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Oksana_A [137]
3 years ago
14

PLEASE HELP 15 POINTS AND BRAINLIEST!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
natta225 [31]3 years ago
5 0
The picture in the attached figure

we know that
area of a circle=pi*r²
diameter=8 in
so
radius r=8/2----> 4 in
area=pi*4²-----> 3.14*4² in²

the answer is
3.14*4² in²

Brilliant_brown [7]3 years ago
3 0

Answer:

3.14*4² in²

Step-by-step explanation:

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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

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

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Answer:

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I need help solving this question please :)
dsp73

Answer:

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Step-by-step explanation:

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