well, A is true, but so is C. So, the answer is D.
If you had solved the quadratic you would have found both solutions
7x+84 = -x^2+40x-6
Play A: y=7x+84
Play B: y=-x^2+40x-6
A. The attendance was the same on day 30. The attendance was 294 at both place that day.****
B.The attendance was the same on day 3. The attendance was 105 at both place that day.
C. The attendance was never the same at both plays
D. The attendance was the same on days 3 and 30. The attendance at both plays on those days was 105 and 294 respectively.
I think the answer is A, please correct me if I'm wrong please.
If you had solved the quadratic you would have found both solutions
7x+84 = -x^2+40x-6
1. A (8/15, 12/5)
2. Infinitely many
3. y < 2
4. 4 days
5. B
6. (5/4, 61/4)
7. B
8. y < 3x - 2 and y > -4x + 2
9. 6 cups of the 10% peanuts and 12 cups of the 25% peanuts
10. A
11. 720 pounds
12. C; B; D; A
13. D < B < A < C
14. B
15. 6n^3 - 19n^2 - n + 2
16. k^2 - 9
17. 4n^2 - 4
18. (x - 9) and (x + 4)
19. (3x + 5) and (-x-6)
20. A
21. Reaches a maximum height of 372.25 feet in 4.63
22. (0, 1)
23. (-2, 3)
24. 208.02 m
25. No solutions
26. The attendance was the same on days 4 and 40...
27. (2,5) and (-5, -9)
28. (0, -1) and (11, 87)
29. 7 sqrt 3
30. 6 sqrt 3
31. $1.80
32. 2x^2 + 2x - 4
33. -60
34. 0
35. B
36. 5d^6
37. 8
38. (p-7)(p+7)
39. 4,2
40. 13
41. 8
42. sqrt 30/6
43. B
44. A
45. A
Here The Answers, Make Sure You Mark Them Right When you Answer Babes!!
Play A: y = 15x + 76
Play B: y = โx2 + 36x โ 4
A. The attendance was the same on day 5. The attendance was 151 at both plays on that day.
B. The attendance was the same on day 16. The attendance was 316 at both plays on that day.
C. The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316 respectively.
D. The attendance was never the same at both plays.
15x + 76 = -x^2 + 36x - 4
Simplifying this equation, we get:
x^2 - 21x + 80 = 0
Factoring this quadratic, we get:
(x - 5)(x - 16) = 0
So, the attendance was the same on days 5 and 16. To find the attendance on those days, we can substitute x = 5 and x = 16 into either equation:
y = 15x + 76
y = -x^2 + 36x - 4
We get:
On day 5, attendance was 151 at both plays.
On day 16, attendance was 316 at both plays.
Therefore, the answer is C.
A. about 2.91 hours
B. about 160.00 hours
C. about 22.00 hours
D. about 5.50 hours
t * r = k
When t = 4 hours and r = 40 miles per hour, we can find the value of k:
4 * 40 = k
k = 160
Now, we can use this value of k to find the time required to drive the same distance at 55 miles per hour:
t * 55 = 160
t = 160 / 55
t โ 2.91 hours (rounded to two decimal places)
Therefore, the answer is A. It will take about 2.91 hours to drive the same distance at 55 miles per hour.
77 38 8 43 9 19 57 5 9 69 9 15 72 32 75 80 92
A. time chartZero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 1. Twenty to 29 hours has a frequency of 1. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 1. Seventy to 79 hours has a frequency of 3. Eighty to 89 hours has a frequency of 1. Ninety to 99 hours has a frequency of 1.
B. time chartZero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 2. Twenty to 29 hours has a frequency of zero. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 2. Seventy to 79 hours has a frequency of 2. Eighty to 89 hours has a frequency of 1. Ninety to 99 hours has a frequency of 1.
C. time chartZero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 2. Twenty to 29 hours has a frequency of zero. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 1. Seventy to 79 hours has a frequency of 3. Eighty to 89 hours has a frequency of 1. Ninety to 99 hours has a frequency of 1.
D. time chartZero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 2. Twenty to 29 hours has a frequency of zero. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 3. Seventy to 79 hours has a frequency of 1. Eighty to 89 hours has a frequency of zero. Ninety to 99 hours has a frequency of 2.
Range of hours Frequency
0-9 5
10-19 2
20-29 0
30-39 2
40-49 1
50-59 1
60-69 1
70-79 3
80-89 0
90-99 1
Therefore, the answer is C. The frequency table that represents the data is: "Zero to 9 hours has a frequency of 5. Ten to 19 hours has a frequency of 2. Twenty to 29 hours has a frequency of zero. Thirty to 39 hours has a frequency of 2. Forty to 49 hours has a frequency of 1. Fifty to 59 hours has a frequency of 1. Sixty to 69 hours has a frequency of 1. Seventy to 79 hours has a frequency of 3. Eighty to 89 hours has a frequency of 0. Ninety to 99 hours has a frequency of 1."
40, 7, 2, 35, 12, 23, 18, 28
A. minimum = 2
first quartile = 7
median = 20.5
third quartile = 35
maximum = 40
B. minimum = 2
first quartile = 7
median = 18
third quartile = 31.5
maximum = 40
C. minimum = 2
first quartile = 9.5
median = 23
third quartile = 35
maximum = 40
D. minimum = 2
first quartile = 9.5
median = 20.5
third quartile = 31.5
maximum = 40
22 / 32
2, 7, 12, 18, 23, 28, 35, 40
The minimum is the smallest value, which is 2.
To find the median, we need to find the middle value in the data set. Since there are 8 values, the median is the average of the fourth and fifth values:
median = (18 + 23) / 2 = 20.5
To find the first quartile, we need to find the median of the lower half of the data set. The lower half of the data set is:
2, 7, 12, 18
Since there are 4 values in the lower half, the first quartile is the average of the second and third values:
first quartile = (7 + 12) / 2 = 9.5
To find the third quartile, we need to find the median of the upper half of the data set. The upper half of the data set is:
23, 28, 35, 40
Since there are 4 values in the upper half, the third quartile is the average of the second and third values:
third quartile = (28 + 35) / 2 = 31.5
The maximum is the largest value, which is 40.
Therefore, the answer is D. The minimum is 2, the first quartile is 9.5, the median is 20.5, the third quartile is 31.5, and the maximum is 40.
(10x2 โ 13x + 12) รท (5x + 1)
A. 2x minus 3 plus 15 over 5x plus 1
B. 2x minus 5 plus 6 over 5x plus 1
C. 2x minus 3 minus 15 over 5x plus 1
D. 2x + 12
```
2x - 5
------------
5x + 1 | 10x^2 - 13x + 12
- (10x^2 + 2x)
---------------
-15x + 12
-15x - 3
---------
15
```
Therefore, the answer is B. (10x^2 - 13x + 12) รท (5x + 1) = 2x - 5 + 15 / (5x + 1) = (2x - 5 + 6) / (5x + 1) = (2x + 1) / (5x + 1)
Play A: y = 15x + 76
Play B: y = โx2 + 36x โ 4
A. The attendance was the same on day 5. The attendance was 151 at both plays on that day.
B. The attendance was the same on day 16. The attendance was 316 at both plays on that day.
C. The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316 respectively.
D. The attendance was never the same at both plays.
15x + 76 = -x^2 + 36x - 4
Simplifying this equation, we get:
x^2 - 21x + 80 = 0
Factoring this quadratic, we get:
(x - 5)(x - 16) = 0
So, the attendance was the same on days 5 and 16. To find the attendance on those days, we can substitute x = 5 and x = 16 into either equation:
y = 15x + 76
y = -x^2 + 36x - 4
We get:
On day 5, the attendance was:
15(5) + 76 = 151
On day 16, the attendance was:
-(16)^2 + 36(16) - 4 = 316
Therefore, the answer is C. The attendance was the same on days 5 and 16. The attendance at both plays on those days was 151 and 316 respectively.