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aniked [119]
3 years ago
6

Irene solved this problem. Tony bought 3 pounds of oranges. He paid $0.95 per pound. How much did Tony pay for the oranges? Iren

e thought the answer was $2.85. Is Irene's answer reasonable? A. Yes, Irene's answer is reasonable. B. No, Irene's answer should be closer to $280. C. No, Irene's answer should be closer to $28. D. No, Irene's answer should be closer to $0.28.
Mathematics
2 answers:
Bad White [126]3 years ago
7 0
To answer this lets solve the question Irene answered and see what our answer will come to. den we will determine whether or not Irene is right, and if she is wrong why she is wrong. lets do it:- 

Lets look at the info we have:-

$0.95 per pound
Bought 3 pounds. 

<span>Lets solve by multiplying:- 
</span>
0.95 × 3 = 2.85

<span>Now lets see what Irene's answer was.
</span>
$2.85 
<span>$2.85 is equal to $2.85 (My answer)

So, Irene's answer is reasonable. 
</span>
So, Your answer to d main question would be <span>A. Yes, Irene's answer is reasonable.
</span>
Hope I helped ya!!!! xD 
allochka39001 [22]3 years ago
5 0
Yes Irene's answer is reasonable. U add up $0.95 three times and you get $2.85
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{\large{\textsf{\textbf{\underline{\underline{Given :}}}}}}

★ Sanya has a piece of land which is in the shape of a rhombus.

★ She wants her one daughter and one son to work on the land and produce different crops, for which she divides the land in two equal parts.

★ Perimeter of land = 400 m.

★ One of the diagonal = 160 m.

{\large{\textsf{\textbf{\underline{\underline{To \: Find :}}}}}}

★ Area each of them [son and daughter] will get.

{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}

Let, ABCD be the rhombus shaped field and each side of the field be x

[ All sides of the rhombus are equal, therefore we will let the each side of the field be x ]

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\longrightarrow  \tt AB+BC+CD+AD=400m

\longrightarrow  \tt x + x + x + x=400

\longrightarrow  \tt 4x=400

\longrightarrow  \tt  \: x =  \dfrac{400}{4}

\longrightarrow  \tt x= \red{100m}

\therefore Each side of the field = <u>100m</u><u>.</u>

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So, For \triangle ABD,

Here,

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\therefore \tt Simi \:  perimeter \:  [S] =  \boxed{ \sf \dfrac{a + b + c}{2} }

\longrightarrow \tt S = \dfrac{100 + 100 + 160}{2}

\longrightarrow \tt S =  \cancel{ \dfrac{360}{2}}

\longrightarrow \tt S = 180m

Using <u>herons formula</u><u>,</u>

\star \tt Area  \: of  \: \triangle = \boxed{\bf{{ \sqrt{s(s - a)(s - b)(s - c) } }}} \star

where

• s is the simi perimeter = 180m

• a, b and c are sides of the triangle which are 100m, 100m and 160m respectively.

<u>Putt</u><u>ing</u><u> the</u><u> values</u><u>,</u>

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(180 - 100)(180 - 100)(180 - 160) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180(80)(80)(20) }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{180 \times 80 \times 80 \times 20 }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{9 \times 20 \times 20 \times 80 \times 80}

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  \tt \sqrt{ {3}^{2} \times  {20}^{2}  \times  {80}^{2}  }

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} =  3 \times 20 \times 80

\longrightarrow \tt  Area_{ ( \triangle \:  ABD)} = \red{   4800  \: {m}^{2} }

Thus, area of \triangle ABD = <u>4800 m²</u>

As both the triangles have same sides

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Area of \triangle BCD = 4800 m²

<u>Therefore, area each of them [son and daughter] will get = 4800 m²</u>

{\large{\textsf{\textbf{\underline{\underline{Note :}}}}}}

★ Figure in attachment.

{\underline{\rule{290pt}{2pt}}}

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