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3 years ago

Steven is comparing the cost of a fresh lobster dinner at two different restaurants. The first

Mathematics

1 answer:

Anarel [89]3 years ago

6 0

Answer:

Step-by-step explanation:

2x+51=4x+19

32=2x

x=16. The weight is 16 pounds

plug it back in to find the cost

32+51=83. 64+19 = 83. The cost is 83 dollars

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3 years ago

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1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

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1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

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$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

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We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

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