Suppose that the amount of algae in a pond triples every 2 hours. If the pond initially contains 80 pounds of algae, how much algae will be in the pond after 8 hours?
After 2 hours, the pond will contain 3 times 80 pounds, or 240 pounds of algae. After another 2 hours, the amount of algae will triple again, giving a total of 3 times 240 pounds, or 720 pounds. After 6 hours, the amount of algae will triple once more, giving 3 times 720 pounds, or 2160 pounds. Finally, after 8 hours, the amount of algae will triple once more, giving a total of 3 times 2160 pounds, or 6480 pounds of algae. Therefore, there will be 6480 pounds of algae in the pond after 8 hours.
What is the order of the numbers from least to greatest?
A = 6.2 × 10–5 B = 12.3 × 10–7 C = 5.2 × 103 D = 3.7 × 10–5
B, D, A, C
A catapult launches a boulder with an upward velocity of 92 m/s. The height of the boulder, h, in meters after t seconds is given by the function h = –5t2+ 92t + 16. How long does it take to reach maximum height? What is the boulder’s maximum height? Round to the nearest hundredth, if necessary.
The maximum height of the boulder occurs at the vertex of the parabolic function given by h = –5t^2 +92t +16. The t-coordinate of the vertex is given by t = –b/2a, where a = –5 and b = 92. Substituting these values, we get:
t = –92/(2 * –5) = 9.2 seconds
This means that the boulder takes 9.2 seconds to reach maximum height.
To find the maximum height, we can substitute this value of t into the given function to get:
h = –5(9.2)^2 + 92(9.2) + 16 ≈ 423.68
So the boulder's maximum height is approximately 423.68 meters.
Suppose that y varies inversely with x, and y = 2 when x = 6. What is an equation for the inverse variation?
If y varies inversely with x, then y is equal to a constant value divided by x. Let k denote this constant of proportionality. Then we have:
y = k/x
To find the value of k, we can use the initial condition given in the problem. We know that when x = 6, y = 2. Substituting these values into the equation above, we get:
2 = k/6
Simplifying this equation, we can multiply both sides by 6 to get:
k = 12
Therefore, the equation for the inverse variation is:
y = 12/x
Simplify the rational expression. State any excluded values.
3x minus 12 over x minus 4
A. x
B. 3; where x ≠ 4
C. 3; where x ≠ 3
D. 0
The given rational expression is:
(3x - 12)/(x - 4)
We can factor the numerator and simplify the expression as follows:
(3x - 12)/(x - 4) = 3(x - 4)/(x - 4)
Notice that the factor of (x - 4) appears in both the numerator and denominator, and therefore can be canceled out. This leaves us with:
3
Therefore, the simplified expression is just 3.
However, we must note that the expression (3x - 12)/(x - 4) is undefined at x = 4, since division by zero is not allowed. Therefore, the excluded value is x = 4.
The correct answer is:
A. x; where x ≠ 4
What are the minimum, first quartile, median, third quartile, and maximum of the data set?
18, 20, 11, 10, 8, 6, 12, 4
First, we need to order the data set from smallest to largest:
4, 6, 8, 10, 11, 12, 18, 20
The minimum is 4 and the maximum is 20.
To find the median, we need to find the middle value. Since there are 8 values in the data set, the middle value is the average of the 4th and 5th values:
median = (10 + 11)/2 = 10.5
To find the quartiles, we first need to find the median of the lower and upper halves of the data set.
Lower half: 4, 6, 8, 10
Upper half: 11, 12, 18, 20
The first quartile (Q1) is the median of the lower half, which is the average of the 2nd and 3rd values:
Q1 = (6 + 8)/2 = 7
The third quartile (Q3) is the median of the upper half, which is the average of the 6th and 7th values:
Q3 = (12 + 18)/2 = 15
Therefore, the minimum is 4, the first quartile is 7, the median is 10.5, the third quartile is 15, and the maximum is 20.
The number of hours a group of contestants spends preparing for a quiz show are listed below. What is the frequency table that represents that data?
8 12 25 32 45 50 62 73 80 99 4 18 9 39 36 67 33
To make a frequency table, we need to count how many times each value appears in the data set. Here are the counts for each value:
- 4: 1
- 8: 1
- 9: 1
- 12: 1
- 18: 1
- 25: 1
- 32: 1
- 33: 1
- 36: 1
- 39: 1
- 45: 1
- 50: 1
- 62: 1
- 67: 1
- 73: 1
- 80: 1
- 99: 1
We can put these counts into a table with two columns: one for the values in the data set and one for the corresponding frequencies. The completed frequency table is:
| Value | Frequency |
|-------|-----------|
| 4 | 1 |
| 8 | 1 |
| 9 | 1 |
| 12 | 1 |
| 18 | 1 |
| 25 | 1 |
| 32 | 1 |
| 33 | 1 |
| 36 | 1 |
| 39 | 1 |
| 45 | 1 |
| 50 | 1 |
| 62 | 1 |
| 67 | 1 |
| 73 | 1 |
| 80 | 1 |
| 99 | 1 |
The time t required to drive a certain distance varies inversely with the speed r. If it takes 4 hours to drive the distance at 35 miles per hour, how long will it take to drive the same distance at 45 miles per hour?
If the time t required to drive a certain distance varies inversely with the speed r, we can write:
tr = k
where k is a constant of proportionality. We can use the given information to find the value of k. If it takes 4 hours to drive the distance at 35 miles per hour, we have:
(4)(35) = k
Simplifying the equation above, we get:
k = 140
So the equation for this inverse variation is:
tr = 140
To find the time required to drive the same distance at 45 miles per hour, we can use this equation and solve for t, when r = 45:
(4)(35) = (t)(45)
Simplifying this equation, we get:
t = (4)(35) / 45 = 3.11 hours
Therefore, it will take approximately 3.11 hours to drive the same distance at 45 miles per hour.
What are the solutions to the system?
y = x2 + 5x – 9
y = 2x + 1
A. (2, –9) and (–5, 5)
B. (2, 5) and (–5, –5)
C. (2, 5) and (–5, –9)
D. no solution
Since both equations are set equal to y, we can set them equal to each other to get:
x^2 + 5x - 9 = 2x + 1
Simplifying this equation by subtracting 2x and 1 from both sides, we get:
x^2 + 3x - 10 = 0
We can factor the left-hand side of this equation to get:
(x + 5)(x - 2) = 0
Therefore, the solutions to the system are the x-values that make this equation true, which are x = -5 and x = 2. To find the corresponding y-values, we can substitute these values into either equation; let's use the second equation:
y = 2x + 1
When x = -5, we get y = -9. When x = 2, we get y = 5. Therefore, the solutions to the system are:
(-5, -9) and (2, 5)
The correct answer is:
C. (2, 5) and (–5, –9)
A model rocket is launched from a roof into a large field. The path of the rocket can be modeled by the equation y = –0.04x2+ 8.3x + 4.3, where x is the horizontal distance, in meters, from the starting point on the roof and y is the height, in meters, of the rocket above the ground. How far horizontally from its starting point will the rocket land?
The rocket will land when y = 0, since at that point the rocket will have returned to ground level. We can use the given equation to find the value of x when y = 0:
0 = –0.04x^2 + 8.3x + 4.3
To solve this quadratic equation, we can either use the quadratic formula or factor out the greatest common factor of –0.04. Factoring gives:
0 = –0.04(x^2 - 207.5x - 107.5)
Now we can find the roots of the equation:
x^2 - 207.5x - 107.5 = 0
Using the quadratic formula, we get:
x = [207.5 ± √(207.5^2 + 4(1)(107.5))]/(2)
Simplifying this expression, we get:
x = [207.5 ± 61.78]/(2)
Therefore, the two possible solutions for x are:
x = 134.64 or x = 72.86
Since x represents a horizontal distance, we can discard the negative solution and conclude that the rocket will land approximately 134.64 meters horizontally from its starting point on the roof.
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 = 16x + 150
Play B: y = –x2 + 60x – 10
A. The attendance was the same on day 40. The attendance was 790 at both plays that day.
B. The attendance was the same on day 4. The attendance was 214 at both plays that day.
C. The attendance was the same on days 4 and 40. The attendance at both plays on those days was 214 and 790 respectively.
D. The attendance was never the same at both plays
To find the day(s) when the attendance was the same at both plays, we need to set the two equations equal to each other and solve for x:
16x + 150 = –x^2 + 60x – 10
Rearranging and simplifying this equation gives:
x^2 + 44x - 160 = 0
We can solve for x by factoring the left-hand side of this equation:
(x + 10)(x - 16) = 0
Therefore, the two solutions are x = -10 and x = 16, but since we are looking for the number of days since opening night, we should discard the negative solution.
So, the attendance was the same on day 16. To find the attendance on that day, we can substitute x = 16 into either of the equations and solve for y:
y = 16x + 150 = 16(16) + 150 = 406
Therefore, the attendance at both plays on day 16 was 406.
The correct answer is:
A. The attendance was the same on day 16. The attendance was 406 at both plays that day.
Good grief:
Suppose that the amount of algae in a pond triples every 2 hours. If the pond initially contains 80 pounds of algae, how much algae will be in the pond after 8 hours?
3^4 = 81
81 * 80 = 6480
The answer is correct.
Starting with the initial amount of 80 pounds of algae, we know that the amount of algae in the pond triples every 2 hours. After 2 hours, the pond will contain 3 times 80 pounds, or 240 pounds of algae. After another 2 hours, the amount of algae will triple again, giving a total of 3 times 240 pounds, or 720 pounds. After 6 hours, the amount of algae will triple once more, giving 3 times 720 pounds, or 2160 pounds. Finally, after 8 hours, the amount of algae will triple once more, giving a total of 3 times 2160 pounds, or 6480 pounds of algae.
Therefore, there will be 6480 pounds of algae in the pond after 8 hours.