Think of a vertical number line. The positive values count down as we move down this number line. The negative numbers count up, so to speak, as we move down the number line. So we have -1, -2, -3, ... as we move down. This indicates that -5 is <u>not</u> smaller than -7.
Diagram is below.
Answer:
I think it's 4/10 which is 0.4
Step-by-step explanation:
Answer:
1/4 divided by 5 = 0.5 or 1/2
Step-by-step explanation:
Answer:
17) MC(x) = 35 − 12/x²
18) R(x) = -0.05x² + 80x
Step-by-step explanation:
17) The marginal average cost function (MC) is the derivative of the average cost function (AC).
AC(x) = C(x) / x
MC(x) = d/dx AC(x)
First, find the average cost function:
AC(x) = C(x) / x
AC(x) = (5x + 3)(7x + 4) / x
AC(x) = (35x² + 41x + 12) / x
AC(x) = 35x + 41 + 12/x
Now find the marginal average cost function:
MC(x) = d/dx AC(x)
MC(x) = 35 − 12/x²
18) x is the demand, and p(x) is the price at that demand. Assuming the equation is linear, let's use the points to find the slope:
m = (40 − 50) / (800 − 600)
m = -0.05
Use point-slope form to find the equation of the line:
p(x) − 50 = -0.05 (x − 600)
p(x) − 50 = -0.05x + 30
p(x) = -0.05x + 80
The revenue is the product of price and demand:
R(x) = x p(x)
R(x) = x (-0.05x + 80)
R(x) = -0.05x² + 80x
Step-by-step explanation:
Given the linear equation, y = ⅔x + 1, where the <u>slope</u>, m = ⅔, and the y-intercept, (0, 1) where<em> b</em> = 1.
<h3><u>Start at the y-intercept:</u></h3>
In order to graph the given linear equation, start by plotting the coordinates of the y-intercept, (0, 1). As we know, the <u>y-intercept</u> is the point on the graph where it crosses the y-axis. It coordinates are (0, <em>b</em>), for which the value of b represents the value of the y-intercept in slope-intercept form, y = mx + b.
<h3><u>Plot other points using the slope:</u></h3>
From the y-intercept, (0, 1), we must use the slope, m = ⅔ (<em>rise</em> 2, <em>run</em> 3) to plot the other points on the graph. Continue the process until you have sufficient amount of plotted points on the graph that you could connect a line with.
Attached is a screenshot of the graphed linear equestion, which demonstrates how I plotted the other points on the graph using the "rise/run" techniques" discussed in the previous section of this post.