Answer:
Read below
Step-by-step explanation:
I don't understand the answer choices, but supplementary angles add up to 180°. It looks like the third option is correct.
For question number 1:The plot H = H(t) is the parabola and it reaches its maximum in the moment when exactly at midpoint between the roots t = 0 and t = 23. At that moment t = 23/2 or 11.5 seconds.
For question number 2:To find the maximal height, just simply substitute t = 11.5 into the quadratic equation. The answer would be 22.9.
For question number 3:H(t) = 0, or, which is the same as -16t^2 + 368t = 0.Factor the left side to get -16*t*(t - 23) = 0.t = 0, relates to the very start of the process, when the ash started its way up.The other root is t = 23 seconds, and it is precisely the time moment when the bit of ash will go back to the ground.
Answer:
Reduced fraction: 14/25
Step-by-step explanation:
To write 0.56 in terms of fractions we need to multiply and divide the number by 100:
(0.56 * 100)/100 = 56/100. Then we simplify the fraction by dividing the numerator and denominator by the common factor 2 as many times as possible:
56/100 = 28/50 = 14/25
The fraction 14/25 cannot be more simplified
![y=x^5-3\\ y'=5x^4\\\\ 5x^4=0\\ x=0\\ 0\in [-2,1]\\\\ y''=20x^3\\\\ y''(0)=20\cdot0^3=0](https://tex.z-dn.net/?f=y%3Dx%5E5-3%5C%5C%20y%27%3D5x%5E4%5C%5C%5C%5C%205x%5E4%3D0%5C%5C%20x%3D0%5C%5C%200%5Cin%20%5B-2%2C1%5D%5C%5C%5C%5C%20y%27%27%3D20x%5E3%5C%5C%5C%5C%0Ay%27%27%280%29%3D20%5Ccdot0%5E3%3D0)
The value of the second derivative for
is neither positive nor negative, so you can't tell whether this point is a minimum or a maximum. You need to check the values of the first derivative around the point.
But the value of
is always positive for
. That means at
there's neither minimum nor maximum.
The maximum must be then at either of the endpoints of the interval
![[-2,1]](https://tex.z-dn.net/?f=%5B-2%2C1%5D)
.
The function
is increasing in its entire domain, so the maximum value is at the right endpoint of the interval.
