At an extracellular of 1.3 mM, a negative slope was observed at positive potentials. Once students start dealing with linear functions in Grade 8 and beyond, they become familiar with the meaning of negative slope and negative rate of change, and they can use those terms to describe functions of the form f ( x) k x with k < 0. In 90 mM Na+, 5 mM K+ solution, Na+/K+ pump current-voltage curves at negative membrane potentials have a positive slope and can be described by a monotonically increasing sigmoidal function. The occurrence of a negative slope can be explained by the voltage dependence of the apparent affinity for activation of the Na+/K+ pump by external K+, consistent with the existence of an external ion well for K+ binding. Similar results are seen in Na(+)- and Ca(2+)-free solutions in the presence of 2 mM Ni2+, an experimental condition designed to prevent Na+/Ca2+ exchange. If external is reduced below 0.5 mM, negative slopes are observed over this entire voltage range. In Na(+)-free solution containing 5 mM K+, Na+/K+ pump current is relatively voltage independent over the potential range from -160 to +40 mV. Its such a short driveway, I can fake it in, but I would like to calculate quantities. I have tried numerous approaches and all cut reverts to positive. For cut, I would like to have a negative slope away from the shoulder edge. Reductions in steady-state current (on the order of 700 nA/cm2) produced by 50 microM ouabain or dihydro-ouabain or by K+ removal, therefore, primarily represent current generated by the Na+/K+ pump. I have a simple driveway corridor along a steep hillside. If, in addition, the Na+/K+ pump is blocked by ouabain, K(+)-sensitive currents no larger than 50 nA/cm2 remain. All solutions contained 5 mM Ba2+ and 20 mM tetraethylammonium (TEA) to block K+ channels. If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s.To investigate the voltage dependence of the Na+/K+ pump, current-voltage relations were determined in prophase-arrested oocytes of Xenopus laevis. If the object is moving with a velocity of -8 m/s, then the slope of the line will be -8 m/s. Find step-by-step Chemistry solutions and your answer to the following textbook question: If a linear graph has a negative slope, what can you say about the. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s. The principle is that the slope of the line on a position-time graph is equal to the velocity of the object. That is, the slope of the line is 0 m/s - the same as the velocity during this time interval.īoth of these examples reveal an important principle. During the last 5 seconds (5 to 10 seconds), the line slopes up 0 meters. However, I do increase the risk-free rate from basically zero to higher values. Thus, the slope of the line on the graph equals the velocity of the car. I have no issues calculating the efficient frontier. time graph has a slope of +5 meters/1 second for the first five seconds. For the first five seconds the line on the graph slopes up 5 meters for every 1 second along the horizontal (time) axis.
If the position-time data for such a car were graphed, then the resulting graph would look like the graph at the right.
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