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3 years ago

Can someone help me find the value of x and the value of y? I need them to show their work, too.

Mathematics

1 answer:

solniwko [45]3 years ago

4 0

$\bf \begin{cases} x-2y=0\\ 2x-5y=-4 \end{cases} \\\\[-0.35em] ~\dotfill\\\\ x-2y=0\implies x=2y \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{substituting the found \underline{x} in the 2nd equation}}{2(2y)-5y=-4}\implies 4y-5y=-4\implies -y=-4 \\\\\\ y=\cfrac{-4}{-1}\implies \blacktriangleright y=4 \blacktriangleleft \\\\[-0.35em] ~\dotfill$

$\stackrel{\textit{substituting the found \underline{y} in the 1st equation}}{x-2(4)=0}\implies x-8=0\implies \blacktriangleright x=8 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (8,4)~\hfill$

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Kryptonite is a material found on the planet Krypton and has various effects, most importantly on Superman. The most common type

Rashid [163]

Answer:

The maximum amount of red kryptonite present is 33.27 g after 4.93 hours.

Step-by-step explanation:

dy/dt = y(1/t - k)

separating the variables, we have

dy/y = (1/t - k)dt

dy/y = dt/t - kdt

integrating both sides, we have

∫dy/y = ∫dt/t - ∫kdt

㏑y = ㏑t - kt + C

㏑y - ㏑t = -kt + C

㏑(y/t) = -kt + C

taking exponents of both sides, we have

$\frac{y}{t} = e^{-kt + C} \\\frac{y}{t} = e^{-kt}e^{C} \\\frac{y}{t} = Ae^{-kt} (A = e^{C})\\y = Ate^{-kt}$

when t = 1 hour, y = 15 grams. So,

$y = Ate^{-kt}\\15 = A(1)e^{-kX1}\\15 = Ae^{-k}$ (1)

when t = 3 hours, y = 30 grams. So,

$y = Ate^{-kt}\\30 = A(3)e^{-kX3}\\30 = 3Ae^{-3k}$ (2)

dividing (2) by (1), we have

$\frac{30}{15} = \frac{3Ae^{-3k}}{Ae^{-k}} \\2 = 3e^{-2k}\\\frac{2}{3} = e^{-2k}$

taking natural logarithm of both sides, we have

-2k = ㏑(2/3)

-2k = -0.4055

k = -0.4055/-2

k = 0.203

From (1)

$A = 15e^{k} \\A = 15e^{0.203} \\A = 15 X 1.225\\A = 18.36$

Substituting A and k into y, we have

$y = 18.36te^{-0.203t}$

The maximum value of y is obtained when dy/dt = 0

dy/dt = y(1/t - k) = 0

y(1/t - k) = 0

Since y ≠ 0, (1/t - k) = 0.

So, 1/t = k

t = 1/k

So, the maximum value of y is obtained when t = 1/k = 1/0.203 = 4.93 hours

$y = 18.36(1/0.203)e^{-0.203t}\\y = \frac{18.36}{0.203}e^{-0.203X1/0.203}\\y = 90.44e^{-1}\\y = 90.44 X 0.3679\\y = 33.27 g$

So the maximum amount of red kryptonite present is 33.27 g after 4.93 hours.

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