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ivann1987 [24]
2 years ago
8

Full explanation please, I will mark brainlist if correct

Mathematics
1 answer:
blagie [28]2 years ago
8 0

Answer:

19.41 units

Step-by-step explanation:

<u>Given coordinates of points;</u>

  • (-10, 9)
  • (-6, -10)

To determine the distance between the two points, the distance formula is required. The distance formula is used where it is not possible to calculate the distance through a straight line. The distance formula is expressed as,

  • ⇒ \sqrt{(x_{2} - x_{1})^{2} + (y_2 - y_1)^{2}  }

The distance between the two points can be determined by plugging the coordinates into the formula;

  • ⇒ \sqrt{(x_{2} - x_{1})^{2} + (y_2 - y_1)^{2}  }
  • ⇒ \sqrt{[-6 - (-10)]^{2} + [-10 - 9]^{2}  }

Then, we can simplify the root as needed to determine the distance;

  • ⇒ \sqrt{[-6 + 10]^{2} + [-19]^{2}  }
  • ⇒ \sqrt{[4]^{2} +361 }
  • ⇒ \sqrt{16 +361 }
  • ⇒ \sqrt{377}

Finally, we need to convert the root into decimals (stated in question). It is impossible to determine the root in decimal form. Therefore, I used a calculator to determine the distance in decimal form.

  • ⇒ \sqrt{377} ≈ 19.41 units [Using calculator]

Therefore, the distance between the two points is about 19.41 units.

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A tank with a capacity of 500 gal originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of
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Let A(t) denote the amount of salt in the tank at time t.

Salt flows in at a rate of

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and flows out at a rate of

(A(t)/(200 + t) lb/gal) * (2 gal/min) = 2 A(t)/(500 + t)

(in case you're unsure about the denominator: the tank starts off with 200 gal of solution, and each minute solution flows in at a rate of 3 gal/min and thus the tank gains (3 gal/min) * (1 min) = 3 gal. At the same time, solution flows out at a rate of 2 gal/min and thus the tank loses 2 gal, giving a net change in volume of (3 - 2)*t = t gal)

Then the net rate of salt flow is given by the ODE,

\dfrac{\mathrm dA(t)}{\mathrm dt}-\dfrac{2A(t)}{200+t}=3

Multiply both sides by (200+t)^{-2}:

(200+t)^{-1}\dfrac{\mathrm dA(t)}{\mathrm dt}-2(200+t)^{-3}A()=3(200+t)^{-2}

\implies\dfrac{\mathrm d}{\mathrm dt}\bigg((200+t)^{-2}A(t)\bigg)=3(200+t)^{-2}

Integrating both sides and solving for A(t) gives

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A(t)=-2(200+t)+C(200+t)^2

The tank starts off with 100 lb of salt in solution, so A(0)=100 and we find

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The tank will begin to overflow once the volume of solution reaches 500 gal; this happens when

500=200+t\implies t=300

or 300 minutes or 5 hours after solution starts flowing. At this point, the tank will contain

A(300)=2125

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Theoretically, the amount of salt in the tank will increase forever, since A(t)\to\infty as t\to\infty.

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Bezzdna [24]

Answer:

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