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Do Now. Solve the system by SUBSTITUTION y = 2x - 7 2x + y = 1 (2, -3). Algebra 1 Released EOC Test Review. Objective . SWBAT review concepts and questions from Algebra 1 Released EOC. Problem 1. Problem 1. Problem 1.
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Do Now Solve the system by SUBSTITUTION y = 2x - 7 2x + y = 1 (2, -3)
Objective • SWBAT review concepts and questions from Algebra 1 Released EOC.
Problem 1 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.
Problem 2 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.
Problem 3 Based on the difference of squares rule, factoring the expression you will get: (t+6)(t-6) Which is answer choice B.
Problem 4 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) = 4x2 – 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.
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 = w2 + 8w+12 So choice D is our answer
Problem 6 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 -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
Problem 7 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 = 260 130 130 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.
Problem 8 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. n2 + n = 5n + 10 – 5 n2 -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.
Problem 9 To see how long it takes the object to hit the ground we need to set our equation equal to zero. 0 = -5t2 + 20t + 60 -5(t2- 4t - 12) = 0 t2 – 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.
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
Problem 11 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.
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.
Problem 13 Using the Pythagorean Theorem we get: x2 + (x+3)2 = (x+6)2 x2 + x2+6x+9 = x2+12x+ 36 2x2+6x+9 = x2+12x+36 x2 -6x -27 = 0 (x – 9)(x+3) =0 x = 9 or x = -3 So x must equal 9 because Measurement can be negative. X + 3 X + 6 x
Problem 14 Katie’s Turns 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 4th turn is when Katie will have more points.
Problem 15 Alex: 1 mi Sally: 3520 yd 15 min 24 min A: 1 mi ∙ 60 min 1 mi = 1760 yd 15 min 1 hr 2 mi = 3520 yd A: 4 mi S: 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.
Problem 16 • = 81/3 ∙x2/3∙y3/3∙z4/3 2x2/3yz4/3 So answer choice B is the correct answer
Problem 17 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.
Problem 18 E = mc2 To solve for m we divide both sides by c2 and get m = _E_ Which is choice D c2
Problem 19 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.
Problem 20 Based on Exponents Property, we will multiply our exponents and get x-1 which simplifies further to 1_. So choice B is correct. x
Problem 21 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
Problem 22 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.
Problem 23 y = .10x + 10 z = 0.20x So y – z is .10x + 10 – 0.20x = -0.10x + 10 Which is choice B.
Problem 24 Method one is neither constant or exponential but Method 2 is exponential because the rate of change is a product.