How to solve system of equation from a table
1 answer:
Answer:
Look for the same entry in both (all) tables.
Step-by-step explanation:
We assume here that the system of equations consists of two equations in two variables. If there are more equations in more variables, the general approach is the same.
A "solution" to a system of equations is a set of variable values that satisfies all equations of the system simultaneously. A table for one equation will generally list sets of variable values that satisfy that equation. <em>When the same set of values appears in the table for each of the equations, then that set of values is the solution</em>.
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<u>Example</u>
The attachment shows tables for two equations:
- y = 2x
- y = x+3
Highlighted are the table entries that are the same for both equations. This is the solution to the system of equations. (x, y) = (3, 6) satisfies both equations:
- 6 = 2·3
- 6 = 3+3
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Answer:
About 0.6548 grams will be remaining.
Step-by-step explanation:
We can write an exponential function to model the situation. The standard exponential function is:
The original sample contained 510 grams. So, a = 510.
Each half-life, the amount decreases by half. So, r = 1/2.
For t, since one half-life occurs every 38 days, we can substitute t/38 for t, where t is the time in days.
Therefore, our function is:
One year has 365 days.
Therefore, the amount remaining after one year will be:
About 0.6548 grams will be remaining.
Alternatively, we can use the standard exponential growth/decay function modeled by:
The starting sample is 510. So, C = 510.
After one half-life (38 days), the remaining amount will be 255. Therefore:
Solving for k:
Thus, our function is:
Then after one year or 365 days, the amount remaining will be about: