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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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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!