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IceJOKER [234]
3 years ago
14

Plz help ill give brainliest

Mathematics
1 answer:
MArishka [77]3 years ago
6 0

Answer:

100% answer B

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(-3,0) and (5, -2)

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y - 0 = -2/8 (x + 3)

y = -2/8 x - 6/8

8y = -2x - 6

2x + 8y = - 6

Answer is

D. 2x + 8y = -6

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The figure shows a right triangle.
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Let g(x) be the reflection of f(x) = x 2 + 3 in the x-axis. What is the function rule for g(x)?
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G(x) will be g(x)=-2x-3 because it is a reflection of f(x) over the x-axis. F(x)'s reflection over the x-axis will become -f(x).
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Your swimming pool containing 60,000 gal of water has been contaminated by 6 kg of a nontoxic dye that leaves a swimmer's skin a
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[a] Dye is removed from the pool at a rate of

(250 gal/min) * (<em>q</em>/60,000 g/gal) = -<em>q</em>/240 g/gal

where <em>q</em> denotes the amount of dye in the pool at time <em>t</em>. Clean water is pumped back into the pool, so no dye is being re-added.

So the net rate of change of the amount of dye in the pool is given by the differential equation,

\dfrac{\mathrm dq}{\mathrm dt}=-\dfrac{q(t)}{240}

with the intial value, <em>q</em>(0) = 6000 g (or 6 kg).

[b] The ODE above is separable as

\dfrac{\mathrm dq}q=-\dfrac{\mathrm dt}{240}

Integrate both sides to get

\ln|q|=-\dfrac t{240}+C

e^{\ln|q|}=e^{-t/240+C}

\implies q(t)=e^{-t/240+C}=e^{-t/240}e^C=Ce^{-t/240}

Now plug in the initial condition:

6000=Ce^0\implies C=6000

so the particular solution to the IVP is

q(t)=6000e^{-t/240}

[c] The acceptable concentration of the dy is 0.03 g/gal, which in a pool containing 60,000 gal of water corresponds to

(0.03 g/gal) * (60,000 gal) = 1800 g = 1.8 kg

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1800=6000e^{-t/240}\implies0.3=e^{-t/240}

\implies\ln0.3=-\dfrac t{240}

\implies t=-240\ln0.3\approx288.953

so the amount of dye in the pool is within the acceptable tolerance after about 289 min have passes, or about 4.82 hrs. So no, the filtration system is not up to the task.

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Answer:

The answer is B

Step-by-step explanation:

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