<span>Thus, the standardized form of a quadratic equation is ax2 + bx + c = 0, where "a" does not ... We said that the graph of y = x2 was a function because it passed the vertical line test. .... Find the vertex of the parabola y = 2x2 - 12x + 7. ... To find the y-value, we would substitute 3 in for x, or find f(3) = 2(3) 2 — 12(3) + 7 = -11.</span>
Answer:
For example, 16/21
Step-by-step explanation:
Are there choices?
There is an infinite number of numbers between any two numbers.
2/3 = 14/21
6/7 = 18/21
One example of a fraction between 2/3 and 6/7 is 16/21
16/21 is between 2/3 and 6/7.
9514 1404 393
Answer:
- maximum: 15∛5 ≈ 25.6496392002
- minimum: 0
Step-by-step explanation:
The minimum will be found at the ends of the interval, where f(t) = 0.
The maximum is found in the middle of the interval, where f'(t) = 0.
![f(t)=\sqrt[3]{t}(20-t)\\\\f'(t)=\dfrac{20-t}{3\sqrt[3]{t^2}}-\sqrt[3]{t}=\sqrt[3]{t}\left(\dfrac{4(5-t)}{3t}\right)](https://tex.z-dn.net/?f=f%28t%29%3D%5Csqrt%5B3%5D%7Bt%7D%2820-t%29%5C%5C%5C%5Cf%27%28t%29%3D%5Cdfrac%7B20-t%7D%7B3%5Csqrt%5B3%5D%7Bt%5E2%7D%7D-%5Csqrt%5B3%5D%7Bt%7D%3D%5Csqrt%5B3%5D%7Bt%7D%5Cleft%28%5Cdfrac%7B4%285-t%29%7D%7B3t%7D%5Cright%29)
This derivative is zero when the numerator is zero, at t=5. The function is a maximum at that point. The value there is ...
f(5) = (∛5)(20-5) = 15∛5
The absolute maximum on the interval is 15∛5 at t=5.
Answer:
About $20.30
Step-by-step explanation:
$466.90 divided by 23.
Answer:
8 hours
Step-by-step explanation:
12/2= 6
He grades 6 exams per hour
48/6= 8
it would take 8 hours to grade 48 exams
Bonus:
it takes him 10 minutes to grade an exam. That doesnt matter for this question but it's good 2 know
