A situation which best describes a student that is engaged in active learning is: C. a student records questions to ask the teacher about a concept.
<h3>What is a question?</h3>
A question can be defined as a group of words or sentence that are developed, so as to elicit an information in the form of answer from an individual such as a student or teacher.
This ultimately implies that, questions are primarily formulated by professionals to test the understanding of an individual. Also, questions should be formulated by a good reader or student so as to enable him or her connect to a story and understand a concept better.
In this context, we can reasonably infer and logically deduce that a situation which best describes a student that is engaged in active learning is when a student records questions to ask the teacher about a concept.
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Complete Question:
Which of the following situations best describes a student engaged in active learning?
A. A student uses only one of the resources provided by the teacher.
B. A student skips parts of the lesson that she finds difficult.
C. A student records questions to ask the teacher about a concept.
D. A student does not discuss math outside of the lesson.
<em>Answer:</em>
<h3><em>1. Check mirrors</em></h3><h3><em>2. Put on your seat belt</em></h3>
<em>Explanation:</em>
<em>1. Checking your mirrors are very important because if someone screwed with them then it can mess up your driving. </em>
<em />
<em>2. Putting on your seat belt is a law so you must put it on and it can save your life one day. </em>
Answer:
248.756 mV
49.7265 µA
Explanation:
The Thevenin equivalent source at one terminal of the bridge is ...
voltage: (100 V)(1000/(1000 +1000) = 50 V
impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω
The Thevenin equivalent source at the other terminal of the bridge is ...
voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V
impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω
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The open-circuit voltage is the difference between these terminal voltages:
(50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage
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The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:
(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA
