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Angelina_Jolie [31]
2 years ago
11

I keep putting in the formula but I keep getting the answer wrong​

Mathematics
1 answer:
Advocard [28]2 years ago
8 0

Answer:

314 in²  (nearest whole number)

Step-by-step explanation:

<u>Radius of a regular polygon</u>: The distance from the <u>center</u> of the polygon to any vertex.  The radius of a hexagon is equal to the length of one side.

Therefore, from inspection of the given diagram:

  • radius = 11 in  ⇒  side length = 11 in

To find the area of a regular polygon, we first need to calculate the apothem.   The <u>apothem</u> is the line drawn from the center of the polygon to the midpoint of one of its sides.

\textsf{Length of apothem (a)}=\dfrac{s}{2 \tan\left(\frac{180^{\circ}}{n}\right)}

where:

  • s = length of one side
  • n = number of sides

Given:

  • s = 11 in
  • n = 6

Substitute the given values into the formula and solve for a:

\implies \textsf{a}=\dfrac{11}{2 \tan\left(\frac{180^{\circ}}{6}\right)}=\dfrac{11\sqrt{3}}{2}

<u>Area of a Regular Polygon</u>

\textsf{A}=\dfrac{n\:s\:a}{2}

where:

  • n = number of sides
  • s = length of one side
  • a = apothem

Given:

  • n = 6
  • s = 11
  • \textsf{a}=\dfrac{11\sqrt{3}}{2}

Substitute the given values into the formula and solve for A:

\implies \sf A=\dfrac{6 \cdot 11 \cdot \dfrac{11\sqrt{3}}{2}}{2}

\implies \sf A=314.3672216...

\implies \sf A=314\:\:in^2\:\:(nearest\:whole\:number)

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Answer:

\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}

Step-by-step explanation:

<u>The Derivative of a Function</u>

The derivative of f, also known as the instantaneous rate of change, or the slope of the tangent line to the graph of f, can be computed by the definition formula

\displaystyle f'(x)=\lim\limits_{\Delta x \rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}

There are tables where the derivative of all known functions are provided for an easy calculation of specific functions.

The derivative of the inverse tangent is given as

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Where u is a function of x as provided:

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u=(x+\sqrt{1+x^2})

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\displaystyle y'=3[tan^{-1}u]'=3\frac{u'}{1+u^2}

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\boxed{\displaystyle y'=3\frac{1+\frac{x}{\sqrt{1+x^2}}}{2+2x^2+2x\sqrt{1+x^2}}}

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3 years ago
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Answer: The solution is the ordered pair (1/2,  -3/4) so x = 1/2 and y = -3/4 pair up together. The two lines, when graphed, cross at this location.

note: 1/2 = 0.5 and -3/4 = -0.75; so the intersection point can be written as (0.5, -0.75)

===================================================

Explanation:

The second equation has y isolated. We can replace the y in the first equation with the expression 1/2x - 1

3x + 2y = 0

3x + 2( y ) = 0

3x + 2( 1/2x - 1) = 0 ... y is replaced with 1/2x-1

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4x - 2 = 0

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Use this x value to find the y value it pairs with

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y = 1/4 - 1

y = 1/4 - 4/4

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SashulF [63]
<h3><u>The value of the larger number, x, is 57.</u></h3><h3><u>The value of the smaller number, y, is equal to 42.</u></h3>

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We can quickly get a temporary value for x by altering the original equation.

x - y = 15

<em><u>Add y to both sides.</u></em>

x = 15 + y

Now that we have a value of x, we can find the exact value of y.

2(15 + y) + 8 = 3y - 4

<em><u>Distributive property.</u></em>

30 + 2y + 8 = 3y - 4

<em><u>Combine like terms.</u></em>

38 + 2y = 3y - 4

<em><u>Subtract 2y from both sides.</u></em>

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<em><u>Add 4 to both sides.</u></em>

y = 42

Now that we know the exact value of y, we can plug it back into the original equation.

x - 42 = 15

<em><u>Add 42 to both sides.</u></em>

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2 years ago
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