Answer:
22572000
Step-by-step explanation: I tried and I'm not trolling ok, I did try.
<span>(y=mx+b) or (ax+by=c) hope this helped
</span>
Answer:
B
Step-by-step explanation:
Using the determinant to determine the type of zeros
Given
f(x) = ax² + bx + c ( a ≠ 0 ) ← in standard form, then the discriminant is
Δ = b² - 4ac
• If b² - 4ac > 0 then 2 real and distinct zeros
• If b² - 4ac = 0 then 2 real and equal zeros
• If b² - 4ac < 0 then 2 complex zeros
Given
f(x) = (x - 1)² + 1 ← expand factor and simplify
= x² - 2x + 1 + 1
= x² - 2x + 2 ← in standard form
with a = 1, b = - 2, c = 2, then
b² - 4ac = (- 2)² - (4 × 1 × 2) = 4 - 8 = - 4
Since b² - 4ac < 0 then the zeros are complex
Thus P(x) has no real zeros
Answer:
C. f(x) = 3(2 x)
Step-by-step explanation: