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Do Now Solve the system by SUBSTITUTION y = 2x - 7 2x + y = 1 (2, -3)

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Algebra 1 Released EOC Test Review

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Objective SWBAT review concepts and questions from Algebra 1 Released EOC.

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Problem 1

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Based on the given data the y-intercept is 5 and the rate of change (slope) is 2/3, so based on the answer choices choice B would be correct.

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Problem 2

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Looking at the graph we Can generate the equation Y ≤ -2x + 5; however the Answer choices are written In words where they are in standard form 2x + y ≤ 5 and the only Answer choice that models This equation is Choice C.

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Problem 3

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Based on the difference of squares rule, factoring the expression you will get: (t+6)(t-6) Which is answer choice B.

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Problem 4

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Based on the equation we can find the vertex by using the equation x = -b/2a so substituting the values in we get: x = -(-8)/2(4) = 1. We then will substitute x into the equation f(x) = 4x 2 – 8x + 7. So f(1) = 4(1) 2 – 8(1) + 7 = 3. So our vertex is (1,3).Knowing this we can now eliminate choices B and C. We can now look at our y Intercept which is 7 and we then can eliminate choice A. So our answer is D.

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Problem 5 4 + w + 2 w + 2 To find the area of the rectangle we use the Formula L∙W = (4+w+2)(w+2) = (w+6)(w + 2). Factoring this expression we get: N = w 2 + 8w+12 So choice D is our answer

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Problem 6

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Shawn walks at a speed of 5 feet per second BUT he begins walking 20 seconds earlier, so An equation to represent each boys walking speed is: Shawn: d = 5t + 100 (at 20 seconds Shawn walked 100 feet) Curtis: d = 6t So solving the system through substitution we get: 6t = 5t + 100 -5t t = 100 So they were walking for 100 seconds when they met BUT Shawn had a 20 second lead so Shawn was walking for 120 seconds

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Problem 7

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For this problem we need to set up a system of equations. Let x = candy bars and Let y = drinks. So 60x + 110y = 265 120x + 90y = 270 To solve this system multiply the first equation by 2 and solve by elimination. 120x + 220y = 530 120x + 90y = 270 130y = 260130 y = 2 y is the number of drinks so we need to substitute to find x. 120x + 90(2) =270 120x + 180 = 270 120x = 90 x = 0.75 So the cost of the candy bars was $0.75.

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Problem 8

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Let n = the first positive integer, so the 3 consecutive numbers are: n, n+1, n+2 Since the product of the two smaller integers is 5 less than the largest integer we will set up our equation: n(n+1) = 5(n+2)-5 To find the smallest integer we need to solve for n. n 2 + n = 5n + 10 – 5 n 2 -4n -5 = 0 (n – 5)(n + 1) = 0 n = 5 and n = -1, but since n has to be a positive integer n = 5 only makes sense.

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Problem 9

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To see how long it takes the object to hit the ground we need to set our equation equal to zero. 0 = -5t 2 + 20t + 60 -5(t 2 - 4t - 12) = 0 t 2 – 4t – 12 = 0 (t – 6) (t + 2) = 0 t = 6 or t = -2 Time can not be negative so at 6 seconds the object will hit the ground.

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Problem 10 Let x = Antonio’s Age Let y = Sarah’s Age 2x + 3y = 34 y = 5x Use Substitution to find Sarah’s age 2x + 3(5x) = 34 2x + 15x = 34 17x = 34 x = 2 y = 5(2) = 10 So Sarah’s age is 10

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Problem 11

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When finding the value of k we need to find the difference from the graph and f(x) = 2(2) x. The y-intercept of the graph is -3 and the y intercept of f(x) = 2(2) x is 2. So the difference between 2 and -3 is -5. So the value of k is -5.

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Problem 12 Vv f(x) = 2x + 12 f(7) = 2(7) + 12 f(7) = 26 So it costs $26 dollars to rent 7 movies. Since Makayla has $10, she now needs $16 to rent 7 movies.

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Problem 13

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x X + 3X + 3 X + 6 Using the Pythagorean Theorem we get: x 2 + (x+3) 2 = (x+6)2 x 2 + x 2 +6x+9 = x 2 +12x+ 36 2x 2 +6x+9 = x 2 +12x+36 x 2 -6x -27 = 0 (x – 9)(x+3) =0 x = 9 or x = -3 So x must equal 9 because Measurement can be negative.

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Problem 14

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End of TurnPoints 0100 1200 2400 3800 41600 Katie’s Turns End of TurnPoints 0100 1300 2500 3700 4900 Jen’s Turns So at the end of turn 3 is when Katie’s points increase but the question said at the beginning of what turn, so at the beginning of the 4 th turn is when Katie will have more points.

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Problem 15

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Alex: 1 mi Sally: 3520 yd 15 min 24 min A: 1 mi ∙ 60 min1 mi = 1760 yd 15 min 1 hr2 mi = 3520 yd A: 4 miS: 2 mi = 1mi 1 hr 24 min 12 min S: 1 mi ∙ 60 min 12 min 1 hr 5 mi 1 hr So Sally walked 1mi/hr faster than Alex.

