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svetlana [45]
3 years ago
13

Laura is mailing packages. Each small package costs her $3.40 to send. Each large package cost her $4.30. How much will it cost

her to send 1 small package and 7 large package?

Mathematics
2 answers:
GenaCL600 [577]3 years ago
7 0

Answer:$33.5

Step-by-step explanation:

Darya [45]3 years ago
6 0

Answer:

<u>$33.5</u>

Step-by-step explanation:

Given:

  • the Cost the the small package is $3.40
  • the Cost of the large package is $4.30

Solution:

To send 1 small package and 7 large packages, it would cost

[the cost of 1 small package] + 7 multiplied by [the cost of 1 large package]

which numerically looks like,

3.40 + 7x(4.30)

= 3.40 added to 30.1

=$33.5

If this answer helped you, please mark it as brainliest

Thanks :)

       

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Solve: [3(1/5)] + 75% - [1(3/10)]​
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Step-by-step explanation:

\large{\underline{\underline{\bf{\purple{Solution\:  : - }}}}}

\begin{gathered}\\  \dashrightarrow{\sf{3\dfrac{1}{5}} + 75\% - {1\dfrac{3}{10}}} \\  \end{gathered}

\red \divideontimes Firstly, converting mixed fractions into improper fraction.

\begin{gathered} \\  {\dashrightarrow{\sf{\dfrac{(3 \times 5) + 1}{5}} + 75\% - {\dfrac{(1 \times 10) + 3}{10}}}} \\ \end{gathered}

\begin{gathered}{\dashrightarrow{\sf{\dfrac{15 + 1}{5}} + 75\% - {\dfrac{10 + 3}{10}}}} \\ \end{gathered}

\begin{gathered}{\dashrightarrow{\sf{\dfrac{16}{5}} + 75\% - {\dfrac{13}{10}}}} \\ \end{gathered}

\red \divideontimes As we know that, 1% = 1/100, then converting 75% into improper fraction.

\begin{gathered}\\{\dashrightarrow{\sf{\dfrac{16}{5}} +  \dfrac{75}{100}  - {\dfrac{13}{10}}}} \\  \end{gathered}

\red \divideontimes According to the BODMAS, Firstly we will perform the addition.

\begin{gathered}\\{\dashrightarrow{\sf{\bigg(\dfrac{16}{5}} +  \dfrac{75}{100} \bigg)  - {\dfrac{13}{10}}}} \\ \end{gathered}

\red \divideontimes Taking the LCM of denominators.

\begin{gathered}\\{\dashrightarrow{\sf{\bigg(\dfrac{(16 \times 20) + (75 \times 1)}{100}} \bigg)  - {\dfrac{13}{10}}}} \\ \end{gathered}

\begin{gathered}{\dashrightarrow{\sf{\bigg(\dfrac{320 + 75 }{100}} \bigg)  - {\dfrac{13}{10}}}} \\ \end{gathered}

\begin{gathered}{\dashrightarrow{\sf{\bigg(\dfrac{395 }{100}} \bigg)  - {\dfrac{13}{10}}}} \\ \end{gathered}

\red \divideontimes Now, converting 395/100 in simplest form.

\begin{gathered}\\  {\dashrightarrow{\sf{\bigg( \cancel{\dfrac{395 }{100}}} \bigg)  - {\dfrac{13}{10}}}} \\ \end{gathered}

\begin{gathered}{\dashrightarrow{\sf{\bigg(\dfrac{79}{20}} \bigg)  - {\dfrac{13}{10}}}} \\ \end{gathered}

\red \divideontimes Now, performing subtraction, according to BODMAS.

\begin{gathered}\\  {\dashrightarrow{\sf{\dfrac{79}{20}}   - {\dfrac{13}{10}}}} \\ \end{gathered}

\red \divideontimes Again, taking LCM of denominators.

\begin{gathered}\\  {\dashrightarrow{\sf{\dfrac{(79 \times 1) - (13 \times 2)}{20}}}} \\ \end{gathered}

\begin{gathered}{\dashrightarrow{\sf{\dfrac{79 -26}{20}}}}  \\ \end{gathered}

\begin{gathered}{\dashrightarrow{\sf{\dfrac{53}{20}}}} \\  \end{gathered}

\begin{gathered}\red{\bigstar{\underline{\boxed{\bf{Answer = \dfrac{53}{20}}}}}}\end{gathered}

<u>∴ The answer is 53/20.</u>

\begin{gathered}\end{gathered}

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