{y|y ≤ 5}. It means the maximum value of the function is 5. It is possible if the vertex of a downward parabola is in the form of (h,5).Answer:
Step-by-step explanation:Answer:
The correct option is 2.
Step-by-step explanation:
The range of a parabola is {y|y ≤ 5}.
It means the maximum value of the function is 5. It is possible if the vertex of a downward parabola is in the form of (h,5).
The vertex form of a parabola is
Where, (h,k) is vertex and a is a constant.
If a<0, then f(x) is a down ward parabola and if a>0, then f(x) is an upward parabola.
In option 1,
It is an upward parabola with vertex (4,5), therefore the range of the function is {y|y ≥ 5}. Option 1 is incorrect.
In option 2,
It is a downward parabola with vertex (4,5), therefore the range of the function is {y|y ≤ 5}. Option 2 is correct.
In option 3,
It is an upward parabola with vertex (5,4), therefore the range of the function is {y|y ≥ 4}. Option 3 is incorrect.
In option 4,
It is a downward parabola with vertex (5,4), therefore the range of the function is {y|y ≤ 4}. Option 4 is incorrect.
hope this helps theirs 3 pictures
9514 1404 393
Answer:
D. y = ±√(25 -x²)
Subtract x² and take the square root to solve for y.
x² +y² = 25 . . . . . . . given
y² = 25 -x² . . . . . . . . subtract x²
y = ±√(25 -x²) . . . . . take the square root
First term (a) =8
Common difference (d)= t2-t1
=12-8
=4
Now, sum of first 31th term (tn31) =n/2{2a+(n-1)d}
= 31/2{2×8+(31-1)4}
=31/2{16+(30×4)
=31/2(16+120)
=31/2×126
=31×63
Similarly use 19 as (n) for the 19th term
