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Gennadij [26K]
3 years ago
9

Two cables were attached to the top of a pole and anchored to the ground on opposite sides of the pole. What is the length of th

e pole? Do not include units. Round to the nearest tenth of a foot.

Mathematics
1 answer:
poizon [28]3 years ago
7 0
This is a 16.2ft tall pole
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Bob wants to create two pens, as shown in the figure. One pen is for a garden and it needs a heavy duty fence to keep out the cr
sattari [20]
\bf \textit{the area of either pen is 2744, thus}\\\\
A=xy\implies 2744=xy\implies \cfrac{2744}{x}=y

\bf \begin{array}{llll}
\textit{perimeter of the garden pen}\\\\
P=2(x+y)\\\\
P=2\left( x+ \cfrac{2744}{x}\right)\\\\
P=2x+\cfrac{5488}{x}\\
----------\\

\textit{cost of it}\\\\
C=15\left( 2x+\cfrac{5488}{x} \right)\\\\
C=30x+\cfrac{82320}{x}
\end{array}\qquad 
\begin{array}{llll}
\textit{perimeter of the dog pen}\\\\
P=2x+y\\\\
P=2x+\cfrac{2744}{x}\\
----------\\
\textit{cost of it}\\\\
C=5\left( 2x+\cfrac{2744}{x} \right)\\\\
C=10x+\cfrac{27440}{x}
\end{array}

notice... the dog's pen perimeter, does not include the side that's bordering the garden's, since that side will use the heavy duty fence, instead of the light one

so, the sum of both of those costs, will be the C(x)

\bf C(x)=\left( 30x+\cfrac{82320}{x} \right)+\left( 10x+\cfrac{27440}{x} \right)
\\\\\\
C(x)=40x+\cfrac{109760}{x}

so, just take the derivative of it, and set it to 0 to find the extremas, and do a first-derivative test for any minimum

8 0
3 years ago
A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?
alexandr1967 [171]

Answer:

The probability that the last card dealt is an ace is \frac{1}{13}.

Step-by-step explanation:

Given : A deck of ordinary cards is shuffled and 13 cards are dealt.

To find : What is the probability that the last card dealt is an ace?

Solution :

There are total 52 cards.

The total arrangement of cards is 52!.

There is 4 ace cards in total.

Arrangement for containing ace as the 13th card is 4\times 51!.

The probability that the last card dealt is an ace is

P=\frac{4\times 51!}{52!}

P=\frac{4\times 51!}{52\times 51!}

P=\frac{4}{52}

P=\frac{1}{13}

Therefore, the probability that the last card dealt is an ace is \frac{1}{13}.

4 0
3 years ago
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