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Problem 16 Calc ACTIVE

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Problem 16 = 8 1/3 ∙x 2/3 ∙y 3/3 ∙z 4/3 2x 2/3 yz 4/3 So answer choice B is the correct answer

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Problem 17

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School Buys: 50x, where x is the candy bars Cost $30 a box to buy School Sells: 50x, where x is the candy bars Want to make $10 profit so they need to make $40. So to find out ho much each candy bar should cost we set up an equation: 50x = 40 x = 40/50 = 0.80 So each candy bar should cost $0.80 which is choice C.

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Problem 18

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E = mc 2 To solve for m we divide both sides by c 2 and get m = _E_ Which is choice D c 2

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Problem 19

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This is a quadratic function and the question is asking for the least which is the minimum value of this function. To find the minimum value we need to find the x value of the vertex, because x equals the number of years since 1964. Vertex formula: x = -b =-(-458.3) = 11.01 2a 2(20.8) So 11 years since 1964 is 1964+11 = 1975 So the year of 1975 is when the car value was at its least. So answer choice C is correct.

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Problem 20

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Based on Exponents Property, we will multiply our exponents and get x -1 which simplifies further to 1_. So choice B is correct. x

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Problem 21

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0.07 – 0.04 = 0.03 0.14 – 0.07 = 0.07 0.25 – 0.14 = 0.11 0.49 – 0.25 = 0.24 So the average rate of change is: 0.03+0.07+0.11+0.24 4 = 0.1125 Or you can find the average rate of change (slope) of (8,0.04) and (12, 0.49) 0.49 – 0.04 = 0.45 = 0.1125 12 – 8 4

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Problem 22

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We know the y intercept of f(x) is 5 so we need to find the y intercept of g(x) and find the difference. The rate of change of g(x) is ½ so to find the y intercept we need to find what g(x) equals when x = 0. So the table at the right shows the extension of the table where x = 0. We now see that the y intercept of g(x) is 5.5. So the difference is 5.5 – 5.0, which is 0.5 and Choice C, is the best answer.

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Problem 23

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y =.10x + 10 z = 0.20x So y – z is.10x + 10 – 0.20x = -0.10x + 10 Which is choice B.

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Problem 24

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Method one is neither constant or exponential but Method 2 is exponential because the rate of change is a product.

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Problem 25

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The slope of the line is 1/3. To find the slope of the 2 nd function we need To use the x and y intercepts to find the slope. x intercept: 4/3 (4/3, 0) y intercept: -2 ( 0, -2) So using the slope formula: y2 – y1 x2 – x1 We get: -2 – 0_ = _-2 _ = -2 ∙-3/4 = 3/2 0 – 4/3 -4/3 3/2 > 1/3 so answer choice B is correct.

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Problem 26

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The Slope -0.0018 is decreasing, so we can eliminate choices A and B. 0.0018 = _18_ = _1.8_ 10,000 1,000 So answer choice D is correct.

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Problem 27

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K is the midpoint: (2+6, 4+8) = (4,6) 2 2 Equation of line JL: J(2,4) and L(6,8) Find the slope: 8-4/6-2 = 1 y = mx + b 4 = 1(2) + b 2 = b Equation of JL is: y = x + 2 The line perpendicular to JL Because it is perpendicular the slope of this line must be the opposite reciprocal of JL. So the opposite reciprocal of 1 is -1. Since the midpoint of K is on the line we can use the point (4,6) to find our line. y = mx + b 6 = -1(4) +b 6 = -4 + b 10 = b So our equation is y = -x + 10 which is answer choice A.

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Do Now! In you composition Notebook Discuss your experience in Algebra 1 this year, and what would you like to see different in Geometry, Be honest!!

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Problem 28

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We need to use the distance formula: √(x 2 -x 1 ) 2 + (y 2 -y 1 ) 2 √(1- -1) 2 + (3- -1) 2 √(4+16) = √20 ≈ 4.47 √(1-2) 2 + (3- -3) 2 √(1+36) = √37 ≈ 6.08 √(-1-2) 2 + (-1- -3) 2 √(9+4) = √13 ≈ 3.61 4.47+6.08+3.61 = 14.16 Perimeter ≈14.16 Which is answer choice B.

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Problem 29

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Without DelawareWith Delaware Mean: 31.375downMean: 28.1 Range: 53upRange: 57

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Problem 29 Without DelawareWith Delaware IQR: 44upIQR: 46.5 IQR: Q3 – Q1: The median of 1 st and 2 nd half 6, 9, 11, 12, 46, 54, 54, 59 Q1: 6, 9, 11, 12 The median is 9+11/2 = 10 Q3: 46, 54, 54, 59 The median 54+54/2 = 54 So Q3 – Q1 = 54 – 11 = 44

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Problem 29 Without DelawareWith Delaware Standard Deviation 22.1822.85 To find SD subtract each value from the mean and then square that value. You then find the mean of those values and find the square root of the mean.

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Problem 29 Without Delaware Mean: 31.375 (6-31.375) 2 = 643.9 (59-31.375) 2 = 763.14 (12 – 31.375) 2 = 375.39 (11 – 31.375) 2 = 415.14 (9 – 31.375) 2 = 500.64 (54 – 31.375) 2 = 511.89 (46 – 31.375) 2 = 213.89 The mean of these values are 491.185. Standard Deviation is: √491.185 ≈ 22.16 You will do the same for With Delaware.

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Problem 30

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Total Number 0.25 0.24 0.27 The difference between Juniors and Seniors is 0.27 – 0.24 is 0.03. But the college surveyed 3,500 students so we must multiply 0.03 by 3,500 which is 105 students.

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Problem 31

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AgeFictionNonfictionTotal 21-30642286 31-407638114 Total14060200 We need to change the Relative frequency values to Actual numbers based on the 200 members in the club. So answer choice D is correct.

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Problem 32

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Finding the Linear Regression Model from the Calculator. (STAT Edit(enter data) STAT Calc 4(linReg) Enter) We get y = 0.8x = 8/10x = 4/5x. So our slope is 4/5 which is 4 miles every 5 minutes which is answer choice D.

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Problem 33

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A = ½ h (b 1 + b 2 ) h = 4, b 1 = x – 3, b 2 = x+7 A = ½ 4 (x – 3) + (x + 7) A = ½ 4 (2x + 4) A = ½ 8x + 16 A = 4x + 8 So Choice B is correct.

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Problem 34

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Model for this scenario: Let pens = x Let pencils = y 100≤x≤240 70≤y≤170 x + y < 300 Total Profit: 1.25x+0.75y We will then make a table from this equation

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Problem 34 Graphing all 3 inequalities on graphing calculator we get this graph:

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Problem 34 Total Profit: 1.25x+0.75y Not possible because 70≤y≤170 And 69 is less than 70.

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Problem 35

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Model for Scenario: 4lbs almonds = 22 Almonds = 22/4 = 5.5 per pound Cashews = 5.50 +.60(5.50) Cashews = 8.80 per pound Combined = 8.8C+5.5(4) = 6.50 (c+4) 8.8c+5.5(4) = 6.0(c+4) =8.8c + 22 = 6.50c +26 2.3c = 4 C = 4/ 2.3 ≈ 1.74 total mixture is 1.74 + 4 = 5.74 And the Cashew % is 1.74/5.74 ≈ 30% which is choice C

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Problem 36

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f(x) = g(x) 10x+5 = 7.5x + 25 2.5x + 5 = 25 2.5x = 20 x = 8 Which is choice C

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Problem 37

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0.05x + 0.10y = 0.65 (Since 0.65 is odd x must be odd also.) So our domain(x values) are 1, 3, 5, 7, 9, 11, 13 which is Choice D.

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Problem 38

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V(x) = 107000(1.009) 2/3x = 107,000(1.009 2/3 ) x = 107,000(1.00599) x = 107,000(1 + 0.00599) x 0.00599 ≈ 0.60% Which is choice C.

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Problem 39

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When n = 0, C = 10.5 C(n) = 10.5 + 1.5n C(n) = 12 – 1.5 + 1.5n C(n) = 12 + 1.5n – 1.5 C(n) = 12 + 1.5(n – 1) Which is choice A.

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Problem 40

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Graph the inequalities 1.75x + 1.25y ≤ 10 2x + 1y ≤ 12 0≤x≤5 0≤y≤8

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Problem 40 Let x = Chocolate Chip Cookies Let y = Peanut Butter Cookies To maximize: 4x+2y So at 5 batches of CC and 1 batch of PB is when its at its Max. So answer choice A is correct. Choc Chip Peanut Butter Profit (4x+2y) 5 batches =8.75 flours + 10 eggs 1 batch 1.25 flour 1 egg 5(4)+1(2 ) = $22 4 batches = 7 flour + 8 eggs 2 batches 2.5 flour 2 eggs 4(4) + 2(2) = $20 3 batches =4.25 flour + 6 eggs 4 batches 5 flour 4 eggs 3(4) + 4(2) = $20

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Problem 41

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Year (NEXT)Trees (NOW) 12 28 332 4128 The sequence is being multiplied by 4 so our equation is NEXT = NOW ∙ 4 Which is choice A.

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Problem 42

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Model for Scenario: m = 178-4 = 174 = 6 29 – 0 29 y = 6x + 4 t(n) = 6n + 4 Which is answer choice C. YearTrees 04 (y-intercept) 29178

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Problem 43

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This a rectangle that is not A square because all sides are Not equal so answer choice C

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Problem 44

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The midpoint is (10,2) So answer choice D is correct

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Problem 45

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Answer choice B is correct.

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Problem 46

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The volume is 16.62 So answer choice A is correct

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Problem 47

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Because the SD And IQR decreased Choice C is correct

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Problem 48

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Problem 49

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Problem 50 Most useful signifies the mean is the best.

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Gooooood LUCK TRY your best Take your time Get some Rest And DON’T worry!

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