### Algebra 1 Released EOC Test Review

```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
at the graph we
Problem 2 Looking
Can generate the equation
Y ≤ -2x + 5; however the
In words where they are
in standard form
2x + y ≤ 5 and the only
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
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+12
So 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 = 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
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
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
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
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
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
Problem 14
Katie’s Turns
Jen’s Turns
End of Turn
Points
End of Turn
Points
0
100
0
100
1
200
1
300
2
400
2
500
3
800
3
700
4
1600
4
900
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 15
Alex: 1 mi
15 min
A: 1 mi ∙ 60 min
15 min 1 hr
A: 4 mi
1 hr
Sally: 3520 yd
24 min
1 mi = 1760 yd
2 mi = 3520 yd
S: 2 mi = 1mi
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 Calc ACTIVE
Problem 16
•
=
81/3 ∙x2/3∙y3/3∙z4/3
2x2/3yz4/3
Problem 17
Problem 17
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
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 multiply
our exponents and get x-1 which simplifies
further to 1_. So choice B is correct.
x
Problem 21
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
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 = 0.20x
So y – z is
.10x + 10 – 0.20x
= -0.10x + 10
Which 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 25
The 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 – 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.
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 27
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.
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!!
Problem 28
Problem 28
We 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 = 14.16
choice B.
Problem 29
Problem 29
Without Delaware
Mean: 31.375 down
Range: 53
up
With Delaware
Mean: 28.1
Range: 57
Problem 29
Without Delaware
With Delaware
IQR: 44
up
IQR: 46.5
IQR: Q3 – Q1: The median of 1st and 2nd 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
Problem 29
Without Delaware
With Delaware
Standard Deviation
22.18
22.85
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.
Problem 29
Without Delaware
You will do the same for With 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
(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
Problem 30
Problem 30
Total Number
0.25
0.24
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.
Problem 31
Problem 31
Age
Fiction
Nonfiction
Total
21-30
64
22
86
31-40
76
38
114
Total
140
60
200
We need to
change the
Relative frequency
values to
Actual numbers
based on the
200 members in the
choice D is correct.
Problem 32
Problem 32
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
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 + 16
A = 4x + 8
So Choice B is correct.
Problem 34
Problem 34
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
Problem 34
Graphing all 3 inequalities on graphing
calculator we get this graph:
Problem 34
Total Profit: 1.25x+0.75y
x
y
1.25x+0.75
y
230
69
339.25
229
70
338.75
228
71
338.25
227
72
337.75
Not possible because 70≤y≤170
And 69 is less than 70.
Problem 35
Problem 35
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
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 = 10.5
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.
Problem 40
Problem 40
Graph the inequalities
1.75x + 1.25y ≤ 10
2x +
1y ≤ 12
0≤x≤5
0≤y≤8
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
Problem 41
Problem 41
Year (NEXT)
Trees (NOW)
1
2
2
8
3
32
4
128
The sequence is being multiplied by 4 so our equation is
NEXT = NOW ∙ 4 Which is choice A.
Problem 42
Problem 42
Model for Scenario:
Year
Trees
0
4 (y-intercept)
29
178
m = 178-4 = 174 = 6
29 – 0 29
y = 6x + 4
t(n) = 6n + 4 Which is answer choice C.
Problem 43
Problem 43
This a rectangle that is not
A square because all sides are
Not equal so answer choice C
Problem 44
Problem 44
The midpoint is (10,2)
So answer choice D is correct
Problem 45
Problem 45
B is correct.
Problem 46
Problem 46
The volume is 16.62
correct
Problem 47
Problem 47
Because the SD
And IQR decreased
Choice C is correct
Problem 48
Problem 48
Problem 49
Problem 49
Problem 50
Most useful signifies the mean is the best.
Gooooood LUCK