The graph which you seek is attached to this answer.
NO LINKS!!! Please help fill in the blanks. Part 7a.
1 answer:
<h3>Answer: Check out the screenshots below.</h3>
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Explanations:
Part 1
For the first box, we use the log rule that log(A)+log(B) = log(A*B)
Then in the second box, we'll convert to exponential form. The logs are assumed to be base 10.
The third box then factors and uses the zero product property.
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Part 2
We check each solution generated in part 1.
Plugging x = 5/3 will lead to the same number on each side. Therefore, x = 5/3 is a true solution.
In contrast, plugging x = -2 leads to a false equation. Recall that the domain of y = log(x) is x > 0. This means we cannot replace x with negative numbers. The value x = -2 is extraneous.
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Part 3
There's not much to explain here that isn't already done so on the screenshot below.
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Answer:
(d) m∠AEB = m∠ADB
Step-by-step explanation:
The question is asking you to compare the measures of two inscribed angles. Each of the inscribed angles intercepts the circle at points A and B, which are the endpoints of a diameter.
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<h3>applicable relations</h3>Several relations are involved here.
- The measures of the arcs of a circle total 360°
- A diameter cuts a circle into two congruent semicircles
- The measure of an inscribed angle is half the measure of the arc it intercepts
In the attached diagram, we have shown inscribed angle ADB in blue. The semicircular arc it intercepts is also shown in blue. A semicircle is half a circle, so its arc measure is half of 360°. Arc AEB is 180°. That means inscribed angle ADB measures half of 180°, or 90°. (It is shown as a right angle on the diagram.)
If Brenda draws angle AEB, it would look like the angle shown in red on the diagram. It intercepts semicircular arc ADB, which has a measure of 180°. So, angle AEB will be half that, or 180°/2 = 90°.
The question is asking you to recognize that ∠ADB = 90° and ∠AEB = 90° have the same measure.
m∠AEB = m∠ADB
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<em>Additional comment</em>
Every angle inscribed in a semicircle is a right angle. The center of the semicircle is the midpoint of the hypotenuse of the right triangle. This fact turns out to be useful in many ways.
