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2 years ago

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Mathematics

1 answer:

Scilla [17]2 years ago

5 0

Step-by-step explanation:

The area of a square is given by :

A = a²

Where

a is the side of the square

For square A, Area = 64 sq units

64 = a²

a = 8 units

For square B, Area = 30 sq units

30 = a²

a = $\sqrt{30}$ units

For square C, Side length = 14 units

A = 14²

A = 196 sq units

For square D, Side length = $\sqrt{30}$ units

A = $(\sqrt{30})^2$ ²

A = 30 sq units

Hence, this is the required solution.

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3 years ago

a grandmother says her grandson is two years older than her granddaughter and that together, they are at least 12 years old. how

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Hello There!

As for this problem, we would have to set up an equation in order to determine the missing age. We would do this because we want to find the missing age and by setting up an equation it will help find the age easier, as opposed to manually calculating it. With that said, let's begin to solve.

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3 years ago

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denis23 [38]

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2 years ago

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t),

pishuonlain [190]

Answer:

S(3)=22

Step-by-step explanation:

The rate of change of the number of squirrels S(t) that live on the Lehman College campus is directly proportional to 30 − S(t).

$\dfrac{dS}{dt}=k(30-S(t))\\ \dfrac{dS}{dt}+kS(t)=30k\\$The integrating factor: e^{\int k dt}=e^{kt}\\$Multiply all through by the integrating factor\\ \dfrac{dS}{dt}e^{kt}+kS(t)e^{kt}=30ke^{kt}$

$(Se^{kt})'=30ke^{kt} dt\\$Integrate both sides\\ Se^{kt}=\dfrac{30ke^{kt}}{k}+C$ (C a constant of integration)\\Se^{kt}=30e^{kt}+C\\$Divide both sides by e^{kt}\\S(t)=30+Ce^{-kt}$

When t=0, S(t)=15

$15=30+Ce^{-k*0}\\C=15-30\\C=-15$

When t = 2, S(t)=20

$20=30-15e^{-2k}\\20-30=-15e^{-2k}\\-10=-15e^{-2k}\\e^{-2k}=\dfrac23\\$Take the natural log of both sides$\\-2k=\ln \dfrac23\\k=-\dfrac{\ln(2/3)}{2}$

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$S(t)=30-15e^{\frac{\ln(2/3)}{2}t}\\$When t=3$\\S(t)=30-15e^{\frac{\ln(2/3)}{2} \times 3}\\S(3)=21.8 \approx 22$

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2 years ago