PARCC Released Items 2016 #1 to #30

    (r – 3)2 = A Place r on the left side. (r – 3)2 = A Isolate r: Divide both sides by    (r – 3)2 = A   (r – 3)2 = A  Undo the square by taking the square root. r – 3 = A   Isolate r: Add 3 to both sides +3 +3 r = + 3 A  

7 7 Factor and use the Zero Product Property: x2 – 49 is a difference of squares (x + 7)(x + 7)(x – 7) = 0 x2 – 49 = (x + 7)(x – 7) x + 7 = 0 or x – 7 = 0 To solve, each factor can equal zero. – 7 – 7 + 7 + 7 x = – 7 or x = 7

   C. (0, 0) is NOT on the line (0, -1) and (2, 3) are on the graph. Test (15, 29): y = 2x – 1 D. Test (-0.5, -2): -2 = 2(-0.5) – 1 29 = 2(15) – 1 -2 = 1 – 1  29 = 29  Test (0.3, -0.4): -0.4 = 2(0.3) – 1.0 B. Test (2000, 1999) -0.4 = 0.6 – 1.0  y = 2x – 1 1999 = 2(2000) – 1 Test (0.5, 0): 0 = 2(0.5) – 1 1999 = 3999 False 0 = 1 – 1  E. Test ( ¼ , -½ ): -½ = 2/1( ¼ ) – 1   -½ = ½ – 2/2   Test (4/5 , 3/5 ): 3/5 = 2/1(4/5 ) – 1 3/5 = 8/5 – 5/5 

1 2x(x2 – 2) – 1(x2 – 2) – x(x2 – x – 2) 2x3 – 4x – 1x2 + 2 – x3 + x2 + 2x 2 2x3 – 1x2 – 4x + 2 1x3 + 1x2 + 2x 2 1x3 –2x + 2 ax3 + bx2 + cx + d

 dotted line solution: area shaded twice (pink area) solid line Solve for y: 2x  y ≥ 1 x + y  3 Solve for y: x x 2x 2x y ≥ 2x + 1 Division by 1 Inequality symbol reverses y  1x + 3 1 1 1 1 y ≤ 2x  1 1  shade above the line ≤ shade below the line

12 = change in feet = f( ½) – f(0) change in sec. ½  0 average rate of ascent = 6 – 0 feet ½  0 sec = 6 ½ = 6/1  1/2 = 6/1  2/1 = 12 feet per second t = 0 when she jumps into air = 16(02) + 20(0) = 16(0) + 0 = 0 + 0 = 0 feet t = ½ when she catches the bone = 16(½)2 + 20(½) =  16/1( ¼ ) + 10 =  4 + 10 = 6 feet

  Equation for a parabola: f(x) = (x – h)2 + k with a vertex of (h, k) f(x) = (x – h)2 + k  f(x) = (x – 2)2 + 1 vertex = (2, 1) 2 right and 1 up from (0, 0) Suppose that f(x) = x2 with a vertex of (0, 0) f(x + 3) = (x + 3)2 f(x  -3) = (x  -3)2 + 0 f(x  h) = (x  h)2 + k  vertex = (3, 0) 3 left from (0, 0) horizontal shift 3 units to left

  + ( ) = 0; 0 is rational  = = 2; 2 is rational 4 2   = (irrational number) 3 6 If you have one irr. #, you need a second irr. # to make a perfect square under the radical.  +  = 2 (still irrational) You need at least 1 irrational number to get an irrational sum. 20   = (still irrational) 2 10  You need two opposite irr. numbers for a rat. sum. If there are no irr. #, the ans. is rat.

To be a function, each x-value must be paired with exactly one y-value. This means that the y-term cannot have an even exponent or an absolute value.  Solved directly for y.  Solved directly for y. Not solved directly for y. The y-term has an absolute value.  Solved directly for y.  Solved directly for y.  Solved directly for y.

    To find time when on level ground, (Find x-intercepts) Solve: 0.005x(x – 18) = 0 0.005x = 0 or x – 18 = 0 0.005 0.005 +18 +18 x = 0 or x = 18 The crown is in the middle (average) of the times the road is on level ground:  (0 + 18)  2 = 18  2 = 9 crown occurs at x = 9  y-coordinate of crown = height above level ground 0.005x(x – 18)  = 0.005(9)(9 – 18) = 0.005(9)(–9)  = 0.405

