Answer:
e=15.8
Step-by-step explanation:
Well we use the law of cosines
e^2=11^2+6^2-2(11)(6)cos(134)
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2 answers:
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Answer:
1. Domain (-∞, ∞)
Range: (-20, ∞)
x-intercepts: (-5, 0), and (5, 0)
y-intercepts: (0, -25)
Interval positive: (-∞, -5), and (5, ∞)
Interval negative: (-5, 5)
Interval increasing: (0, ∞)
Interval decreasing: (-∞, 0)
Given that the shape is that of a quadratic graph, we have;
f(x) = (x - 5)×(x + 5) = x² - 25 = 0
Where;
The 5s are from the x-intercept
The slope = dy/dx = d(x² - 25)dx = 2·x
The slope at the point (-2, 0), therefore, x is -2, we have;
Slope = 2·x = 2 × x = 2 × (-2) = -4
The slope at the point (-2, 0) = -4
The rate of change over the interval = (-20 - (-25))/(-2 - 0) = -2.5
The rate of change over the interval = -2.5
2) Domain (-∞, ∞)
Range: (3, ∞)
x-intercepts: N/A
y-intercepts: (0, 7)
Interval positive: (-∞, ∞)
Interval negative: NA
Interval increasing: (-2, ∞)
Interval decreasing: (-∞, -2)
Given that the shape at (-2, 0) is that of a straight line graph, we have;
Rate of change = dy/dx = (7 - 3)/(0 - (-2)) = 2
The rate of change over the interval (-2, 0) = 2
3) Please find attached, the required graph drawn using Microsoft Excel
4) The key features are;
Domain: (0, 50/0.5) = (0, 100)
Range: (0, 50)
x-intercepts: (0, 0)
y-intercepts: (0, 0)
Interval positive: (0, 100)
Interval negative: N/A
Interval increasing: (0, 100)
Interval decreasing: N/A
The rate of change over the interval (-2, 0) = 0.5 ft³/min
Step-by-step explanation:
The domain is the range of values of the input (x) variables
The range is the range of values of the output (y) variables
The x-intercepts. are the points where the graph crosses the x-axis
The y-intercepts. are the points where the graph crosses the y-axis
The interval positive is the range of values over which the y-values are larger than 0
The interval negative is the range of values over which the y-values are lesser than 0
The interval increasing is the range of values over which the y-values are continuously increasing
The interval decreasing is the range of values over which the y-values are continuously decreasing
The rate of change over an interval is the ratio of the change in the y-values to the corresponding change in the x-values, over the range.
