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Annette [7]
3 years ago
13

A direct variation function contains the points (–8, –6) and (12, 9). Which equation represents the function?

Mathematics
1 answer:
mars1129 [50]3 years ago
6 0

Answer:

y=\frac{3}{4}x

Step-by-step explanation:

If it is a direct variation function, then \frac{y}{x}=k.

k can be found by choosing a point and doing y divided by x.

So we can find k by doing -6 divided by -8 or 9 divided by 12.

Both of these values are equal to 0.75 or 3/4 as a fraction.

So you can say any point in this relation will follow the equation:

\frac{y}{x}=\frac{3}{4}

Multiply both sides by x:

y=\frac{3}{4}x

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Answer:

0.84 square in

Step-by-step explanation:

Since the capacity of the cable is proportional to its​ cross-sectional area. If a cable that is 0.3 sq in can hold 2500 lb then per square inch it can hold

2500 / 0.3 = 8333.33 lb/in

To old 7000 lb it the cross-sectional area would need to be

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A department store sells a pair of shoes with a 80% markup. If the store bought the pair of shoes for $55.25. Round the price to
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Use the Commutative Property to write an expression equivalent to –5d + 1/2.
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2 years ago
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PLS TRY TO ANSWER ASAP.
Harrizon [31]

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.

4 0
3 years ago
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