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algol [13]
3 years ago
14

Plz help me with this

Mathematics
2 answers:
forsale [732]3 years ago
5 0

Answer:

\frac{31x}{30}

Step-by-step explanation:

Before adding we require the fractions to have a common denominator

The lowest common denominator of 2, 5 and 3 is 30, thus

\frac{15x}{15(2)} + \frac{6x}{6(5)} + \frac{10x}{10(3)}

= \frac{15x}{30} + \frac{6x}{30} + \frac{10x}{30}

Add the numerators leaving the denominator

= \frac{15x+6x+10x}{30} = \frac{31x}{30}

wariber [46]3 years ago
3 0

Answer: \bold{c)\quad \dfrac{31x}{30}}

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

\dfrac{x}{2}+\dfrac{x}{5}+\dfrac{x}{3}\\\\\text{The LCD of 2, 5, and 3 is }2\cdot 5\cdot 3=30\\\text{Multiply each term so their denominator is 30:}\\\\\\\dfrac{x}{2}\bigg(\dfrac{15}{15}\bigg)+\dfrac{x}{5}\bigg(\dfrac{6}{6}\bigg)+\dfrac{x}{3}\bigg(\dfrac{10}{10}\bigg)\\\\\\=\dfrac{15x}{30}+\dfrac{6x}{30}+\dfrac{10x}{30}\\\\\\=\dfrac{15x+6x+10x}{30}\\\\\\=\large\boxed{\dfrac{31x}{30}}

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Answer:

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Step-by-step explanation:

This is a problem of optimization.

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The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

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\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\  \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\

x

As d_b=x, the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

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