [0 + 45 + (-6)] - [-9 + 0 + (-6)]
Dx = 39 + 15
Dx = 54
x = Dx ÷ D
x = 54 ÷ 18
x = 3
By cramer's rule,
y = Determinant of y ÷ D
Let determinant of y = Dy
By Sarrus' rule,
Dy = [(0 × 2 × -3) + (0 × 5 × 3) + (-½ × 1 × 9)] - [(-½ × 2 × 3) + (5 × 9 × 0) + (-3 × 0 × 1)
Dy = [0 + 0 - 4½] - [-3 + 0 + 0]
Dy = -4½ + 3
Dy = -1½
y = Dy ÷ D
y = -1½ ÷ 18
y ≈ 0.0833 (3 significant figure)
Solving z using equation rule,
y = ½z
0.0833 = ½z
z = 0.0833 ÷ ½
z ≈ 0.167 (3 significant figure)
∴ x = 3, y = 0.0833 and z = 0.167.