On the 1st place he can put 6 textbooks, on the 2nd 5 textbooks etc...
6 · 5 · 4 · 3 = 360
Answer: D) He can arrange textbooks in 360 ways.
Given:
A circle of radius r inscribed in a square.
To find:
The expression for the area of the shaded region.
Solution:
Area of a circle is:

Where, r is the radius of the circle.
Area of a square is:

Where, a is the side of the square.
A circle of radius r inscribed in a square. So, diameter of the circle is equal to the side of the square.

So, the area of the square is:


Now, the area of the shaded region is the difference between the area of the square and the area of the circle.




Therefore, the correct option is (a).
xn+yn=zn where n>2 and n is an integer
<h2>bolded letters are exponents!!!!</h2>
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
<h3>application</h3>
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.
Given

Recall that the general geometric series representation is given by

Let

Thus,

Multiply both sides by t, to get

Integrate both sides to get