26. Since opening night, attendance at Play A has increased steadily, while attendance at Play B first rose and then fell. Equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. On what day(s) was the attendance the same at both plays? What was the attendance?
Play A: y = 15x + 76
Play B: y = –x2 + 36x – 4
The attendance was the same on day 5. The attendance was 151 at both plays on that day.
The attendance was the same on day 16. The attendance was 316 at both plays on that day.
The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316, respectively.
The attendance was never the same at both plays.
27. What are the solutions to the system?
y = x2 + 5x – 9
y = 2x + 1
(2, –9) and (–5, 5)
(2, 5) and (–5, –59)
(2, 5) and (–5, –9)
no solution
28. Find the solutions to the system.
y = x2 – 2x –2
y = 4x + 5
(–1, 1) and (–7, –23)
(–1, 1) and (7, 33)
(–1, 33) and (7, 1)
no solution
29. Simplify the radical expression.
(Image: Start Root 147 End Root)
(Image)
(Image)
–(Image)
21
30. Simplify the expression.
(Image: square root 3 + 4 square root 3)
(Image: 5 square root 6)
(Image: 5 square root 3)
(Image: 3 square root 6)
(Image: 3 square root 3)
31. Kendra owns a toy store. She charges $5.00 for two cars and one piece of candy. She charges $3.20 for one car and one piece of candy. How much does Kendra charge per car?
$1.40
$3.20
$3.60
$1.80
32. Divide.
(3x3 + 10x2 – 12x – 5) ÷ (3x + 1)
x2 – 3x + 5
x2 + 3x + 5
x2 – 3x – 5
x2 + 3x – 5
33. What is the value of 12x–3y1 for x = –1 and y = 5?
60
–60
300
(Image)
34. What is the degree of the monomial?
5
0
1
–1
5
35. Simplify.
–(3xy4)–4 (1 point)
–81x4y16
(Image)
(Image)
(Image)
36. What is the simplified form of the expression?
5d–6 · d12
5d–72
5d6
6d6
6d–72
37. What is the degree of the monomial?
6x10
6
60
16
10
38. What is the factored form of the expression?
p2 – 49
(p + 7)(p + 7)
(p – 7)(p + 7)
(p – 7)(p – 7)
(p – 30)(p – 19)
39. Solve the equation by completing the square. Round to the nearest hundredth if necessary.
x2 + 5x = 6
6, 1
–6, 1
6, –1
–6, –1
40. What is the length of the hypotenuse of the right triangle shown below?
(Image: A triangle is shown with a horizontal leg labeled 5, a vertical leg labeled 12, a small square in the lower left corner, and a diagonal leg extending from the top of the vertical segment to the right end of the horizontal segment labeled lower c.)
13
7
4.1
17
41. What is the side length of b in the triangle shown below?
(Image: The figure shows a right triangle. The left leg equals 11. The lower leg equals b. The hypotenuse equals 61.)
60
50
72
62.95
42. Simplify the radical expression by rationalizing the denominator.
(Image: Start Fraction 3 over Start Root 15 End Root End Fraction)
(Image)
(Image)
(Image)
(Image)
43. A box has a volume given by the trinomial x3 + 10x2 – 24x. What are the possible dimensions of the box? Use factoring.
x, x – 1, x + 4
x, x – 6, x + 4
x, x – 1, x + 24
x, x – 2, x + 12
44. Divide.
(Image: left parenthesis 20 x squared minus 12 x plus 8 right parenthesis divided by left parenthesis 2 x minus 8 right parenthesis)
(Image)
(Image)
(Image)
(Image)
45. What is the simplified form of (Image: the quantity of 4 times t squared minus 5 times t plus three divided by the quantity of 2 times t squared minus t plus 1)?
(Image: 2 plus the quotient of the quantity negative 3 times t plus 1 over the quantity 2 times t squared minus t plus 1)
(Image: 2 plus the quotient of the quantity negative 7 times t plus 5 over the quantity 2 times t squared minus t plus 1)
(Image: 2 times t minus 5)
(Image: 2 times t plus 3)
27. (2, 5) and (-5, -9)
28. (-1, 1) and (7, 33)
29. $7\sqrt{3}$
30. $5\sqrt{3}$
31. $1.80$
32. x² + 3x - 5
33. -60
34. 0
35. $\frac{-1}{81x^4y^{16}}$
36. $5d^6$
37. 10
38. (p + 7)(p - 7)
39. 1.24 or -6.24
40. 13
41. 60
42. $\frac{3\sqrt{15}}{5}$
43. x, x - 2, x + 12
44. 10x + 4
45. $2 - \frac{7t - 5}{2t^2 - t + 1}$
39. Solve the equation by completing the square. Round to the nearest hundredth if necessary.
x2 + 5x = 6
A. 6, 1
B. –6, 1
C. 6, –1
D. –6, –1
The correct answer is B. -6, 1.
