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yarga [219]
2 years ago
12

Krystal and Mary are selling fruit for a school fundraiser. Customers can buy small boxes of tangerines and large boxes of tange

rines. Krystal sold 7 small boxes of tangerines and 14 large boxes of tangerines for a total of $266. Mary sold 14 small boxes of tangerines and 3 large boxes of tangerines for a total of $182. Find the cost each of one small box of tangerines and one large box of tangerines.​
Mathematics
1 answer:
irina1246 [14]2 years ago
3 0

Using a system of equations, it is found that a small box of tangerines costs $10 and a large box costs $14.

<h3>What is a system of equations?</h3>
  • A system of equations is when two or more variables are related, and equations are built to find the values of each variable.

In this problem, the <em>variables </em>are:

  • Variable x: Cost of a small box.
  • Variable y: Cost of a large box.

Krystal sold 7 small boxes of tangerines and 14 large boxes of tangerines for a total of $266, hence:

7x + 14y = 266

Simplifying by 7:

x + 2y = 38

x = 38 - 2y

Mary sold 14 small boxes of tangerines and 3 large boxes of tangerines for a total of $182, hence:

14x + 3y = 182

Since x = 38 - 2y:

14(38 - 2y) + 3y = 182

25y = 350

y = \frac{350}{25}

y = 14

Then

x = 38 - 2y = 38 - 2(14) = 10

A small box of tangerines costs $10 and a large box costs $14.

To learn more about system of equations, you can take a look at brainly.com/question/24342899

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\bf 512~~,~~\stackrel{512\cdot \frac{1}{2}}{256}~~,~~...4

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\bf n^{th}\textit{ term of a geometric sequence}\\\\a_n=a_1\cdot r^{n-1}\qquad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\a_n=+4\end{cases}

\bf 4=512\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{4}{512}=\left( \cfrac{1}{2} \right)^{n-1}\\\\\\\cfrac{1}{128}=\left( \cfrac{1}{2} \right)^{n-1}\implies \cfrac{1}{2^7}=\left( \cfrac{1}{2} \right)^{n-1}\implies 2^{-7}=\left( 2^{-1}\right)^{n-1}\\\\\\(2^{-1})^7=(2^{-1})^{n-1}\implies 7=n-1\implies \boxed{8=n}

so is the 8th term, then, let's find the Sum of the first 8 terms.

\bf \qquad \qquad \textit{sum of a finite geometric sequence}\\\\S_n=\sum\limits_{i=1}^{n}\ a_1\cdot r^{i-1}\implies S_n=a_1\left( \cfrac{1-r^n}{1-r} \right)\quad \begin{cases}n=n^{th}\ term\\a_1=\textit{first term's value}\\r=\textit{common ratio}\\----------\\r=\frac{1}{2}\\a_1=512\\n=8\end{cases}

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Part A: 1 solution

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Step-by-step explanation:

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