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Yakvenalex [24]

2 years ago

14

A shop produces 39 wet suits every 2 weeks. How long will it take to the shop to produce 429 suits?

1 answer:

Whitepunk [10]2 years ago

7 0

It would take about 11 weeks

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You want to get from a point A on the straight shore of the beach to a buoy which is 54 meters out in the water from a point B o

anyanavicka [17]

Answer:

$x =\dfrac{45 \sqrt{6}}{ 2}$

Step-by-step explanation:

From the given information:

The diagrammatic interpretation of what the question is all about can be seen in the diagram attached below.

Now, let V(x) be the time needed for the runner to reach the buoy;

∴ We can say that,

$\mathtt{V(x) = \dfrac{70-x}{7}+\dfrac{\sqrt{54^2+x^2}}{5}}$

In order to estimate the point along the shore, x meters from B, the runner should stop running and start swimming if he want to reach the buoy in the least time possible, then we need to differentiate the function of V(x) and relate it to zero.

i.e

The differential of V(x) = V'(x) =0

= $\dfrac{d}{dx}\begin {bmatrix} \dfrac{70-x}{7} + \dfrac{\sqrt{54^2+x^2}}{5} \end {bmatrix}= 0$

$-\dfrac{1}{7}+ \dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}=0$

$\dfrac{1}{5}\times \dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{5x}{\sqrt{54^2+x^2}}= \dfrac{1}{7}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{1}{\dfrac{7}{5}}$

$\dfrac{x}{\sqrt{54^2+x^2}}= \dfrac{5}{7}$

squaring both sides; we get

$\dfrac{x^2}{54^2+x^2}= \dfrac{5^2}{7^2}$

$\dfrac{x^2}{54^2+x^2}= \dfrac{25}{49}$

By cross multiplying; we get

$49x^2 = 25(54^2+x^2)$

$49x^2 = 25 \times 54^2+ 25x^2$

$49x^2-25x^2 = 25 \times 54^2$

$24x^2 = 25 \times 54^2$

$x^2 = \dfrac{25 \times 54^2}{24}$

$x =\sqrt{ \dfrac{25 \times 54^2}{24}}$

$x =\dfrac{5 \times 54}{\sqrt{24}}$

$x =\dfrac{270}{\sqrt{4 \times 6}}$

$x =\dfrac{45 \times 6}{ 2 \sqrt{ 6}}$

$x =\dfrac{45 \sqrt{6}}{ 2}$

8 0

2 years ago

Does anyone know the answers to this ? Pls help me out

Reptile [31]

Answer:

C. 4

Step-by-step explanation:

2x³+2x¹+2x+5+4/x-1

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