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pickupchik [31]
2 years ago
10

Three students, Alicia, Benjamin, and Caleb, are constructing a square inscribed in a circle with center at point C. Alicia draw

s two diameters and connects the points where the diameters intersect the circle, in order, around the circle. Benjamin tells Alicia that she was not being very accurate. He says that the diameters must be perpendicular to each other. Then she can connect the points, in order, around the circle. Caleb tells Alicia and Benjamin that he doesn't need to draw the second diameter. He says that because a triangle inscribed in a semicircle is a right triangle, he will simply draw two such triangles, one in each semicircle. Together the two triangles will make a square. Who is correct?​
Mathematics
1 answer:
True [87]2 years ago
6 0

Benjamin is correct about the diameter being perpendicular to each other and the points connected around the circle.

<h3>Inscribing a square</h3>

The steps involved in inscribing a square in a circle include;

  • A diameter of the circle is drawn.
  • A perpendicular bisector of the diameter is drawn using the method described as the perpendicular of the line sector. Also known as the diameter of the circle.
  • The resulting four points on the circle are the vertices of the inscribed square.

Alicia deductions were;

Draws two diameters and connects the points where the diameters intersect the circle, in order, around the circle

Benjamin's deductions;

The diameters must be perpendicular to each other. Then connect the points, in order, around the circle

Caleb's deduction;

No need to draw the second diameter. A triangle when inscribed in a semicircle is a right triangle, forms semicircles, one in each semicircle. Together the two triangles will make a square.

It can be concluded from their different postulations that Benjamin is correct because the diameter must be perpendicular to each other and the points connected around the circle to form a square.

Thus, Benjamin is correct about the diameter being perpendicular to each other and the points connected around the circle.

Learn more about an inscribed square here:

brainly.com/question/2458205

#SPJ1

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aleksandr82 [10.1K]
Since it says m1 is 120 and it says find m3 don’t you need to multiply or I’m wrong
6 0
3 years ago
The curve
kherson [118]

Answer:

Point N(4, 1)

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Algebra I</u>

  • Coordinates (x, y)
  • Functions
  • Function Notation
  • Terms/Coefficients
  • Anything to the 0th power is 1
  • Exponential Rule [Rewrite]:                                                                              \displaystyle b^{-m} = \frac{1}{b^m}
  • Exponential Rule [Root Rewrite]:                                                                     \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

<u>Calculus</u>

Derivatives

Derivative Notation

Derivative of a constant is 0

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Derivative Rule [Chain Rule]:                                                                                    \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle y = \sqrt{x - 3}<u />

<u />\displaystyle y' = \frac{1}{2}<u />

<u />

<u>Step 2: Differentiate</u>

  1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                                   \displaystyle y = (x - 3)^{\frac{1}{2}}
  2. Chain Rule:                                                                                                        \displaystyle y' = \frac{d}{dx}[(x - 3)^{\frac{1}{2}}] \cdot \frac{d}{dx}[x - 3]
  3. Basic Power Rule:                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{\frac{1}{2} - 1} \cdot (1 \cdot x^{1 - 1} - 0)
  4. Simplify:                                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}} \cdot 1
  5. Multiply:                                                                                                             \displaystyle y' = \frac{1}{2}(x - 3)^{-\frac{1}{2}}
  6. [Derivative] Rewrite [Exponential Rule - Rewrite]:                                          \displaystyle y' = \frac{1}{2(x - 3)^{\frac{1}{2}}}
  7. [Derivative] Rewrite [Exponential Rule - Root Rewrite]:                                 \displaystyle y' = \frac{1}{2\sqrt{x - 3}}

<u>Step 3: Solve</u>

<em>Find coordinates</em>

<em />

<em>x-coordinate</em>

  1. Substitute in <em>y'</em> [Derivative]:                                                                             \displaystyle \frac{1}{2} = \frac{1}{2\sqrt{x - 3}}
  2. [Multiplication Property of Equality] Multiply 2 on both sides:                      \displaystyle 1 = \frac{1}{\sqrt{x - 3}}
  3. [Multiplication Property of Equality] Multiply √(x - 3) on both sides:            \displaystyle \sqrt{x - 3} = 1
  4. [Equality Property] Square both sides:                                                           \displaystyle x - 3 = 1
  5. [Addition Property of Equality] Add 3 on both sides:                                    \displaystyle x = 4

<em>y-coordinate</em>

  1. Substitute in <em>x</em> [Function]:                                                                                \displaystyle y = \sqrt{4 - 3}
  2. [√Radical] Subtract:                                                                                          \displaystyle y = \sqrt{1}
  3. [√Radical] Evaluate:                                                                                         \displaystyle y = 1

∴ Coordinates of Point N is (4, 1).

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e

4 0
2 years ago
Question below ------------------------------
Ludmilka [50]

Answer:

<h2>The first one is the answer</h2>

Step-by-step explanation:

Hope im correct can you brainliest me?

5 0
2 years ago
Fin the area of the polygons
german

Answer:45

Step-by-step explanation:

First, you need to calculate the area of the square.

(LxW)= 25.

Then, calculate the area of the triangle

(1/2BH)=5x(8/2)=5x4=20

Now, just add 20+25= 45

Please mark me brainliest :)

5 0
2 years ago
A box has five items, three good and two defective. three items are selected at random without replacement. let x be the number
user100 [1]
This problem can be solved from first principles, case by case.  However, it can be solved systematically using the hypergeometric distribution, based on the characteristics of the problem:
- known number of defective and non-defective items.
- no replacement
- known number of items selected.

Let
a=number of defective items selected
A=total number of defective items
b=number of non-defective items selected
B=total number of non-defective items
Then 
P(a,b)=C(A,a)C(B,b)/C(A+B,a+b)
where 
C(n,r)=combination of r items selected from n,
A+B=total number of items
a+b=number of items selected

Given:
A=2
B=3
a+b=3
PMF:
P(0,3)=C(2,0)C(3,3)/C(5,3)=1*1/10=1/10
P(1,2)=C(2,1)C(3,2)/C(5,3)=2*3/10=6/10
P(2,0)=C(2,2)C(3,1)/C(5,3)=1*3/10=3/10
Check: (1+6+3)/10=1 ok
note: there are only two defectives, so the possible values of x are {0,1,2}

Therefore the 
PMF:
{(0, 0.1),(1, 0.6),(2, 0.3)}

7 0
2 years ago
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