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3 years ago

5/9 on a number line

Mathematics

1 answer:

finlep [7]3 years ago

5 0

Given:

The number is $\dfrac{5}{9}$ .

To find:

The location of the given number on a number line.

Solution:

We have,

$\dfrac{5}{9}$

Here $0$ , so $0$ .

Mark 8 equidistant small lines between 0 to 1 on the number line to divide the number line in 9 equal parts. Then, the 5th mark represent the value $\dfrac{5}{9}$ .

The location of number $\dfrac{5}{9}$ on a number line is shown below.

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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

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Answer:

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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

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taking exponents of both sides, we have

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

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-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

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

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

<u>So the maximum amount of red kryptonite present is 33.27 g after 4.93 hours.</u>

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