Put x=5 in above function. J(x)= 39x. J(5) = 39x5 = 195.
Answer:
domain: {x | x is a real number}
range: {y l y> -8}
Step-by-step explanation:
f(x) = 4x² – 8 is a parabola, a U shape.
Since the stretch factor, 4, is positive, it opens up, there it will have a minimum value, the lowest point in the parabola.
y > -8 because the minimum is -8.
Parabolas do not have restricted "x" values. "4" does not restrict x because it is the stretch factor, which determines how wide the parabola is.
Quadratic standard form:
f(x) = ax² + bx + c
"a" represents how wide the graph is. If it's negative it opens down, if it's positive it opens up.
"b", if written, tells you it is not centred on the y-axis. It is not written, so the vertex is on the y-axis.
"c" is the y-intercept. In this case, since b = 0, it is also the minimum value.
Answer:
6 5/36
Step-by-step explanation:
MARK ME BRAINLIEST?
Answer:
its the 3 one
Step-by-step explanation:
