The number of bacteria in a sample increases at a constant rate during an experiment. After 1 minute of the experiment, there are 50 bacteria. Six minutes later, there are 86 bacteria. What is the initial number of bacteria?
We can use the formula for exponential growth to solve this problem:
N = N0 * e^(rt)
where N is the final number of bacteria, N0 is the initial number of bacteria, r is the growth rate, and t is the time elapsed.
We know that after 1 minute, there are 50 bacteria, so we can write:
50 = N0 * e^(r*1)
Similarly, after 7 minutes (1 + 6), there are 86 bacteria:
86 = N0 * e^(r*7)
Dividing the second equation by the first, we get:
86/50 = e^(r*7) / e^(r*1)
Simplifying, we get:
1.72 = e^(r*6)
Taking the natural logarithm of both sides, we get:
ln(1.72) = 6r
Solving for r, we get:
r = ln(1.72) / 6 ≈ 0.093
Now we can use either of the first two equations to solve for N0. Let's use the first one:
50 = N0 * e^(0.093*1)
Solving for N0, we get:
N0 = 50 / e^(0.093*1) ≈ 46.36
Therefore, the initial number of bacteria was approximately 46.36.
The slope of a line is . If y = 5 when x = 4, what is the value of y when x = -4?
We can solve this problem using the point-slope form of the equation of a line:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is a point on the line. We are given that m = , and that the line passes through the point (4, 5):
y - 5 = (1/4)(x - 4)
Simplifying, we get:
y = (1/4)x + 4
Now we can substitute x = -4 and solve for y:
y = (1/4)(-4) + 4
y = 3
Therefore, when x = -4, y = 3.
Two lines have the same initial value of y = 10. One line has a slope of , and the other line has a slope of . What is the difference in the values of the two lines when x = 8?
We can write the equations of the two lines using the point-slope form:
y - 10 = (1/3)(x - 0) (line 1)
y - 10 = (1/2)(x - 0) (line 2)
Simplifying, we get:
y = (1/3)x + 10 (line 1)
y = (1/2)x + 10 (line 2)
To find the difference between the values of the two lines when x = 8, we can simply evaluate each equation at x = 8 and take the difference:
(line 1) y = (1/3)(8) + 10 = 13.67
(line 2) y = (1/2)(8) + 10 = 14
Difference = 14 - 13.67 = 0.33
Therefore, when x = 8, the two lines differ by 0.33 in their y-values.
Which of the following most accurately describes the graph of the line ?
(A)The y-intercept is above 0, and the line is rising as you move from left to right on the graph.
(B)The y-intercept is above 0, and the line is falling as you move from left to right on the graph.
(C)The y-intercept is below 0, and the line is rising as you move from left to right on the graph.
(D)The y-intercept is below 0, and the line is falling as you move from left to right on the graph.
The equation can be written in slope-intercept form as y = 2x - 3, where the slope is 2 and the y-intercept is -3.
Therefore, the y-intercept is below 0, which eliminates choices (A) and (B).
Since the slope is positive, the line is rising as you move from left to right on the graph. This eliminates choice (D).
The only option remaining is (C): "The y-intercept is below 0, and the line is rising as you move from left to right on the graph."
Find the equation of the line graphed on the coordinate plane.
(A)
(B)
(C)
(D)
To write the equation of a line, we need to know its slope and y-intercept. Looking at the graph, we can see that the line passes through the points (0, 2) and (4, 8). We can use these points to find the slope of the line:
slope = (change in y) / (change in x) = (8 - 2) / (4 - 0) = 6 / 4 = 3 / 2
We also know that the line passes through the y-axis at y = 2. This is the y-intercept.
Now we can use the slope-intercept form of the equation of a line, y = mx + b, where m is the slope and b is the y-intercept:
y = (3/2)x + 2
Therefore, (A) is the correct answer.