1x68
2x34
4x17
Hope that helps
Answer and Explanation:
A function is said to be increasing, if the derivative of function is f’(x) > 0 on each point. A function is said to be decreasing if f”(x) < 0.
Let y = v (z) be differentiable on the interval (a, b). If two points z1 and z2 belongs to the interval (a, b) such that z1 < z2, then v (z1) ≤ v (z2), the function is increasing in this interval.
Similarly, the function y = v(z) is said to be decreasing, when it is differentiable on the interval (a , b).
Two points z1 and z2 Є (a, b) such that z1 > z2, then v (z1) ≥ v(z2). The function is decreasing on this interval.
The function y = v (z)
The derivative of function Y’ = v’(z) is positive, then the function is increasing.
The function y = v (z)
The derivative of function y’ is negative, then the function is decreasing.
4 multiple choice questions
3 answers per question
4 × 3 = 12
<em>12 possible answers total</em>
12 - 4 = 8
<em>8 incorrect answers
</em><em>4 correct answers
</em><em>
</em>3 × 3 × 3 × 3 = 81
<em>
</em><em>81 possible ways they can answer
</em><em>
</em>Only one of those ways is 100% correct.
<em>
</em>81 - 1 = 80
80 ways
Answer:
B
Step-by-step explanation:
Given a quadratic function in standard form
y = ax² + bx + c ( a ≠ 0 )
• If a > 0 then the graph opens upwards
• If a < 0 then the graph opens downwards
y = x² + 5x + 6 ← is in standard form
with a = 1 > 0
The graph therefore opens upward
Answer: yall i just need points i’m sorry
Step-by-step explanation:
