Answer:
The highest altitude that the object reaches is 576 feet.
Step-by-step explanation:
The maximum altitude reached by the object can be found by using the first and second derivatives of the given function. (First and Second Derivative Tests). Let be
, the first and second derivatives are, respectively:
First Derivative

Second Derivative

Then, the First and Second Derivative Test can be performed as follows. Let equalize the first derivative to zero and solve the resultant expression:


(Critical value)
The second derivative of the second-order polynomial presented above is a constant function and a negative number, which means that critical values leads to an absolute maximum, that is, the highest altitude reached by the object. Then, let is evaluate the function at the critical value:


The highest altitude that the object reaches is 576 feet.
Q55.
UTA+ATS=UTS
x+18+120=12x+17
x+138=12x+17
12x-x=138-17
11x=121
x=121/11
x=11
mUTA=x+18
put the value of x
11+18
mUTA=29
Q57.
AQP+RQA=RQP
9x+2+75=1+28x
9x+77=1+28x
28x-9x=77-1
19x=76
x=76/19
x=4
mRQP=1+28x
put the value of x
1+28(4)
1+112
mRQP=113
2.5=2 dozen cookies
x=15 dozen cookies
2/5 / 2= 1.25 (unit rate)
15*1.25=18.75, or 18 and 3/4 cups of flour.
Hope I helped, and if I did feel free to vote brainliest!
Answer: His class has approximately 150 textbooks in total
Explanation: Because it is given that each student carries about 5 textbooks, with 30 students in the class, the equation would be 5 • 30 = 150. Some students may have more, while others may have less, so the average is given as 5 in order to simplify the equation.