Let C = pounds of cheap tea (the one that sells for $2.60/lb), and Let E = pounds of expensive tea (the one that sells for $2.85/lb).

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1 Chapter Quiz Part - Solutions to Most-Misse Problems 5. A merchant blens tea that sells for $.85 per poun with tea that sells for $.6 per poun to prouce 9 lb of a mixture that sells for $.75 per poun. How many pouns of each type of tea oes the merchant use in the blen? Let C = pouns of cheap tea (the one that sells for $.6/lb), an Let E = pouns of expensive tea (the one that sells for $.85/lb). Then the info given in the problem suggests the linear system: E + C = 9.85E +.6C =.75( 9 ) = 47.5 One can solve these two equations with two unknowns in various ways (substitution, etc). The aition metho works like this (I notice that multiplying the secon equation by woul eliminate all the ecimals; then I multiplie the top equation by 5 to cancel the C s ): ( 5) E + C = 9 5E 5 C = = + =.85 E.6 C E 5 C Aing these two equations: 5E = 7 E = = 54, 5 so we nee 54 pouns of the tea that sells for $.85 per poun, an 9 54 = 6 pouns of the tea that sells for $.6 per poun.. Solve 4. Express the solution using interval notation an graph the solution set. x x ( ) ( x ) x( x ) x( x ) ( x ) ( ) ( ) ( x ) 4 4 x x x x x x x x x x x 4 ( ) x( x ) ( + x)( x) x( x ) Next we make our little + chart: x 4x+ 4 x + x 4 x x x

2 + x x Q = Notice that we are looking at where x x, i.e., we are looking in the above chart for where Q is + an where Q is. Thus the solution set for our nonlinear inequality Q, is [,) (, ].. Solve the inequality 5x + 4<9. Express the solution using interval notation. 5x+ 4 < 9 5x< 5 x< Thus the solution is all x in the interval (, ). 4. In the vicinity of a bonfire, the temperature T in at a istance of x meters from the center of the fire was given by T = 765. At what range of istances from the fire's center was the temperature less than C? x We ll solve the inequality, 5 >. Notice that the enominator is ALWAYS positive. So we can x + multiply both sies by this enominator, thus obtaining 5( x + ) > 765. Diviing both sies by 5 yiels: x + > 55, an then subtracting from both sies results in x >5. Taking the square root of both sies, realizing that x must be positive, we get: x > 5. Points more than 5 meters away from the fire will be less than C. 5. Solve the inequality 4x 6. Express the solution using interval notation. 4x 6 4x 6 4x 6 x This solution set is the interval [ ) 9,.

3 x 7. Solve the nonlinear inequality > x. Express the solution using interval notation an graph the solution set. x + ( x + ) ( x + ) ( x ) x x x x x x x+ > x x> > > x+ x+ x+ x+ Next we make our little + chart: x x x x x x + > > > x + x + x + Notice that we are looking at where ( x ) x + Q = > x +, i.e., we are looking in the above chart for where Q is +. Thus the solution set for our nonlinear inequality Q, is,,. ( ) 8. Solve the inequality 5 x + + 9>8. Express the answer using interval notation. 5 x+ + 9 > 8 5 x+ > Notice that the left sie of the equal sign is positive (or zero), which thus HAS to be larger than negative one. Thus, any real value for x will result in the inequality being true. The solution set is thus all real numbers, which in interval notation is written as (, ) 9. Express the phrase All real numbers x at least units from 4 as an inequality involving an absolute value. We seek all real numbers x that are at a istance of or more units from 4, or in other wors, the istance from x to 4 is greater than or equal to : x 4 ( Remember that the istance from A to B is given by A - B. )

4 . Determine the values of the variable for which the expression x x is efine as a real number. To be a real number, the enominator must be positive (that ½ power means the same as square root ): Making our little +- chart: < x x ( x 4)( x 5) < + = Prouct x x 5 + Prouct So x x is efine as a real number whenever that prouct is positive, on the set: ( ) (, 4 5, ). Fin a point on the y-axis that is equiistant from the points (, 7) an (5, ). Points on the y-axis are all of the form (, y), an so we seek a y such that The istance from (, 7) to (, y), is equal to the istance from (5, ) to (, y). It s easier to square both sies to eliminate the square roots ; ( ) + ( 7 y) = ( 5 ) + ( y) y+ y = 5+ + y+ y 5 + 4y = 6 + y y = 4 y = 6. Fin the area of the region that lies outsie of the circle but insie of the circle to two ecimal places. The first equation can be written x + y =, the circle centere at the origin with raius.

