Answer:
your answers obviously 210
Answer:
the pay one
Step-by-step explanation:
i thinks sorry if i get it wrong
Answer:
(d) f(x) = -x²
Step-by-step explanation:
For the vertex of the quadratic function to be at the origin, both the x-term and the constant must be zero. That is, the function must be of the form ...
f(x) = a(x -h)² +k . . . . . . . . . . vertex form; vertex at (h, k)
f(x) = a(x -0)² +0 = ax² . . . . . vertex at the origin, (h, k) = (0, 0)
Of the offered answer choices, the only one with a vertex at the origin is ...
f(x) = -x² . . . . . a=-1
Manipulate the first equation to
y = 22-7z
Then substitute in the second to find z
8(22-7z) + 7z = 127
176-56z+7z = 127
-49z = -49
z = 1
So y = 22-7(1) = 22-7 = 15
y = 15; z = 1
