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andreev551 [17]
4 years ago
12

A baseball player hits a ball towards the left field wall of a ball park that is 330 feet from the home plate. Using the equatio

n
y =  - 0.008 {x}^{2} + 3x + 3
did the player hit the ball over the wall (homerun)? how far did the ball land(when is y=0)? What is the maximum height of the baseball? What is the height of the ball the moment it is hit?

Mathematics
1 answer:
Paha777 [63]4 years ago
5 0

Answer:

  • yes, it is a home run
  • 376 ft from home plate
  • 284.25 ft high
  • 3 ft high when hit

Step-by-step explanation:

a) The height of the ball at x = 330 is ...

  y = (-0.008·330 +3)330 +3 = 0.36·330 +3 = 118.80 +3 = 121.80

The ball is 121.8 ft high at the distance of the wall. If the wall is higher than that, the ball did not go over the wall. We expect the wall is not that high, so the ball clears the wall for a home run.

___

b) When y=0, the value(s) of the distance x can be found using the quadratic formula:

  0 = -0.008x^2 +3x +3

  x = (-3 ±√(3² -4(-0.008)(3)))/(2·(-0.008))

  = (3 ±√9.96)/0.016 ≈ 376.0 . . . feet (positive solution)

The ball landed 376 ft from home plate.

___

c) The maximum of the quadratic can be found where x is -b/(2a) = -3/(2·(-.008)) = 3000/16 = 187.5.

The value of y at that distance is ...

  y = (-0.008·187.5 +3)187.5 +3 = 284.25 . . . feet

The maximum height is 284.25 ft.

___

d) The height of the baseball at home plate (where x=0) is

  y = -0.008·0 +3·0 +3 = 3 . . . feet

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1.
Slope-300.
Y int: 40,000
Y=300x+40000
2.
S: -5
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Y=-5x+250
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Y=2x+5
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5 0
3 years ago
Suppose brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt.
Oliga [24]

Answer:

(a) 0.288 kg/liter

(b) 0.061408 kg/liter

Step-by-step explanation:

(a) The mass of salt entering the tank per minute, x = 0.2 kg/L × 5 L/minute = 1 kg/minute

The mass of salt exiting the tank per minute = 5 × (5 + x)/500

The increase per minute, Δ/dt, in the mass of salt in the tank is given as follows;

Δ/dt = x - 5 × (5 + x)/500

The increase, in mass, Δ, after an increase in time, dt, is therefore;

Δ = (x - 5 × (5 + x)/500)·dt

Integrating with a graphing calculator, with limits 0, 10, gives;

Δ = (99·x - 5)/10

Substituting x = 1 gives

(99 × 1 - 5)/10 = 9.4 kg

The concentration of the salt and water in the tank after 10 minutes = (Initial mass of salt in the tank + Increase in the mass of the salt in the tank)/(Volume of the tank)

∴ The concentration of the salt and water in the tank after 10 minutes =  (5 + 9.4)/500 = (14.4)/500 = 0.288

The concentration of the salt and water in the tank after 10 minutes = 0.288 kg/liter

(b) With the added leak, we now have;

Δ/dt = x - 6 × (14.4 + x)/500

Δ = x - 6 × (14.4 + x)/500·dt

Integrating with a graphing calculator, with limits 0, 20, gives;

Δ = 19.76·x -3.456 = 16.304

Where x = 1

The increase in mass after an increase in = 16.304 kg

The total mass = 16.304 + 14.4 = 30.704 kg

The concentration of the salt in the tank then becomes;

Concentration = 30.704/500 = 0.061408 kg/liter.

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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3 years ago
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Liono4ka [1.6K]

Given : y = -25 and x = -5

We can notice that if we Multiply 'x' with 5 we get 'y'

That is (-5) × 5 = -25

⇒ x multiplied by 5 = y

(a) The Equation that relates x and y is 5x = y

(b) We found that the Equation of this Direct variation is : 5x = y

The Question is to find the value of y when x = 3

The Value of y at x = 3 can be found by simply substituting x = 3 in the equation we found.

⇒ y = 5x

⇒ y = 5 × 3

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So, the Value of y when x = 3 is 15

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