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klemol [59]
3 years ago
5

Devon is trying to find the unit price on a 6-pack of drinks on sale for $2.99. his sister says that at that price,each drink wo

ykd cost just over $2.00 . is she correct,and how do you now? if she is not, how would devon's sister find the correct price?
Mathematics
1 answer:
tresset_1 [31]3 years ago
6 0
It is not true

6pack = 6 pieces ? 

6 pieces for $2.00 = $12   not $2.99

Devon should divide $2.99/6  ~~  $0.49


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o-na [289]

so hmmm let's get the area of the whole hexagon, and then get the area of the circle inside it, then <u>subtract the area of the circle from that of the hexagon's</u>, what's leftover is what we didn't subtract, namely the shaded part.

\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\cot\stackrel{\stackrel{degrees}{\downarrow }}{\left( \frac{180}{n} \right)}~ \begin{cases} n=\textit{number of sides}\\ s=\textit{length of a side}\\[-0.5em] \hrulefill\\ n=\stackrel{hexagon}{6}\\ s=\frac{9}{2} \end{cases}\implies A=\cfrac{1}{4}(6)\left( \cfrac{9}{2} \right)^2 \cot\left( \cfrac{180}{6} \right)

A=\cfrac{1}{4}(6)\cfrac{9^2}{2^2} \cot(30^o)\implies A=\cfrac{243}{8}\cot(30^o)\implies A=\cfrac{243\sqrt{3}}{8} \\\\[-0.35em] ~\dotfill\\\\ \textit{area of circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=\frac{4}{5} \end{cases}\implies A=\pi \left( \cfrac{4}{5} \right)^2\implies A=\cfrac{16\pi }{25} \\\\[-0.35em] ~\dotfill

\stackrel{\textit{area of the hexagon}}{\cfrac{243\sqrt{3}}{8}}~~ - ~~\stackrel{\textit{area of the circle}}{\cfrac{16\pi }{25}}\implies \cfrac{6075\sqrt{3}-128\pi }{200}

5 0
2 years ago
In the pulley system shown in this figure, MQ = 30 mm, NP = 10 mm, and QP = 21 mm. Find MN.
ankoles [38]
<h2>Answer:</h2>

\boxed{\overline{MN}=37.96}

<h2>Step-by-step explanation:</h2>

For a better understanding of this problem, see the figure below. Our goal is to find \overline{MN}. Since:

\angle MRS=\angle MQP=90^{\circ} \\ \\ \overline{MQ}=\overline{MR}=30mm

and \overline{MN} is a common side both for ΔMRN and ΔMQN, then by SAS postulate, these two triangles are congruent and:

\overline{RN}=\overline{QN}

By Pythagorean theorem, for triangle NQP:

\overline{QN}=\sqrt{\overline{NP}^2+\overline{QP}^2} \\ \\ \overline{QN}=\sqrt{10^2+21^2} \\ \\ \overline{QN}=\sqrt{541}

Applying Pythagorean theorem again, but for triangle MQN:

\overline{MN}=\sqrt{\overline{MQ}^2+\overline{QN}^2} \\ \\ \overline{MN}=\sqrt{30^2+(\sqrt{541})^2} \\ \\ \boxed{\overline{MN}=37.96}

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