Answer:
Interest: $50 | Total: $250
Step-by-step explanation:
You are going to want to use the simple interest formula for this. The one below is modified for solving the interest earned.
<em>I = interest amount </em>
<em>P = principal amount </em>
<em>r = interest rate (decimal form) </em>
<em>t = time</em>
<em />
Now we can plug in the values into the equation:
This means that Ariane is charged $50 worth of interest. She has to pay back $250 total dollars.
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:
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:
I don't understand the pattern
