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jeka57 [31]
4 years ago
14

Tanya kicks a ball into the air. The function that models this is f(t) = -16t^2 + 96t, where h is the height of the ball in feet

and t is the time seconds
Mathematics
1 answer:
Mkey [24]4 years ago
3 0

The question is incomplete. The complete question is here.

Tanya kicks a ball into the air. The function that models this scenario is h(t) = -16t² + 96t, where h is the height of the ball in feet and t is the time seconds. When will the ball hit the ground?

Answer:

The ball will hit the ground after 6 seconds

Step-by-step explanation:

∵ h(t) = -16t² + 96t

∵ h is the height of the ball in feet for t seconds

- That means at the ground h = 0

∵ h(t) = 0 when the ball hit the ground

∴ 0 = -16t² + 96t

Let us solve it by adding 16t² to both sides

∴ 16t² = 96t

- Subtract 96t from both sides

∴ 16t² - 96t = 0

Factorize it by taking 16t as a common factor from each term

∵ 16t² ÷ 16t = t

∵ 96t ÷ 16t = 6

∴ 16t(t - 6) = 0

Equate each factor by 0

∵ 16t = 0

- Divide both sides by 16

∴ t = 0 ⇒ initial position

∵ t - 6 = 0

- Add 6 to both sides

∴ t = 6 ⇒ final position

∴ The ball will hit the ground after 6 seconds

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Elena-2011 [213]

Answer:

Step-by-step explanation:

Since the land is a round enclosure, and it has 2.5 foot wide enclosure of pebbles surrounding, then the circumference of the garden is 2.5ft.

First we need to get the diameter of the garden using the formula;

Circumference= πd

2.5 = πd

d = 2.5/π

d = 2.5/3.142

d =0.796 feet

The total area of a round section of the land = πd²/4

The total area of a round section of the land = π(0/796)²/4

The total area of a round section of the land  = 1.9892/4

<em>The total area of a round section of the land  = 0.4973 ft²</em>

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4 years ago
What is .015 as fraction
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3 0
3 years ago
a. Verify that the given point lies on the curve. b. Determine an equation of the line tangent to the curve at the given point.
xxMikexx [17]

Answer:

y = 13*( -x/9 + 1/5)

Step-by-step explanation:

Given:

- The curve has an equation as follows:

                               44 = 5x^2 + 3xy + 3y^2

Find:

a. Verify that the given point (2​,2​) lies on the curve.

b. Determine an equation of the line tangent to the curve at the given point.

Solution:

- To verify whether the point lies on the given curve we will substitute the coordinates of the point into the equation as follows:

                              44 = 5*(2)^2 + 3*(2)(2) + 3*(2)^2

                              44 = 20 + 12 + 12

                              44 = 44 ......Hence proven.

- The equation of the line tangent to the curve is expressed as a linear function as follows:

                              y = m*x + C

Where, m is the gradient of the line.

            C is the y-intercept.

                              m = Δy / Δx = dy/dx

- We will take the derivative of the given curve with respect to x as follows:

                             0 = 10x + 3*( y + xy' )  + 6y*y'\\\\-10x - 3y = y' ( 3x + 6y)\\\\ y' = - \frac{10x + 3y}{3x + 6y}

- Evaluate y' at the point (2,2) we get:

                            y' = - ( 10(2) + 3(2) ) / ( 3(2) + 6(2) )

                            y' = - ( 26 ) / (18)

                            y'= m = - 13/9

- To evaluate C, we will use the point (2,2) for linear expression above with m as follows:

                            y = -13*x/9 + C

                            2 =-13*(2)/9 + C

                            C = 13 / 5

- The equation of the tangent is as follows:

                            y = 13*( -x/9 + 1/5)  

8 0
3 years ago
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Given function:

       f(x) = 2 - 5x

Table;

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         -1                          7

         0                          2

          1                          -3

          2                        -8

The function is f(x) = 2 -5x

For;   x = -2;        f(x) = 2 - 5(-2)  = 12

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                            f(x) = 2 - 5(1) = -3

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5 0
4 years ago
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