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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. - PowerPoint PPT Presentation
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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
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
Problem 2 Looking at the graph weCan generate the equationY ≤ -2x + 5; however the Answer choices are writtenIn words where they are in standard form2x + y ≤ 5 and the only Answer choice that models This equation is Choice C.
Problem 3
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
Problem 4
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+12So choice D is our answer
Problem 6
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 = 6tSo solving the system through substitution we get:6t = 5t + 100-5t -5t t = 100So 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
Problem 7For 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 = 270To solve this system multiply the first equation by 2 and solve by elimination.
120x + 220y = 530120x + 90y = 270130y = 260130 130 y = 2
y is the number of drinks so we need to substitute to find x.120x + 90(2) =270120x + 180 = 270 120x = 90 x = 0.75So the cost of the candy bars was $0.75.
Problem 8
Problem 8Let n = the first positive integer, so the 3 consecutive
numbers are: n, n+1, n+2Since 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)-5To 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
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 = -2Time can not be negative so at 6 seconds the
object will hit the ground.
Problem 10
Let x = Antonio’s AgeLet y = Sarah’s Age 2x + 3y = 34 y = 5xUse Substitution to find Sarah’s age 2x + 3(5x) = 342x + 15x = 3417x = 34 x = 2
y = 5(2) = 10So Sarah’s age is 10
Problem 11
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 + 12f(7) = 2(7) + 12f(7) = 26So it costs $26 dollars to rent 7 movies.Since Makayla has $10, she now needs $16 to rent 7 movies.
Problem 13
Problem 13
x
X + 3
X + 6
Using the Pythagorean Theorem we get: x2 + (x+3)2 = (x+6)2x2 + x2+6x+9 = x2+12x+ 362x2+6x+9 = x2+12x+36x2 -6x -27 = 0(x – 9)(x+3) =0 x = 9 or x = -3So x must equal 9 because Measurement can be negative.
Problem 14
Problem 14 End of Turn Points
0 100
1 200
2 400
3 800
4 1600
Katie’s TurnsEnd of Turn Points
0 100
1 300
2 500
3 700
4 900
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
Problem 15Alex: 1 mi Sally: 3520 yd 15 min 24 minA: 1 mi ∙ 60 min 1 mi = 1760 yd 15 min 1 hr 2 mi = 3520 ydA: 4 mi S: 2 mi = 1mi 1 hr 24 min 12 min
S: 1 mi ∙ 60 min 12 min 1 hr5 mi 1 hr
So Sally walked 1mi/hr faster than Alex.
Problem 16 Calc ACTIVE
Problem 16
• =81/3 x∙ 2/3 y∙ 3/3 z∙ 4/3
2x2/3yz4/3
So answer choice B is the correct answer
Problem 17
Problem 17School Buys:50x, where x is the candy barsCost $30 a box to buySchool Sells:50x, where x is the candy barsWant 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
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
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
Problem 20
Based on Exponents Property, we will multiplyour exponents and get x-1 which simplifiesfurther to 1_. So choice B is correct.
x
Problem 21
Problem 21
0.07 – 0.04 = 0.030.14 – 0.07 = 0.070.25 – 0.14 = 0.110.49 – 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
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
Problem 23
y = .10x + 10 z = 0.20xSo y – z is .10x + 10 – 0.20x = -0.10x + 10Which is choice B.
Problem 24
Problem 24
Method one is neither constant or exponential but Method 2 is exponential because the rate of change is a product.
Problem 25
Problem 25The slope of the line is 1/3.To find the slope of the 2nd 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 – x1We get: -2 – 0_ = _-2 _ = -2 -3/4 = 3/2∙ 0 – 4/3 -4/3
3/2 > 1/3 so answer choice B is correct.
Problem 26
Problem 26
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.
Problem 27
Problem 27K is the midpoint: (2+6 , 4+8) = (4,6)
2 2Equation 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 = bEquation 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 = bSo our equation is y = -x + 10 which is answer choice A.
Do Now!
In you composition NotebookDiscuss your experience in Algebra 1 this year, and what would you like to see different in Geometry, Be honest!!
Problem 28
Problem 28We need to use the distance formula:√(x2-x1)2 + (y2-y1)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 = 14.16Perimeter ≈14.16 Which is answer choice B.
Problem 29
Problem 29
Without Delaware With DelawareMean: 31.375 down Mean: 28.1Range: 53 up Range: 57
Problem 29Without Delaware With DelawareIQR: 44 up IQR: 46.5IQR: Q3 – Q1: The median of 1st and 2nd half6, 9, 11, 12, 46, 54, 54, 59Q1: 6, 9, 11, 12The median is 9+11/2 = 10Q3: 46, 54, 54, 59The median 54+54/2 = 54So Q3 – Q1 = 54 – 11 = 44
Problem 29
Without Delaware With DelawareStandard Deviation
22.18 22.85To 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.