To solve by completing the square:
1. Move the constant term to the right side:
x^2 + 5x = 6
x^2 + 5x - 6 = 0
2. Take half of the coefficient of x (which is 5) and square it:
(5/2)^2 = 6.25
3. Add and subtract this value inside the left side of the equation:
x^2 + 5x + 6.25 - 6.25 - 6 = 0
4. Factor the first three terms inside the left side:
(x + 2.5)^2 - 12.25 = 0
5. Add 12.25 to both sides:
(x + 2.5)^2 = 12.25
6. Take the square root of both sides (remembering to include both the positive and negative square roots):
x + 2.5 = ±3.5
7. Solve for x:
x = -2.5 ± 3.5
x = -6 or x = 1
Rounding to the nearest hundredth, the solutions are -6 and 1.
45. What is the simplified form of (Image: the quantity of 4 times t squared minus 5 times t plus three divided by the quantity of 2 times t squared minus t plus 1)?
(Image: 2 plus the quotient of the quantity negative 3 times t plus 1 over the quantity 2 times t squared minus t plus 1)
(Image: 2 plus the quotient of the quantity negative 7 times t plus 5 over the quantity 2 times t squared minus t plus 1)
(Image: 2 times t minus 5)
(Image: 2 times t plus 3)
The answer is:
(Image: 2 plus the quotient of the quantity negative 7 times t plus 5 over the quantity 2 times t squared minus t plus 1)
To simplify the expression, we need to divide the numerator by the denominator:
$\frac{4t^2 - 5t + 3}{2t^2 - t + 1}$
Using long division, we get:
(Image: 2 plus the quotient of the quantity negative 7 times t plus 5 over the quantity 2 times t squared minus t plus 1)
Therefore, the simplified form of the expression is:
(Image: 2 plus the quotient of the quantity negative 7 times t plus 5 over the quantity 2 times t squared minus t plus 1)
26. Since opening night, attendance at Play A has increased steadily, while attendance at Play B first rose and then fell. Equations modeling the daily attendance y at each play are shown below, where x is the number of days since opening night. On what day(s) was the attendance the same at both plays? What was the attendance?
Play A: y = 15x + 76
Play B: y = –x2 + 36x – 4
The attendance was the same on day 5. The attendance was 151 at both plays on that day.
The attendance was the same on day 16. The attendance was 316 at both plays on that day.
The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316, respectively.
The attendance was never the same at both plays.
The answer is:
The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316, respectively.
To find the attendance on each day, we set the two equations equal to each other and solve for x:
15x + 76 = -x^2 + 36x - 4
Rewriting the equation in standard form:
x^2 - 21x + 80 = 0
Factoring this quadratic:
(x - 5)(x - 16) = 0
Therefore, the attendance was the same on days 5 and 16. Plugging in x = 5 and x = 16 into either equation, we get:
At day 5: y = 15(5) + 76 = 151
At day 16: y = 15(16) + 76 = 316
So on day 5, the attendance at both plays was 151, and on day 16, the attendance at both plays was 316.
42. Simplify the radical expression by rationalizing the denominator.
(Image: Start Fraction 3 over Start Root 15 End Root End Fraction)
To rationalize the denominator, we multiply the top and bottom of the fraction by the conjugate of the denominator:
$\frac{3}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}} = \frac{3\sqrt{15}}{15}$
Simplifying the numerator:
$\frac{3\sqrt{3} \cdot \sqrt{5}}{15} = \frac{\sqrt{3}}{5}\cdot\sqrt{5}$
Therefore, the simplified form of the expression is:
$\frac{\sqrt{3}}{5}\cdot\sqrt{5}$
26. To find the day(s) when the attendance was the same at both plays, we need to solve the equation y = 15x + 76 for Play A and y = –x^2 + 36x – 4 for Play B.
To find the day(s) when the attendance was the same, we need to find the values of x that make the equations equal.
Setting the two equations equal to each other, we have:
15x + 76 = –x^2 + 36x – 4.
Rearranging the equation, we get:
x^2 + 21x - 80 = 0.
Now, we can solve this quadratic equation. Factoring or using the quadratic formula, we find that the solutions are x = -16 and x = 5.
Therefore, the attendance was the same on day 5 and day 16.
To find the attendance at those days, we substitute the values of x back into either of the equations.
For day 5, when x = 5, the attendance is:
y = 15(5) + 76 = 151.
So, the attendance at both plays on day 5 was 151.
For day 16, when x = 16, the attendance is:
y = 15(16) + 76 = 316.
So, the attendance at both plays on day 16 was 316.
Therefore, the attendance was the same on days 5 and 16, and the attendance at both plays on those days was 151 and 316, respectively.
The correct answer is: The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316, respectively.