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Demonstration of Learning

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by

Shantal Gerken

on 29 April 2014

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Transcript of Demonstration of Learning

The Problem
Theo's mother has given him at most \$150 to buy clothes for school. The pants cost \$30 each and the shirts cost \$15 each. Write a linear inequality to describe the situation. Graph the solutions and give three combinations of pants and shirts that Theo could buy.
Two Stars, and a Wish
The Word Problem!

Stars:
1. The graph I had to make.
2. The translation and solving
Wish:
1. The reasons behind the steps.

Thank you for watching!
Step 3/4
Next, time to graph. We solved for y, which as explained before, makes the equation solve for point slope form (y=mx+b). Were 10 is the y intercept (b), 2 is the slope (m), and x is our variable (x). Let's graph!
For the sake of time, I am making a really dumb looking graph.
She can have any of the points in the blue area.
Step 4/4
Now, we just need to find three points that complete this inequality. Your answer is any point in the blue area, or on the line (This is only because of the fact that the line is equal too.)
That means that the points can be:
1. (1, 1)
2. (2.999999, 0)
3. (3, 1.11)
Or what ever is in the blue area.
Step 1/4
First, we need to 'translate' the word problem. Start with the numbers: 150, 30, 15. Next, read the words around the numbers. "At most \$150" That indicates that it is going to be at either less than, or equal too. "30 dollars for a pair of pants, and 15 for shirts." We will use x for the number of pants, and y for the number of shirts. The inequality?:
150>(or equal) 30x+15y
Step 2/4
Next we take our inequality that we just made, and solve for y. This way it will be solved for point slope form (y = mx+b). Then it will make our lives a lot easier to graph. First though we need to solve for y. The first step subtract 30x from both sides.
150_>30x+15y
30x- 30x-
150 + 30x-_>15y
Next, divide both side by 15
(150)/15+(30x)/15_> (15y)/15
10+2x_>y
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0 1 5 10 15 20 25 30 35 40 45 50
Points include:
(1, 1)
(2, 3)
(10, 0)
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