Problem 29Without DelawareMean: 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(54 – 31.375)2 = 511.89(46 – 31.375)2 = 213.89The mean of these values are 491.185. Standard Deviation is: √491.185 ≈ 22.16
You will do the same for With Delaware.
Problem 30
Problem 30Total Number
0.250.240.240.27
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.
Problem 31
Problem 31Age Fiction Nonfiction Total
21-30 64 22 86
31-40 76 38 114
Total 140 60 200
We need tochange the Relative frequency values toActual numbers based on the 200 members in the club. So answer choice D is correct.
Problem 32
Problem 32
Finding the Linear Regression Model from the Calculator. (STATEdit(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.
Problem 33
Problem 33
A = ½ h (b1 + b2) h = 4, b1 = x – 3, b2 = x+7 A = ½ 4 (x – 3) + (x + 7)A = ½ 4 (2x + 4)A = ½ 8x + 16A = 4x + 8 So Choice B is correct.
Problem 34
Problem 34
Model for this scenario:Let pens = xLet pencils = y100≤x≤24070≤y≤170 x + y < 300Total Profit: 1.25x+0.75yWe will then make a table from this equation
Problem 34
Graphing all 3 inequalities on graphing calculator we get this graph:
Problem 34
Total Profit: 1.25x+0.75yx y 1.25x+0.75
y230 69 339.25
229 70 338.75
228 71 338.25
227 72 337.75
Not possible because 70≤y≤170And 69 is less than 70.
Problem 35
Problem 35Model for Scenario:4lbs almonds = 22Almonds = 22/4 = 5.5 per poundCashews = 5.50 + .60(5.50)Cashews = 8.80 per poundCombined = 8.8C+5.5(4) = 6.50
(c+4)8.8c+5.5(4) = 6.0(c+4)=8.8c + 22 = 6.50c +262.3c = 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
Problem 36
Problem 36
f(x) = g(x) 10x+5 = 7.5x + 25 2.5x + 5 = 25 2.5x = 20 x = 8 Which is choice C
Problem 37
Problem 37
0.05x + 0.10y = 0.65 (Since 0.65 is odd x must be odd also.)
x y
1 6
3 5
5 4
7 3
9 2
11 1
13 0
So our domain(x values) are 1, 3, 5, 7, 9, 11, 13 which is Choice D.
Problem 38
Problem 38
V(x) = 107000(1.009)2/3x
= 107,000(1.0092/3)x
= 107,000(1.00599)x
= 107,000(1 + 0.00599)x
0.00599 ≈ 0.60%Which is choice C.
Problem 39
Problem 39
When n = 0 , C = 10.5C(n) = 10.5 + 1.5nC(n) = 12 – 1.5 + 1.5nC(n) = 12 + 1.5n – 1.5C(n) = 12 + 1.5(n – 1) Which is choice A.
Problem 40
Problem 40
Graph the inequalities1.75x + 1.25y ≤ 10 2x + 1y ≤ 120≤x≤5 0≤y≤8
Problem 40Let x = Chocolate Chip CookiesLet y = Peanut Butter CookiesTo maximize: 4x+2ySo 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 batch1.25 flour1 egg
5(4)+1(2) = $22
4 batches= 7 flour + 8 eggs
2 batches2.5 flour2 eggs
4(4) + 2(2) = $20
3 batches=4.25 flour + 6 eggs
4 batches5 flour4 eggs
3(4) + 4(2) = $20
Problem 41
Problem 41Year (NEXT) Trees (NOW)
1 2
2 8
3 32
4 128
The sequence is being multiplied by 4 so our equation isNEXT = NOW 4 Which is choice A.∙
Problem 42
Problem 42
Model for Scenario:
m = 178-4 = 174 = 6 29 – 0 29 y = 6x + 4 t(n) = 6n + 4 Which is answer choice C.
Year Trees
0 4 (y-intercept)
29 178
Problem 43
Problem 43
This a rectangle that is notA square because all sides areNot equal so answer choice C
Problem 44
Problem 44The midpoint is (10,2)So answer choice D is correct
Problem 45
Problem 45
Answer choiceB is correct.
Problem 46
Problem 46The volume is 16.62So answer choice A is correct
Problem 47
Problem 47
Because the SD And IQR decreasedChoice C is correct
Problem 48
Problem 48
Problem 49
Problem 49
Problem 50
Most useful signifies the mean is the best.
Gooooood LUCK
TRY your bestTake your timeGet some RestAnd DON’T worry!