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Jlenok [28]
2 years ago
7

Some students performed an experiment in which they dropped a rubber ball from a height of 920 centimeters. They noticed that af

ter each bounce, it reached 56% of its previous height. Which equation models the height, H, for n bounces?
Mathematics
1 answer:
sweet [91]2 years ago
8 0

The equation models the height, H, for n bounces, will be H = 920 (0.56)ⁿ.

<h3>What is an exponent?</h3>

Consider the function:

y = a (1 ± r) ˣ

Where n is the number of times this growth/decay occurs, a = initial amount, and r = fraction by which this growth/decay occurs.

If there is a plus sign, then there is exponential growth happening by r fraction or 100r %

If there is a minus sign, then there is exponential decay happening by r fraction or 100r %

Some students performed an experiment in which they dropped a rubber ball from a height of 920 centimeters.

They noticed that after each bounce, it reached 56% of its previous height.

Then the equation models the height, H, for n bounces will be

H = 920 (0.56)ⁿ

More about the exponent link is given below.

brainly.com/question/5497425

#SPJ1

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Answer:

Step-by-step explanation:

Your final answer is the standard form of a parabola.  Since your equation has an x-squared term in it and not a y-squared term, your form will be

(x-h)^2=4p(y-k)

To get it into this form we will solve the quadratic for y and then set it equal to 0 so we can complete the square on it.  Solving for y then setting y equal to 0:

x^2-2x-23=8y so

\frac{1}{8}x^2-\frac{2}{8}x-\frac{23}{8}=0

We only need to complete the square on the x-terms, so we will move the constant back over to the other side of the equals sign:

\frac{1}{8}x^2-\frac{2}{8}x=\frac{23}{8}

The rule for completing the square is that the leading coefficient HAS to be a 1.  Ours is 1/8, so we have to factor it out.  When we do that we are left with:

\frac{1}{8}(x^2-2x)=\frac{23}{8}

To complete the square on the left, we take half the linear term, square it, and add it onto both sides.  Our linear term is 2 (the number stuck to the x-term).  Half of 2 is 1, and 1 squared is 1.  So we add it into the parenthesis on the left.  BUT we cannot discount the 1/8 sitting out front there.  It is a multiplier.  So what we actually added on the left is 1/8(1).  That looks like this:

\frac{1}{8}(x^2-2x+1)=\frac{23}{8}+\frac{1}{8}

Now we will write the left side into its perfect square binomial (which was the whole reason for doing this!) and simplify the right at the same time:

\frac{1}{8}(x-1)^2=\frac{24}{8}

Now we will set the whole thing back to equal y:

\frac{1}{8}(x-1)^2-3=y

That's one form.  But you need it in vertex form, so we add 3 to both sides:

\frac{1}{8}(x-1)^2=y+3 and then multiply both sides by 8:

(x-1)^2=8(y+3)

If you need to break it down further to include what your p value is, then:

(x-1)^2=4(2)(y+3)

Either that one or the one right above it should work.

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