  Type each of these equations into the TI-84: First, we need to make the right side equal to zero. Y1 = 2(x – 3)2 1 real solution (1 x-intercept)  Solve 2(x + 3)2 + 3 = 0 1 1 Y1 = 2(x + 3)2 + 3 no real solution (no x-intercept) 8 8  Solve (x  1)2  4 = 0 2 2 Y1 = (x  1)2  4 2 real solutions (2 x-intercepts) +15 +15 Solve (x + 1)2 + 2 = 0 Y1 = (x + 1)2 + 2 no real solution (no x-intercept) Solve x2 + 8x + 15 = 0 Y1 = x2 + 8x + 15 2 real solutions (2 x-intercepts)

Type the equation into the TI-84: Y1 = | 6 – 3x | + 6 Use the table (2nd, Graph) to find the vertex (it looks like x = 2) Graph the vertex at (2, 6) Graph one point on each side of the vertex: Graph (1, 9) and (3, 9) Connect the vertex to each of the points.

f(x) = 3x2 + 18x – 21 a = 3 f(x) = ax2 + bx + c 3 3 6 f(x) = a (x  h)2 + k Type into the TI-84: Y1 = 3x2 + 18x – 21 Use the table (2nd, Graph) to find the vertex (it looks like x = 3) Vertex = (3, 6) Vertex = (h, k) Place h and k into the boxes.

1 7 1 7 Type into the TI-84: Y1 = 2x2 + 4x + 5 y = a (x  h)2 + k Use the table (2nd, Graph) to find the vertex (it looks like x = 1) y = a (x + h)2 + k Vertex = (1, 7) Vertex = (h, k)

ax + c = bx + d Bring x-terms together. d – c bx bx ax  bx + c = d Isolate the x-terms. c c a – b ax – bx = d – c Factor out x. x(a – b) = d – c Isolate x. (a – b) = a – b

150(1.02h) time in hours initial number of bacteria Factor rate of growth = factor – 1 150 0.02 150(1.02) = 1.02 – 1 = 0.02 number of bacteria when h = 1 = 150(1.02h) = 150(1.02)1 = 150(1.02)

 total girls/total people = 16/31 = 0.516 total pizza/total people = 13/31 = 0.419 girls with pizza/total girls = 5/16 = 0.313 girls with pizza/total pizza = 5/13 = 0.385 

  This is an exponential graph, not linear. The function should be exponential. For the nonlinear equations, use the TI-84 and x in place of d to see if (3, 64) and (5, 256) are on the graph: Choice A is linear: c(d) = md + b where m is the slope, b is the y-intercept Press the “Y =“ key to enter the equation, Press the key to make an exponent, Choice B is linear: c(d) = md + b where m is the slope, b is the y-intercept To view the table, press “2nd” then “GRAPH”

  r2 = 2(18x – A)  A = 18x – 0.5r2 Place r on the left side to keep a positive coefficient. + 0.5r2 + 0.5r2 A + 0.5r2 = 18x Isolate r. r = 2(18x – A)   A A Replace 0.5 with ½ . ½ r2 = 18x – A 2(½ r2) = 2(18x – A) Multiply by denom. r2 = 2(18x – A)   

    2 hrs. more on Tuesday = 2 hrs. less on Wednesday # hours pay Tues. c P(c) = 15c  Wed. c – 2 P(c – 2) = 15(c – 2) Wed. pay = 15(c – 2)  = 15c – 30 Tuesday – Wednesday = P(c) – P(c – 2)

5(x)(x)(x) + 30(x)(x) + 35(x) 6 7 = x2  6x  7 7 The product of the 2 last terms in the parentheses should be 7 (____)(1) = 7 (x)(1) = 7 (1)(x) = 7 x = 7

 Maximum point: (12 sec, 19.4 ft) False Max. point: (12 sec, 19.4 ft) Reaches max. altitude after 12 sec. True y-intercept: Projectile launched from 5 feet, not the ground. False From t = 5 to t = 20, the function increases, then decreases. False

13  13 False Vertex (middle) of parabola occurs at t = 12, not t = 10 False  hits ground when height = 0 (y-coordinate = 0); hits ground at t = 25.93 False Let a = 4 h(12 – 4) = h(12 + 4) Let a = 4 h(12 – 12) = h(12 + 12) h(8) = h(16) h(0) = h(24) 17.8 = 17.8 True 5 = 5 True

1 1 10 2 n – = change in y = n – 5 change in x 7  (-3) slope = 0.1n – 0.5 = n – 5 10 = 1n – 5 10 = 1 5 10 10 n – B E D = 1 1 10 2 n – = 1 1 2 10 – + n

Function Z: f(n) = 0.1n – 0.5 Domain = n can be any real number: positive or negative; E is false B C D F Z is linear: slope = 0.1; y-intercept = -0.5 A. n can be any real number A is false B. slope = 0.1; positive slope indicates an increasing function C. slope = constant rate of change = 0.1 D. 0.1(natural number) = rational number; rational number – 0.5 = rational number

