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Find the slope, if it exists, of the line containing the pair of points (-2, -10) and (-13, -12)
The slope m = ? Simplify your answer type an integer or a fraction, type N if the slope is undefined.
Slope = (y2 – y1)/(x2 – x1) = (-12 + 10)/(-13 + 2) = -2/-11 = 2/11
Use the intercepts to graph the equation x – 5 = y
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y=x–5
When y = 0, x = 5  The x-intercept is (5, 0)
When x = 0, y = -5  The y-intercept is (0, -5)
Graph:
Find an equation of the line having the given slope and containing the given point
m = 6, (9,2) The equation of the line is y = ? Simplify your answer, use integers or fractions for any numbers
in the expression.
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The line is y – y1 = m(x – x1)
 y – 2 = 6(x – 9)  y = 6x - 52
Find an equation of the line containing the given pair of points (-5,-7) and (-2,-9)
The equation of the line is y = ? Simplify your answer, use integers or fractions for any numbers in the
expression.
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Slope of the line = (y2 – y1)/(x2 – x1) = (-9 + 7)/(-2 + 5) = -2/3
The line is y – y1 = m(x – x1)
y + 7 = (-2/3)(x + 5)  y = (-2/3)x – (31/3)
Media services charges $40 for a phone and $30/month for its economy plan, find a model that determines the
total cost, C(t), of operating a media services phone for t months. C(t)= ?
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The model is C(t) = 30t + 40, where C(t) is the cost in dollars.
The table lists data regarding the average salaries of several professional athletes in the years 1991 and 2001.
a) Use the data points to find a linear function that fits the data.
b) Use the function to predict the average salary in 2005 and 2010
Year
Average salary
1991
$253,000
2001
$1,360,000
A linear function that fits the data is S(x) = ? (Let x = the number of years since 1990, and let S = the average
salary x years from 1990.)
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(a) Let the linear function be y = mx + b, where x = the number of years after 1991
[Thus, for 1991, x = 0 and for 2001, x = 10]
253000 = m(0) + b  b = 253000
1360000 = m(10) + b
1360000 = 10m + 253000
10m = 1360000 – 253000 = 1107000
m = 110700
The function is S(x) = 110700x + 253000
In 1920, the record for a certain race was 45.2 sec. In 1940, it was 44.8 sec.,
Let R(t) = the record in the race and t = the number of years since 1920.
a) Find a linear function that fits the data.
b) Use the function in (a) to predict the record in 2003 and in 2006.
c) Find the year when the record will be 43.3 sec
Find a linear function that fits the data R(t) = ? (Round to the nearest hundredth)
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(a) Let the linear function be R(t) = mt + b, where t = the number of years after 1920
[Thus, for 1920, t = 0 and for 1940, t = 20]
45.2 = m(0) + b  b = 45.2
44.8 = m(20) + b
44.8 = 20m + 45.2
20m = 44.8 – 45.2 = -0.4
m = -0.4/20 = -0.02
The function is R(t) = -0.02t + 45.20
(b) t = 83; R(83) = -0.02(83) + 45.20 = 43.54 s and t = 86; R(86) = -0.02(86) + 45.20 = 43.48 s
(c) 43.3 = -0.02t + 45.20  t = (43.3 – 45.20)/-0.02 = 95 years after 1920, which is year 2015.
R(t) = -0.02t + 45.20
In 1991, the life expectancy of males in a certain country was 68.9 years, In 1998 it was 72.1 years. Let E
represent the life expectancy in year t and let t represent the number of years since 1991. The linear function
E(t) that fits the data is E(t) = ?t + ? (Round to the nearest tenth)
Let E(t) = at + b
By data, E(0) = 68.9 = 0 + b  b = 68.9
E(7) = 72.1 = a(7) + b = 7a + 68.9
a = (72.1 – 68.9)/7 = 0.457
The linear function is E(t) = 0.457t + 68.9.