A rectangular piece of land is 73/3 meter long and 61/4 meter wide.Find it's perimeter. Please solve it
2 answers:
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First, let's write the given equation in slope-intercept form: y = mx + b
In slope-intercept form, the slope of the line is m, and the y-intercept is b. The slope is a measure of how steep the graph is at any point and is found by doing rise over run. This means the change in y values divided by the change in x values. Next, y-intercept is just where the graph crosses the y axis.
All we need to do to get the equation in slope-intercept form is to divide each term by 3. This will isolate the y.
As you can see, the slope of the line is 2/3, and the y-intercept is -2.
To graph the line, plot a point at (0,-2). This is the point where the graph crosses the y axis. Then from that point, count up two and right 3. Plot a point here as well. Lastly, connect the two points with a straight line.
See attached picture for the graph.
In slope-intercept form, the slope of the line is m, and the y-intercept is b. The slope is a measure of how steep the graph is at any point and is found by doing rise over run. This means the change in y values divided by the change in x values. Next, y-intercept is just where the graph crosses the y axis.
All we need to do to get the equation in slope-intercept form is to divide each term by 3. This will isolate the y.
As you can see, the slope of the line is 2/3, and the y-intercept is -2.
To graph the line, plot a point at (0,-2). This is the point where the graph crosses the y axis. Then from that point, count up two and right 3. Plot a point here as well. Lastly, connect the two points with a straight line.
See attached picture for the graph.
If you're asking for extrema, like the previous posting
well
like the previous posting, since this rational is identical, just that the denominator is negative, the denominator yields no critical points
and the numerator, yields no critical points either, so the only check you can do is for the endpoints, of 0 and 4
f(0) = 0 <---- only maximum, and thus absolute maximum
f(4) ≈ - 0.19 <---- only minimum, and thus absolute minimum
well
like the previous posting, since this rational is identical, just that the denominator is negative, the denominator yields no critical points
and the numerator, yields no critical points either, so the only check you can do is for the endpoints, of 0 and 4
f(0) = 0 <---- only maximum, and thus absolute maximum
f(4) ≈ - 0.19 <---- only minimum, and thus absolute minimum