Answer:
jacky 12 809 n-t
Step-by-step explanation:
Answer:
he said it not me
Answer:
E, 3.
Step-by-step explanation:
Since both f(x) and g[f(x)] are quadratic polynomials, g(x) must also be a linear polynomial.
Let g(x) = Ax + B, where A and B are constants to be determined.
Then we have A[2x² - 3x + 1] + B ≡ x² - (3/2)x + 3.
=> A = 1/2 and B = 5/2.
Hence, f[g(-1)] = f[(1/2)(-1) + (5/2)] = f(2) = 2(2)² - 3(2) + 1 = 3. (E)
line y= 2x - 3
The required point for the given line is (0, -3). The correct option is (D).
A linear equation in two variable has the general form as y = ax + by + c, where a, b and c are integers and a, b ≠ 0.
It can be represented as a straight line on a graph.
The equation of the given line is y = 2x - 3.
In order to find the point lying on it, consider each of the options one by one as follows,
(a) (2 , 3)
Substitute x = 2 and y = 3 in the given equation to obtain LHS and RHS as,
LHS = y
= 7
RHS = 2x - 3
= 2 × 2 - 3
= 1
Since, LHS ≠ RHS, the given point is not the solution.
(b) (-2, -1)
Substitute x = 2 and y = 3 in the given equation to obtain LHS and RHS as,
LHS = y
= 7
RHS = 2x - 3
= 2 × -1 - 3
= -6
Since, LHS ≠ RHS, the given point is not the solution.
(c) (4, 1)
Substitute x = 2 and y = 3 in the given equation to obtain LHS and RHS as,
LHS = y
= 4
RHS = 2x - 3
= 2 × 1 - 3
= -1
Since, LHS ≠ RHS, the given point is not the solution.
(d) (0, -3)
Substitute x = 2 and y = 3 in the given equation to obtain LHS and RHS as,
LHS = y
= -3
RHS = 2x - 3
= 2 × 0 - 3
= -3
Since, LHS = RHS, the given point is the solution.
Hence, the point (0, -3) is the solution of the given equation.
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Answer: 5,
Step-by-step explanation: I got it right, and I hope this helps you! ✨
Answer:
the evaluation of the equation is 5
Step-by-step explanation:
To add fractions, find the LCD and then combine.