1.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-2);y=-x-2
D.y=-x
2.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (2,-1);y=-3/2x-6
C.y=-3/2x+2
3.Write an equation in slope- intercept form of the line that passes through the given point and is parallel to the graph of the given equation. (4,2);x=-3
D.y=4
4.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (-2,3);y=1/2x-1
B.y=-2x-1
5.Write an equation in slope- intercept form of the line that passes through the given point and is perpendicular to the graph of the given equation. (5,0);y+1=2(x-3)
D.y=-1/2x+5/2
Answer:
ok what is it
Explanation:
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Answer:
ΔT = ΔT0 e^-K T
As I understand Newton's Law of Cooling
ΔT at any time is the difference between the temperature and the surroundings
Originally ΔT0 = 95 - 22 difference between 95 and room temperature
65 - 22 = 33 = 73 e^-KT where t is time to cool to 65 deg
ln (33/73) = -KT K = .794 / 5 = .159 where 5 is time to cool to 65 deg
40 - 22 = 73 e^-.159 T where t is time to cool to 40 deg
18 = 73 e^-.159 T
ln (18 / 73) = -.159 T
T = 8.8 min
It would take 8.8 min for the object to cool to 40 deg C
Suppose the object cooled from 95 to 90 deg, then
ln 68 / 73 = -.159 T and T = .45 min
