Answer: ![sds\\ \\ x^{2} \geq \int\limits^a_b {x} \, dx \lim_{n \to \infty} a_n \geq \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \pi \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \lim_{n \to \infty} a_n \int\limits^a_b {x} \, dx \left \{ {{y=2} \atop {x=2}} \right. x^{2} \lim_{n \to \infty} a_n \pi \neq \sqrt{x} \neq](https://tex.z-dn.net/?f=sds%5C%5C%20%5C%5C%20x%5E%7B2%7D%20%5Cgeq%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%5Cgeq%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%5Cpi%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%262%263%5C%5C4%265%266%5C%5C7%268%269%5Cend%7Barray%7D%5Cright%5D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%5Cint%5Climits%5Ea_b%20%7Bx%7D%20%5C%2C%20dx%20%5Cleft%20%5C%7B%20%7B%7By%3D2%7D%20%5Catop%20%7Bx%3D2%7D%7D%20%5Cright.%20x%5E%7B2%7D%20%20%5Clim_%7Bn%20%5Cto%20%5Cinfty%7D%20a_n%20%5Cpi%20%5Cneq%20%5Csqrt%7Bx%7D%20%5Cneq)
Step-by-step explanation:i need the think points
To find the maximum or minimum value of a function, we can find the derivative of the function, set it equal to 0, and solve for the critical points.
H'(t) = -32t + 64
Now find the critical numbers:
-32t + 64 = 0
-32t = -64
t = 2 seconds
Since H(t) has a negative leading coefficient, we know that it opens downward. This means that the critical point is a maximum value rather than a minimum. If we weren't sure, we could check by plugging in a value for t slightly less and slighter greater than t=2 into H'(t):
H'(1) = 32
H'(3) = -32
As you can see, the rate of change of the object's height goes from increasing to decreasing, meaning the critical point at t=2 is a maximum.
To find the height, plug t=2 into H(t):
H(2) = -16(2)^2 +64(2) + 30 = 94
The answer is 94 ft at 2 sec.
Mode = number which appears the most
The mode for your question is 5
I believe it is b and hmu?
