When we have consecutive values, f(10), f(11), f(12), f(13), f(14), we can make a difference table to determine the degree of f as a polynomial. A quadratic will have a constant second difference:
x 10 11 12 13 14
f(x) 50 71 94 119 146
1st diff 21 23 25 27
2nd diff 2 2 2
We got a constant second difference, so f is a polynomial of degree two.
Answer: This function is quadratic
Answer:
±5i
Step-by-step explanation:
sqrt(-25)
We know that sqrt(ab) = sqrt(a) sqrt(b)
sqrt(-1) sqrt(25)
±i 5
±5i
A: {-4, -2 -1, 2}<br>
B: {-4, 4}<br>
C: {-3, -2, 0, 2}<br>
D: {-4, -2, -1, 0, -2}
Temka [501]
Answer:this is soooo compli
Step-by-step explanation:
Answer:
5, 1, 5, 1.
Step-by-step explanation:
take it one step at a time.
for the first one f(g(1))
first do whats inside of the parenthesis.
> g(1), go to g(x) graph, go to x= 1 and your y-value is your new value (3)
> your new equation is now f(3)
→ go to f(x) graph, go to x= 3, your y-value is 5. Your final answer is 5.
Answer:
Step-by-step explanation:49
