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Kazeer [188]
4 years ago
6

TexFormula1" title=" \frac{2}{15y} + \frac{3}{5y {}^{3} }" alt=" \frac{2}{15y} + \frac{3}{5y {}^{3} }" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Xelga [282]4 years ago
3 0

Answer:

2y²   +   9

---------------

      15y³

Step-by-step explanation:

Start by identifying the LCD, and then change each fraction so that its denominator is the LCD.

Here the LCD is 15y³, which is evenly divisible by 15y and 5y³.

Focus now on the first fraction:  2 / (15y).  Multiplying numerator and denominator of this fraction by y² results in

y²·2          2y²

--------- → ----------

y²·15y       15y³           ←This is the correct LCD

Multiplying numerator and denominator of the second fraction by 3 results in:

   3·3            9

------------ → ---------

 3·5y³         15y³          ←This is the correct LCD

So now those two original terms look like:

 2y²         9

--------- + --------

 15y³       15y³

and this can be written in simpler form as:

2y²   +   9

---------------

      15y³

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Answer:

The length on the blueprint for an actual length of 65 feet is 16\frac{1}{4}\ inches

Step-by-step explanation:

Given:

Scale of the blueprint is \frac{1}{4}\ in=1\ foot

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Now, as per given data;

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Therefore, using unitary method, we can find find the length on the blueprint for an actual length of 65 feet by multiplying 65 and \frac{1}{4}. Therefore,

Length on the blueprint for 65 feet is given as:

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Therefore, the length on the blueprint for an actual length of 65 feet is 16\frac{1}{4}\ inches

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3 years ago
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