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umka21 [38]
3 years ago
6

Helpeoepeoeoeo show work

Mathematics
1 answer:
alukav5142 [94]3 years ago
4 0

<em>IMPORTANT THIGNS TO REMEMBER:</em>

  • Has 20 lb. dog food!
  • Gives dog 1 3/5 everyday!
  • Find how much eaten after 2 days!
  • Find how much is left!

<em>ANSWER:</em>

Dog has eaten 3 1/5 after 2 days!

There is 16 4/5 left in bag!

<em>EXPLANATION:</em>

Since she gives 1 3/5 to her dog everyday and their asking for 2 days you would multiply 1 3/5 × 2 OR add 1 3/5 + 1 3/5

Which would give you 3 1/5

So, the dog has eaten 3 1/5 lb. of dog food in 2 days.

However then, you would subtract 20 lb. - 3 1/5 lb. because there is 20 lb. dog food in all and the dog has eaten 3 1/5 of it. This would give you 16 4/5 lb.

Meaning there is 16 4/5 lb. left in the bag of dog food!

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Answer:

  • <u>0.075 m/min</u>

Explanation:

You need to use derivatives which is an advanced concept used in calculus.

<u>1. Write the equation for the volume of the cone:</u>

      V=\dfrac{1}{3}\pi r^2h

<u />

<u>2. Find the relation between the radius and the height:</u>

  • r = diameter/2 = 5m/2 = 2.5m
  • h = 5.2m
  • h/r =5.2 / 2.5 = 2.08

<u>3. Filling the tank:</u>

Call y the height of water and x the horizontal distance from the axis of symmetry of the cone to the wall for the surface of water, when the cone is being filled.

The ratio x/y is the same r/h

  • x/y=r/h
  • y = x . h / r

The volume of water inside the cone is:

        V=\dfrac{1}{3}\pi x^2y

        V=\dfrac{1}{3}\pi x^2(2.08)\cdot x\\\\\\V=\dfrac{2.08}{3}\pi x^3

<u>4. Find the derivative of the volume of water with respect to time:</u>

            \dfrac{dV}{dt}=2.08\pi x^2\dfrac{dx}{dt}

<u>5. Find x² when the volume of water is 8π m³:</u>

       V=\dfrac{2.08}{3}\pi x^3\\\\\\8\pi=\dfrac{2.08}{3}\pi x^3\\\\\\  11.53846=x^3\\ \\ \\ x=2.25969\\ \\ \\ x^2=5.1062m²

<u>6. Solve for dx/dt:</u>

      1.2m^3/min=2.08\pi(5.1062m^2)\dfrac{dx}{dt}

      \dfrac{dx}{dt}=0.03596m/min

<u />

<u>7. Find dh/dt:</u>

From y/x = h/r = 2.08:

        y=2.08x\\\\\\\dfrac{dy}{dx}=2.08\dfrac{dx}{dt}\\\\\\\dfrac{dy}{dt}=2.08(0.035964m/min)=0.0748m/min\approx0.075m/min

That is the rate at which the water level is rising when there is 8π m³ of water.

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