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Margarita [4]
3 years ago
6

the October meeting of the yearbook club had 21 people attend. That was 3 times as many people as the number of people who came

to the september meeting. How many people were at the September meeting?
Mathematics
2 answers:
kolbaska11 [484]3 years ago
7 0
October meeting is 3 times more than the number of people at September meeting.
Sept meeting = 21 ÷ 3 = 7

7 people were at the September meeting.
frosja888 [35]3 years ago
4 0
You have to do 21x3=63
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20.2057 Units.

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Distance between point (x,0) and Springfield(0,7) is:

\sqrt{(7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}

Distance between point (x,0) and Shelvyfield(0,-7) is:

\sqrt{(-7-0)^2+(0-x)^{2}}=\sqrt{(7)^2+(x)^{2}}

Therefore the Length of the Cable L(x)

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To find the critical point, we set the derivative of L(x)=0

L^{'}(x)=-1+\frac{2x}{\sqrt{\left( 49 - x^{2}\right) }}

\frac{-\sqrt{\left( 49 - x^{2}\right)}+2x}{\sqrt{\left( 49 - x^{2}\right)}}=0\\-\sqrt{\left( 49 - x^{2}\right)}+2x=0\\\sqrt{\left( 49 - x^{2}\right)}=2x\\(\sqrt{\left( 49 - x^{2}\right)})^2=(2x)^2\\49 - x^{2}=4x^2\\49=5x^2\\x^2=\frac{49}{5}\\x= 3.1305

To verify that L(x) has a minimum at this critical number we compute the second derivative L″(x) and find its value at the critical number.

L^{''}(x)=\frac{\left( 98\right) \,\sqrt{\left( 49 - x^{2}\right) }}{{\left( 7 - x\right) }^{2}\,{\left( 7+x\right) }^{2}}\\At \:x=3.1305, L^{''}=0.3993

Since L^{''}(x)  is positive, the minimum point of L(x) exists.

Next, we find the minimum length by substituting z=3.1305 into L(x)

L(3.1305)=(8-3.1305)+2\sqrt{(7)^2+(3.1305)^{2}}

Minimum Length, L=20.2057

The minimum length of the cable is 20.2057 Units.

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3 years ago
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