A pan is heated to 393°F, then removed from the heat and allowed to cool in a kitchen where the room temperature is a constant 6
1 answer:
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.
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Answer:
Because the
The integral converges to
Step-by-step explanation:
For this case we want to find the following integral:
And we can solve the integral on this way:
And if we evaluate the integral using the fundamental theorem of calculus we got:
Because the
The integral converges to
Answer:
FD≈25.94.. rounded = 26
Step-by-step explanation:
FD²=12²+(4x+11)²
FD²=144+16x²+88x+121
FD²=265+16x²+88x
also
FD²=12²+(13x-16)²
FD²=144+169x²-416x+256
FD²=400+169x²-416x
thus
265+16x²+88x = 400+169x²-416x
16x²-169x²+88x+416x+265-400 = 0
-153x²+504x-135 = 0
we will solve this quadratic equation by suing the quadratic formula to find x
x=(-504±sqrt(504²-4(-153)(-135)))/2(-153)
x=(-504±)/2(-153)
x=(-504±)/-306
x=(-504±)/-306
x=(-504±414)/-306
x=(-504+414)/-306 and x=(-504-414)/-306
x=-90/-306 and x=-918/-306
x= 5/17 , 3
substituting x by the roots we found
check for 5/17:
4x+11 = 4×(5/17)+11 = (20/17)+11 = (20+187)/17 = 207/17 ≈ 12.17..
13x-16 = 13×(5/17)-16 = (65/17)-16 = (65-272)/17 = -207/17 ≈ -12.17..
check for 3:
4x+11 = 4×3+11 = 12+11 = 23
13x-16 = 13×3-16 = 23
thus 3 is the right root
therfore
ED=23 and CD=23
FD²=FE²+ED² or FD²=FC²+CD²
FD²=12²+23²
FD²=144+529
FD²=673
FD=√673
FD≈25.94.. rounded = 26