Answer:
1.
2.543.6
Step-by-step explanation:
We are given that
y(0)=200
Let y be the number of bacteria at any time
=Number of bacteria per unit time
Where k=Proportionality constant
2.
,y'(0)=100
Integrating on both sides then, we get
We have y(0)=200
Substitute the values then , we get
Substitute the value of C then we get
Differentiate w.r.t
Substitute the given condition then, we get
Substitute t=2
Then, we get 
e=2.718
Hence, the number of bacteria after 2 hours=543.6
I can’t really graph on this but place a dot on the four on the y axis and go up one and over two and place a dot there. just keep repeating it till u get to the end of the graph :)
The answer is B because if you get 5 on the first test then you will have 10 already. if you get at least 1 point on the second then you will have 11 or higher. since the 1st is doubled then the shaded region will retreat by 2 squares for every point less then the maximum you get on the first test.
<span>let:
X = the distance of the bottom of the ladder from the wall at any time
dX/dt = rate of travel of the bottom of the ladder = 1.1 ft/sec
A = the angle of the ladder with the ground at anytime
dA/dt = rate of change of the angle in radians per second
X = 10 cos A
dX/dt= -10 sin A dA/dt = 1.1
dA/dt = -1.1/(10 sinA)
When X = 6; cosA = 6/10; sinA = 8/10
Therefore:
dA/dt = -1.1/(10 x 0.8) = -0.1375 radiant per second. </span>
