Answer:
Part a) to find the maximum height of the snowball, you have to differentiate the function. Therefore you get ----> dh/dt= -32x-8 . Now equate this to zero and solve for x. x= (-1)/4 now sub this value in to find h(x) [note: i'm talking about the original function] . I got h max = h((-1)/4) = 9 which is the max height.
Step-by-step explanation:
I'm not too sure about the other questions. Sorry
The answer is c Dhsbnsndgdh
Answer:
Step-by-step explanation:
xy = 300
x + y = 50 Solve for y
y = 50 - x substitute into xy = 300
x(50 - x) = 300 Remove the brackets.
50x - x^2 = 300 Bring the left to the right.
0 = x^2 - 50x + 300
This is a quadratic. It will have 2 solutions.
a=1
b = - 50
c = 300
Put these into the quadratic equation.
It turns out that x has two values -- both plus
x1 = 42.03
x2 = 6.97
x1 + y1 = 50
42.03 + y = 50
y = 50 - 42.03
y = 7,97
(42.03 , 7.97)
====================
x2 + y2 = 50
6.97 + y2 = 50
y2 = 50 - 6.97
y2 = 43.03
(6.97 , 43.03)
Answer:

Step-by-step explanation:
see the attached figure with letters to better understand the problem
step 1
Find the measure of arc BC
we know that
----> by complete circle
substitute the given values



step 2
Find the measure of angle 2
we know that
The inscribed angle is half that of the arc it comprises.
so
![m\angle 2=\frac{1}{2}[arc\ BC]](https://tex.z-dn.net/?f=m%5Cangle%202%3D%5Cfrac%7B1%7D%7B2%7D%5Barc%5C%20BC%5D)
substitute
![m\angle 2=\frac{1}{2}[116^o]=58^o](https://tex.z-dn.net/?f=m%5Cangle%202%3D%5Cfrac%7B1%7D%7B2%7D%5B116%5Eo%5D%3D58%5Eo)