An electric device delivers a current of [tex]$15.0 A$[/tex] for 30 seconds. How many electrons flow through it?
Answer (2)
Expand the squared term: ( x + 5 ) 2 = x 2 + 10 x + 25 .
Substitute the expanded term back into the equation: y = ( x 2 + 10 x + 25 ) + 49 .
Combine the constant terms: y = x 2 + 10 x + 74 .
The standard form of the equation is y = x 2 + 10 x + 74 .
Explanation
Understanding the Problem We are given the vertex form of a parabola's equation: y = ( x + 5 ) 2 + 49 . Our goal is to convert this equation into standard form, which is y = a x 2 + b x + c .
Expanding the Squared Term To convert from vertex form to standard form, we need to expand and simplify the given equation. First, let's expand the squared term ( x + 5 ) 2 :
( x + 5 ) 2 = ( x + 5 ) ( x + 5 ) = x 2 + 5 x + 5 x + 25 = x 2 + 10 x + 25
Substituting Back into the Equation Now, substitute this back into the original equation: y = ( x 2 + 10 x + 25 ) + 49
Simplifying the Equation Finally, combine the constant terms to simplify the equation: y = x 2 + 10 x + ( 25 + 49 ) = x 2 + 10 x + 74 So, the standard form of the equation is y = x 2 + 10 x + 74 .
Identifying the Correct Option Comparing our result to the given options, we see that it matches option C.
Examples
Understanding quadratic functions and their transformations is crucial in various fields. For instance, engineers use parabolas to design arches and bridges, ensuring structural stability. Similarly, in physics, projectile motion follows a parabolic path, allowing us to predict the trajectory of objects, like a ball thrown in the air. By converting between vertex and standard forms, we can easily identify key features of the parabola, such as the vertex and y-intercept, which are essential for these applications. For example, knowing the vertex helps determine the maximum height of a projectile or the optimal placement of a satellite dish.
When a current of 15.0 A flows for 30 seconds, a total charge of 450.0 C flows through the device. This charge corresponds to approximately 2.81 × 1 0 21 electrons. Thus, 2.81 × 1 0 21 electrons flow through the device in that time.
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