Determine the 28th term of the sequence -242, -251, -260
1 answer:
We can figure this out using the explicit formula.
n represents the term we are looking for.f(1) represents the first term in the sequence, which in this case, is -242.d represents the common difference, which in this case, is -9.
f(n) = -242 + -9(n - 1)
f(n) = -242 - 9n + 9
f(n) = -233 - 9n
Now, we can input 28 for n and solve.
f(28) = -233 - 9(28)
f(28) = -233 - 252
f(28) = -485
The 28th term of the sequence is -485.
n represents the term we are looking for.f(1) represents the first term in the sequence, which in this case, is -242.d represents the common difference, which in this case, is -9.
f(n) = -242 + -9(n - 1)
f(n) = -242 - 9n + 9
f(n) = -233 - 9n
Now, we can input 28 for n and solve.
f(28) = -233 - 9(28)
f(28) = -233 - 252
f(28) = -485
The 28th term of the sequence is -485.
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<u>Given</u>:
Given that the triangle ABC has vertices A(-2,3), B(0,3) and C(-1,-1).
We need to determine the coordinates of the image after a reflection over the x - axis.
Let A'B'C' denote the coordinates of the triangle after a reflection over the x - axis.
<u>Coordinates of the point A':</u>
The general rule to reflect the coordinate across the x - axis is given by
Substituting the point A(-2,3), we get;
Thus, the coordinates of the point A' is (-2,-3)
<u>Coordinates of the point B':</u>
The general rule to reflect the coordinate across the x - axis is given by
Substituting the point B(0,3), we get;
Thus, the coordinates of the point B' is (0,-3)
<u>Coordinates of the point C':</u>
The general rule to reflect the coordinate across the x - axis is given by
Substituting the point C(-1,-1), we get;
Thus, the coordinates of the point C' is (-1,1)
Hence, the coordinates of the image after a reflection over the x - axis is A'(-2,-3), B(0,-3) and C(-1,1)