Answer:
The approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.
Step-by-step explanation:
The formula for D, the difference in temperature between the pan and the room after t minutes is:

Compute the approximate difference in temperature between the pan and the room after 9 minutes as follows:


Then the approximate temperature of the pan after it has been away from the heat for 9 minutes is:
D = P - R
206.59 = P - 69
P = 206.59 + 69
P = 275.59°F
Thus, the approximate temperature of the pan after it has been away from the heat for 9 minutes is 275.59°F.
Answer:
77
Step-by-step explanation:
The range of the data is largest number minus smallest number.
98-21
77
Given :
A graph with a straight line on it.
To Find :
The equation of the line.
Solution :
From the given graph we can see that the line passes through point ( 0,6 ) and ( 4,10 ) .
Also, equation of line passes through two points
and
is :

Putting all given values, we get :

Therefore, the equation of line is y - x = 6 .
Answer:
15
Step-by-step explanation:
Applying,
The angle bisector theorem of triangle
From the diagram,
Since ΔAMT is an issoceless triangle,
Then,
Line OA divides Line MT into two equal parts.
Therefore,
Line MO = Line OT.............. Equation 1
From the diagram,
Line MO = 4x-1, Line OT = 3x+3
Substitute into equation 1
4x-1 = 3x+3
Collect like terms
4x-3x = 3+1
x = 4.
Therefore,
OT = 3(4)+3
OT = 12+3
OT = 15
Set



so that the volume element is

The integral is then


and so evaluates to 0.