This is exponential decay; the height of the ball is decreasing exponentially with each successive drop. It's not going down at a steady rate. If it was, this would be linear. But gravity doesn't work on things that way. If the ball was thrown up into the air, it would be parabolic; if the ball is dropped, the bounces are exponentially dropping in height. The form of this equation is

$y=a(b)^x$ , or in our case:

$A(n)=a(b)^{n-1}$ , where

a is the initial height of the ball and

b is the decimal amount the bounce decreases each time. For us:

a = 1.5 and

b = .74

Filling in,

$A(n)=1.5(.74)^{n-1}$

If ww want the height of the 6th bounce, n = 6. Filling that into the equation we already wrote for our model:

$A(6)=1.5(.74)^{6-1}$ which of course simplifies to

$A(6)=1.5(.74)^5$ which simplifies to

$A(6)=1.5(.22190066)$

So the height of the ball is that product.

A(6) = .33 cm

A is your answer

3 0

4 years ago

Write an Equation using the following arithmetic sequence: 5, _____, ______, 32

Shalnov [3]

Answer:

The sequence is:

5, 14, 23, 32

Step-by-step explanation:

We know that an Arithmetic sequence with a₁ and the commo difference 'd' has the nth term such as:

aₙ = a₁ + (n-1)d

Given the arithmetic sequence

5, _____, ______, 32

here:

a₁ = 5

n = 4

aₙ

so substituting a₁ = 5, n = 4 and aₙ = 32 in the nth term

aₙ = a₁ + (n-1)d

32 = 5 + (4-1)d

32 = 5 + 3d

3d = 32-5

3d = 27

divide both sides by 3

3d/3 = 27/3

d = 9

Therefore, the common difference: d = 9

Determining the 2nd term:

Using the formula

aₙ = a₁ + (n-1)d

substitute n = 2, a₁ = 5, d = 9

a₂ = 5 + (2-1)9

a₂ = 5 + 1(9)

a₂ = 5 + 9

a₂ = 14

Determining the 3rd term:

Using the formula

aₙ = a₁ + (n-1)d

substitute n = 3, a₁ = 5, d = 9

a₃ = 5 + (3-1)9

a₃ = 5 + 2(9)

a₃ = 5 + 18

a₃ = 23

Thus, the sequence becomes:

5, 14, 23, 32

4 0

3 years ago

Find the value of the variable y, for which:

schepotkina [342]

The value of the variable y is $y=6$

Explanation:

The given equation is $\frac{6}{y-4}-\frac{y}{y+2}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)$

Taking LCM on LHS of the equation, we get,

$\frac{6(y+2)-y(y-4)}{(y-4)(y+2)}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)$

Simplifying the term on RHS of the equation, we get,

$\frac{6(y+2)-y(y-4)}{(y-4)(y+2)}=\frac{6 y}{(y-4)(y+2)}$

Since, both the sides of the equation have the same denominator, we can cancel them.

Thus, we have,

$6(y+2)-y(y-4)=6 y$

Multiplying the terms within the bracket, we get,

$6y+12-y^2-4y=6 y$

Adding the like terms, we have,

$2y+12-y^2=6 y$

Subtracting both sides by $6 y$ , we have,

$-4y+12-y^2=0$

Factoring the equation, we get,

$(y+2)(y-6)=0$

$y=-2, y=6$

The values of the variable y are $y=-2, y=6$

At the point $y=-2$ , the equation $\frac{6}{y-4}-\frac{y}{y+2}=\left(\frac{6}{y-4}\right)\left(\frac{y}{y+2}\right)$ becomes undefined.

Thus, the value of the variable y is $y=6$

8 0

4 years ago