5 Next lets fin the stanar form equation for the secon circle: x y y + + = 48+ x x ( y ) + = 49 ( y ) + = 7 So this is the circle centere at the point (, ) with raius 7. Notice that the first smaller circle is entirely within the larger circle, so the area we want is simply: circle"area of big circle" "area of small circle" = π 7 π = 49π 9π = 4π Rouning off to the hunreth place, we get: 4π A city has streets that run north an south, an avenues that run east an west, all equally space. Streets an avenues are numbere sequentially, as shown in the figure. Fin the walking istance an the straight-line istance between the corner of 4th St. an n Ave. an the corner of 9th St. an 4th Ave. In other wors, fin the istance between the point (4, ) an the point (9, 4): Distance = = 5 + = = 69 =. Determine the correct equation for the line passing through the point (, 7) an the point (, 4). First we fin the slope: y y m x x = = = 4 7 So our line is of the form, y = mx + b = x + b, an we nee to fin b. But we know the line goes through the point (, 7), which means when x= then we have y=7. Substituting into our last equation, we can solve for b: 7 y = x+ b 7 = + b 7 + = b b= 7 Thus, in slope-intercept form our line s equation is y = x+. To get this line into same form as the answer selections given, multiply both sies by, an get everything on the same sie of the equal sign: y = x+ 7 x+ y 7 =

6 . Determine the correct equation for the line with an x-intercept of an y-intercept 8. Our line goes through the points (-, ) an (, 8). The slope is thus: y y m x x = = = = We are tol the y-intercept is 8. So, the slope-intercept equation for this line is: y = mx+ b y = 8x Determine the equation that expresses A is proportional to G an inversely proportional to z. Symbols a, b, an c are constants. kg A = is an equation expressing this proportionality, with k as the constant of proportionality. z This may also be written as G A= k. If we replace the constant k with one of the constants z given in the problem, it becomes: G A= c z 7. Express the statement as a formula: s is inversely proportional to the square root of t. Use the information that if s = then t = 64 to fin the constant of proportionality. k k k s = = = 6 = k t The resistance R of a wire varies irectly as its length L an inversely as the square of its iameter. Fin the constant of proportionality K if a wire 6.7 m long an.7 m in iameter has a resistance of 9 ohms. Fin the resistance R of a wire mae of the same material that is m long an has a iameter of. m. (.7) (.) kl k R = 9 = k = k = L.7 R= R = R = 7.5 Ω ( Note: Ω is the symbol (a Greek letter) use to represent ohms. ) 9. In the short growing season of the Canaian arctic territory of Nunavut, some gareners fin it possible to grow gigantic cabbages in the minight sun. Assume that the final size of a cabbage is proportional to the amount of nutrients it receives, an inversely proportional to the number of other cabbages surrouning it. A cabbage that receive oz of nutrients an ha 6 other cabbages aroun it grew to 7 lb.

7 What size woul it grow to if it receive oz of nutrients an ha only cabbage neighbors? Let S be the final cabbage size, an let N be the amount of nutrients it receives, an let C be the number of neighboring cabbages. Then we are tol that: kn k S = 7 = k = = k = 8. C 6 8.N 8.( ) 8 S = S = = S = 7 C 4. The heat experience by a hiker at a campfire is proportional to the amount of woo on the fire, an inversely proportional to the cube of his istance from the fire. If he is ft from the fire, an someone oubles the amount of woo burning, approximately how far from the fire woul he have to be so that he feels the same heat as before? Let H be the initial heat, an let W be the initial amount of woo on the fire, an let be the istance aske for in the problem. Let H, W, an D be any possible heat, woo, an istance kw kw kw We are tol that H =. This is true of the original situation, H D = H =, 67 k( W ) kw an is also true of the new istance aske for in the problem: H = H = We use the same initial heat for both equations, since that s what the pose question calls for.. Since kw H =, an also 67 H kw =, we must have that kw kw =. Cross-multiplying, we get: 67 = 67 = 44 = kw kw kw kw kw kw kw 44kW = 44 = ft