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3 years ago

-What is the arc length when _ = 2 pi over 3 and the radius is 8 cm?

Mathematics

1 answer:

Brut [27]3 years ago

5 0

The answer is X= Rx Teta, X is the arc length
so the answer is <span>16 pi over 3cm

</span>

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a 60 foot tree casts a shadow 85ft long. the sine of the angle between the ground and the line that connects the tip of the shad

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The shadow to the top of the tree is 60 ft it all makes sense

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3 years ago

Help me please if you can

makvit [3.9K]

Answer:

-2

Step-by-step explanation:

Add them up: -14

Divide by how many there are: -14 / 7 = -2

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3 years ago

Anyoneee?? Mean Median and mode

juin [17]

Answer:

mode is the most occuring number which is in this case 11

you get the median by putting them in order from least to greatest and them marking them out until you get to the middle but if there are two numbers in the middle you add them and then divide by two and that is your median but in this case your median is 11

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3 years ago

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Burka [1]

Answer:

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Step-by-step explanation:

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3 years ago

You have D dollars to buy fence to enclose a rectangular plot of land (see figure at right). The fence for the top and bottom co

alex41 [277]

The perimeter of the rectangular plot of land is given by the expression below

$P=2x+2y$

On the other hand, since the available money to buy fence is D dollars,

$\begin{gathered} D=4(2x)+3(2y) \\ \Rightarrow D=8x+6y \\ D\rightarrow\text{ constant} \end{gathered}$

Furthermore, the area of the enclosed land is given by

$A=xy$

Solving the second equation for x,

$\begin{gathered} D=8x+6y \\ \Rightarrow x=\frac{D-6y}{8} \end{gathered}$

Substituting into the equation for the area,

$\begin{gathered} A=(\frac{D-6y}{8})y \\ \Rightarrow A=\frac{D}{8}y-\frac{3}{4}y^2 \end{gathered}$

To find the maximum possible area, solve A'(y)=0, as shown below

$\begin{gathered} A^{\prime}(y)=0 \\ \Rightarrow\frac{D}{8}-\frac{3}{2}y=0 \\ \Rightarrow\frac{3}{2}y=\frac{D}{8} \\ \Rightarrow y=\frac{D}{12} \end{gathered}$

Therefore, the corresponding value of x is

$\begin{gathered} y=\frac{D}{12} \\ \Rightarrow x=\frac{D-6(\frac{D}{12})}{8}=\frac{D-\frac{D}{2}}{8}=\frac{D}{16} \end{gathered}$ <h2>Thus, the dimensions of the fence that maximize the area are x=D/16 and y=D/12.</h2><h2>As for the used money,</h2> $\begin{gathered} top,bottom:\frac{8D}{16}=\frac{D}{2} \\ Sides:\frac{6D}{12}=\frac{D}{2} \end{gathered}$ <h2>Half the money was used for the top and the bottom, while the other half was used for the sides.</h2>

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1 year ago