Exam 1: Graphs and Models

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 Evaluate the expression arcsin(12) without using a calculator. \text { Evaluate the expression } \arcsin \left( \frac { 1 } { 2 } \right) \text { without using a calculator. }

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 Solve the following equation for x\text { Solve the following equation for } x \text {. } arccos(10xπ)=12\arccos ( 10 x - \pi ) = \frac { 1 } { 2 }

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horsepower H required to overcome wind drag on a certain automobile is approximated by H(x)=0.002x2+0.005x0.027,10x100H ( x ) = 0.002 x ^ { 2 } + 0.005 x - 0.027,10 \leq x \leq 100 where xx is the speed of the car in miles per hour. Find H(x1.1)H \left( \frac { x } { 1.1 } \right) . Round the numerical values in your answer to five decimal places.

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 Find an equation of the line that passes through the point (11,9) and has the slope \text { Find an equation of the line that passes through the point } ( - 11 , - 9 ) \text { and has the slope } m=92.m = \frac { 9 } { 2 } .

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Estimate the slope of the line from the graph. Estimate the slope of the line from the graph.

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 If a line has slope m=4 and passes through the point (4,8), through which of the \text { If a line has slope } m = - 4 \text { and passes through the point } ( 4,8 ) \text {, through which of the } following points does the line also pass?

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Determine which type of function would be most appropriate to fit the given data. Determine which type of function would be most appropriate to fit the given data.

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The following ordered pairs represent temperatures in degrees Fahrenheit taken each hour from 1:00pm1 : 00 \mathrm { pm } until 5:00 pm. Let TT be temperature, and let tt be time, where t=1t = 1 corresponds to 1:00pm,t=21 : 00 \mathrm { pm } , t = 2 corresponds to 2:00pm2 : 00 \mathrm { pm } , and so on. Plot the data. Visually find a linear model for the data and find its equation. From the visual linear model that you created, determine which of the models that follow appears to best approximate the data. (1:00pm,67.4),(2:00pm,71.6),(3:00pm,73.4),(4:00pm,77.6),(5:00pm,79.4)\left( 1 : 00 \mathrm { pm } , 67.4 ^ { \circ } \right) , \left( 2 : 00 \mathrm { pm } , 71.6 ^ { \circ } \right) , \left( 3 : 00 \mathrm { pm } , 73.4 ^ { \circ } \right) , \left( 4 : 00 \mathrm { pm } , 77.6 ^ { \circ } \right) , \left( 5 : 00 \mathrm { pm } , 79.4 ^ { \circ } \right)

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Write the following expression in algebraic form. tan(arcsec(x8))\tan \left( \operatorname { arcsec } \left( \frac { x } { 8 } \right) \right)

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Match the graph of the function given below with the graph of its inverse function. Match the graph of the function given below with the graph of its inverse function.

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 Use the result, "the line with intercepts (a,0) and (0,b) has the equation \text { Use the result, "the line with intercepts } ( a , 0 ) \text { and } ( 0 , b ) \text { has the equation } xa+yb=1,a0,b0\frac { x } { a } + \frac { y } { b } = 1 , a \neq 0 , b \neq 0 ", to write an equation of the line with xx -intercept: (8,0)( 8,0 ) and yy intercept: (0,7)( 0,7 ) .

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Each ordered pair gives the exposure index x of a carcinogenic substance and the cancer mortality yy per 100,000 people in the population. Use the model y=9.2x+108.4y = 9.2 x + 108.4 to approximate yy if x=7x = 7 . Round your answer to one decimal place. (3.50,150.1),(3.58,133.1),(4.42,132.9),(2.26,116.7),(2.36,140.7),(4.85,165.5) (12.65,210.7),(7.42,181.0),(9.35,213.4)

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all intercepts: y=x2x12y = x ^ { 2 } - x - 12

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Fine the slope of the line passing through the pair of points. (3,6),(0,11)( - 3 , - 6 ) , ( 0 , - 11 )

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for symmetry with respect to each axis and to the origin. x2y2=8x ^ { 2 } y ^ { 2 } = 8

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 Write an equation of the line that passes through the point (54,58) and is parallel to \text { Write an equation of the line that passes through the point } \left( \frac { 5 } { 4 } , \frac { 5 } { 8 } \right) \text { and is parallel to } the line 7x3y=07 x - 3 y = 0 .

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 Evaluate the expression cos(arcsin35) without using a calculator. \text { Evaluate the expression } \cos \left( \arcsin \frac { 3 } { 5 } \right) \text { without using a calculator. }

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 Find an equation of the line that passes through the points (18,7) and (18,23)\text { Find an equation of the line that passes through the points } ( 18 , - 7 ) \text { and } ( - 18,23 ) \text {. }

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Write an equation of the line that passes through the given point and is parallel to the given line. Point Line (3,-4) -2x-5y=9

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Use the Horizontal Line Test to determine whether the following statement is true or false. The function f(x)=319x+3f ( x ) = \frac { 3 } { 19 } x + 3 is one-to-one on its entire domain and therefore has an inverse function.

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