    Discriminant Values: D is neg; no real roots D is not perf. sq.; 2 non-rat. roots D = 0; one real root D is perf. sq.; 2 rat. roots Use the discriminant b2 – 4ac y = 1x2 – 7x + 9 (7)2 – 4(1)(9) = 49 – 36 = 13 not a perf. sq.  y = 1x2 – 0x + 9  (0)2 – 4(1)(9)  = 0 – 36 = –36 negative #  y = 1x2 – 6x + 9 (6)2 – 4(1)(9) = 36 – 36 = 0 y = (x – h)2 + k y = 1x2 – 10x + 9 y = x2  1x – 1x + 1 + 8 (10)2 – 4(1)(9) 2 y = (x – 1)2 + 8 y = x2  2x + 9 = 100 – 36 y = (x – 1)(x – 1) + 8 = 64 perf. sq.

 Community A: More of a spread of prices (less consistent); Most of them less expensive Community B: Less of a spread of prices; (more consistent) Most of them more expensive 

Community B: 156, 157, 157, 158, 158, 159, 159, 159, 159, 160, 160, 160, 160, 160, 160, 160, 160, 161, 161, 161, 161, 161, 161, 162, 162, 163, 163, 164, 165, 165, 166, 166, 167, 167, 168, 168, 169, 170, 170, 172, 172, 174, 174, 175, 175  45 prices: median = middle = 23rd term = 161  $161, 000 third quartile: median of terms after the circled number middle = between 167 and 168  between 167,000 and 168,000

39 39 43

 Begin with 256 pennies Trial 1: P(heads) = ½ , so approximately ½ of 256 = 128 pennies will return to the bag The number of pennies in the bag keeps dividing by 2 (multiplying by ½) This represents an exponential function. Trial 2: P(heads) = ½ , so approximately ½ of 128 = 64 pennies will return to the bag Trial 3: P(heads) = ½ , so approximately ½ of 64 = 32 pennies will return to the bag

 The function is exponential, begins with an initial value of 256 pennies, and the value keeps multiplying by ½ The only exponential function is B.

 Begin with 256 pennies; keep dividing by 2 for each trial. Trial 1: 128 pennies Trial 2: 64 pennies Trial 3; 32 pennies Trial 4: 16 pennies Trial 5: 8 pennies Trial 6: 4 pennies Trial 7: 2 pennies Trial 8: 1 penny

 Total number of pennies = 256 # pennies in bag + # pennies on floor = 256 f(n) + g(n) = 256

  40 hour week  4 preparation workers = 160 hours of preparation 40 hour week  7 packaging workers = 280 hours of packaging preparing tomato salsa + preparing corn salsa ≤ 160 2t + 2.5c ≤ 160  packaging tomato salsa + packaging corn salsa ≤ 280 4t + 3c ≤ 280

   2t + 2.5c ≤ 160 4t + 3c ≤ 280 2(20) + 2.5(45) = 152.5 true 2(45) + 2.5(30) = 165 false 2(50) + 2.5(25) = 162.5 false  2(60) + 2.5(10) = 145 true 4(60) + 3(10) = 270 true

2t + 2.5c ≤ 160 4t + 3c ≤ 280 2t + 2.5(20) ≤ 160 4t + 3(20) ≤ 280 2t + 50 ≤ 160 4t + 60 ≤ 280 -50 -50 -60 -60 55 2t ≤ 110 2 2 4t ≤ 220 4 4 t ≤ 55 t ≤ 55 2.5c ≤ 160 2.5 2.5 3c ≤ 280 3 3 64 c ≤ 64 c ≤ 93.3

y-coordinate is positive: sum of coordinates is greater than 2: y  0 x + y  2 x + y  2

All points in the first quadrant have a positive x-coordinate and a positive y-coordinate. This proves that y  0. All points in the first quadrant have a positive x-coordinate and a positive y-coordinate. x + y  2 x + y = positive # + positive # = positive # positive #  0 and 0  2, so therefore positive #  2 This proves that x + y  2

Make an equation for Cars A and B: 70t = 65t + 45 -65t -65t slope = 350 – 0 5 – 0 t = 9 t = 10 Car A will pass Car B 9 hours after noon, 9:00 PM. = 350  5 Car A will pass Car C 10 hours after noon, 10:00 PM. = 70 D = mt + b D = 70t (y-int = 0) 0 110 – 65 = 45 +65 (0, 45) and (1, 110) D = mt + b +65 slope = 110 – 45 1 – 0 D = 65t + 45 = 65