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Mashcka [7]
3 years ago
11

Alexis loves to work out on the treadmill. On Monday, she ran 1.52 miles. On Tuesday she ran 0.75 miles. On Wednesday she ran 2.

65 miles. On Friday, she ran more than she did on Tuesday but less than she did on Monday. What could the possible amount of miles that she ran on Friday? Explain your answer.
Mathematics
1 answer:
poizon [28]3 years ago
6 0

Answer:

1.08 Miles

How I Found The Answer:

2.65 + 1.52 + 0.75 = 4.92

6 - 4.92 = 1.08

I Said Alexis Wished To Have An Even Number So She Ran 1.08 Miles.

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Write an equivalent expression for 1/3/4
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2/6/8

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12. Which of the following is the value of x needed to make the equation
patriot [66]

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it will be c, x=8.5

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I've done every problem I only need with the help this problem
bulgar [2K]

Answers:

\displaystyle \lim_{x \to 2^{+}} f(x) = 1\\\\\displaystyle \lim_{x \to 2^{-}} f(x) = 1\\

Both result in the same limit value. This allows us to say \displaystyle \lim_{x \to 2} f(x) = 1 without the plus or minus over the 2.

The left and right hand limits may not always match like this.

==================================================

Explanation:

The notation \displaystyle \lim_{x \to 2^{+}} f(x) means that we are approaching x = 2 from the right hand side. This is from the positive direction. So we start at say x = 3 and move to x = 2.5 then to x = 2.1 then to x = 2.01 and so on.

Because we started with values x > 2, we will use the third definition of the piecewise function

if x > 2, then f(x) = 3x-5

Plug in x = 2 to get

f(x) = 3x-5

f(2) = 3(2)-5

f(2) = 6-5

f(2) = 1

This shows \displaystyle \lim_{x \to 2^{+}} f(x) = 1

-----------------------------

For the other limit, we're approaching x = 2 from the negative side. So we could start at say x = 0, then move to x = 1, then to x = 1.5 then to x = 1.9 then to x = 1.99, and so on.

We're using x values such that x < 2 now.

So we'll be using the first definition of the piecewise function

If x < 2, then f(x) = x^2 - 3

f(x) = x^2-3

f(2) = 2^2-3

f(2) = 4-3

f(2) = 1

We end up with \displaystyle \lim_{x \to 2^{-}} f(x) = 1

---------------------------------

Both right hand limit and left hand limit result in the same value

Because \displaystyle \lim_{x \to 2^{+}} f(x) = \displaystyle \lim_{x \to 2^{-}} f(x) = 1

We can shorten that to \displaystyle \lim_{x \to 2^{}} f(x) = 1 meaning we can approach x = 2 from either direction to arrive at the same limiting value.

A thing to notice is that f(2) is not equal to 1. Instead the second line of the piecewise function says f(2) = 3.

The fact that the limit as x approaches 2 and f(2) don't agree means this function is not continuous at x = 2.

The graph shows this. We have a removable discontinuity where we effectively picked the point off the graph and move it upward.

See the diagram below.

3 0
3 years ago
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