The problem is : on january 22, 1943, the temp. T in degrees fahrenheit, in spear fish, south dakota, could be approximated by T= -2.15M + 54, where M is the number of minutes, since 9 a.m that morning.
Find the temp. @ 9:01am, 9:08am, and 9:20am
the ploblem is a slope problem i have to put the answers on a graph
thanks alot, matt
2006-11-08 12:44:56 · 5 answers · asked by matt_sorrentino 1 in Science & Mathematics ➔ Mathematics
Just plug in values of M = 1, M = 8, M = 20...
M = 1 (1 minute past = 9:01am):
T = -2.15(1) + 54
T = 54 - 2.15
T = 51.85
M = 8 (8 minutes past = 9:08am)
T = -2.15(8) + 54
T = 54 - 17.2
T = 36.8
M = 20 (20 minutes past = 9:20am)
T = -2.15(20) + 54
T = 54 - 43
T = 11
So graph the points (1, 51.85), (8, 36.8) and (20, 11)
It's a decreasing slope (m = -2.15) and the y-intercept is 54.
2006-11-08 12:47:09 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
T=-2.15M+54
9:01 M=1 T=-2.15+54=51.85 F
9:08 M=8 T=-2.15*8+54= 36.8 F
9:20 M=20 T=-21.5*20+54=11
Glad I wasn't there!
2006-11-08 20:55:53 · answer #2 · answered by yupchagee 7 · 0⤊ 0⤋
Read the question. M is the number of minutes since 9.
Then your first problem is to figure out what is M?
for the 9:08 example -
Subtact the original time from the new time (9:08 - 9:00) and you have M is equal 8.
Substitute that value into the equation -
T = -2.15 * 8 + 54.
Then solve for T.
Process is the same for other times.
2006-11-08 20:50:30 · answer #3 · answered by Trailcook 4 · 0⤊ 0⤋
Time along the bottom of the graph (x axis), Temp on the vertical (y axis). Plot T for 1 minute, 8 minutes and 20 minutes. 51.85, 36.8 and 11 degrees. Getting cold fast isn't it.
2006-11-08 20:52:41 · answer #4 · answered by Gerry R 2 · 0⤊ 0⤋
??????what grade u in??????
2006-11-08 20:48:17 · answer #5 · answered by xoluvergrlox 1 · 0⤊ 0⤋