So if you do 8x14=112 is the area. So 112 divided by 40= 2.8. He will need 2.8 pints of paint.
<h3>perimetr:216</h3><h3>area:60 </h3>
In order to find out the association, you need to plot the data. Given that m is x, and n is y, after plotting, the graph shows a positive linear association. The graph rises from left to right. So the answer is letter D.
In Functions and Function Notation, we were introduced to the concepts of domain and range. In this section, we will practice determining domains and ranges for specific functions. Keep in mind that, in determining domains and ranges, we need to consider what is physically possible or meaningful in real-world examples, such as tickets sales and year in the horror movie example above. We also need to consider what is mathematically permitted. For example, we cannot include any input value that leads us to take an even root of a negative number if the domain and range consist of real numbers. Or in a function expressed as a formula, we cannot include any input value in the domain that would lead us to divide by 0.
Explanation:
Any calibration scale consists of markings indicating calibrated values. The ‘space’ between the marks (lines) is the area of uncertainty with respect to the calibration.
Thus, the possible error is always one-half of the value between the markings, because ON either one you have a calibrated value. In between, no matter how close you think you can “judge” the distance, there is no calibrated reference point, so the ‘error’ of stating a value is +/- the value of half of the calibration accuracy. 0.991 is accurate (assuming that is the calibration limit), and 0.992 or 0.990 would also be “accurate”. The possible error is the +/- 0.0005 beyond that third digit that might be more to one side or the other.
That means the measured value of 0.991g could be between 0.9905g and 0.9915g.
