Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 2 By Herbert I. Gross and Richard A. Medeiros next.

Answer: f(0) = 3 Solution for #1a: In the notation f(x), x represents the input; f, the “program”; and f(x), the output. So starting with… f(x) = 2x + 3 next © 2007 Herbert I. Gross every place we see “x”, we replace it by 0. Doing this we obtain… f(0) = 2(0) + 3 next from which we see that… next f(0) = 0 + 3 = 3

next Notes on #1a When we multiply any number by 0, we obtain 0 as the product. That is, for any number x, x(0) = 0. © 2007 Herbert I. Gross next The fact that x(0) = 0 often “tricks” beginning students into thinking that f(0) = 0. Keep in mind that f(0) represents the image of f if the input is 0.

Notes on #1a © 2007 Herbert I. Gross In terms of a practical application suppose you pay $5 in order to be eligible to order an item which costs $4. If you then buy x items, the cost in dollars, C, is represented by… C = f(x) = 4x + 5 next Notice that the cost is $5 even before you buy one item. To verify this mathematically, we replace x by 0 in the equation and obtain… C = f (x) = 4 x + 5 (0) = 0 + 5 = 5 next

Notes on #1a If you are uncomfortable with the notation “f(x)”, you might want to translate “f(x) = 2x + 3” into the more verbal “program” format… © 2007 Herbert I. Gross next Input x Step 1Multiply by 2Step 2Add 3 Output f(x)

Solution for #1b: We proceed just as we did in Problem #1a; only this time we replace x by - 1. That is, starting with… f(x) = 2x + 3 next © 2007 Herbert I. Gross next every place we see “x”, we replace it by - 1. Doing this we obtain… f( - 1) = 2( - 1) + 3 from which we see that… f( - 1) = - 2 + 3 = 1

next Notes on #1b One aim of this part of the problem is to emphasize that the rule defined by the function f applies to negative numbers as well as to non-negative numbers. © 2007 Herbert I. Gross next If we had wanted f to apply only to non-negative numbers, we would have had to incorporate this information into the domain of f.

Notes on #1b f(x) = 2x + 3 and x ≥ 0 © 2007 Herbert I. Gross next For example, we might have written… Also observe that while the input was negative the output was positive. Of course, f(x) can also be negative. For example, with respect to this problem f( - 4) = 2( - 4) + 3 = - 8 + 3 = - 5

Solution for #1c: Again we proceed just as we did in Problems #1a and #1b; only this time we replace x by 1 / 2. That is, starting with f(x) = 2x + 3 next © 2007 Herbert I. Gross next every place we see “x”, we replace it by 1 / 2. Doing this we obtain… f( 1 / 2 ) = 2( 1 / 2 ) + 3 from which we see that… f( 1 / 2 ) = 1 + 3 = 4

next Notes on #1c When the domain of a function is not stated specifically, we assume that it contains all possible numbers. © 2007 Herbert I. Gross next In the present example, notice that if x is any real number, so also is 2x; and so also is 2x + 3. Hence, if f(x) = 2x + 3, all real numbers are in the domain of f.

Notes on #1c On the other hand, knowing that we can’t divide by 0 (for example, is undefined), suppose that a function g is defined by g(x) =. © 2007 Herbert I. Gross That is, in that case g( - 1.5) = next 1 2( - 1.5) + 3 1010 = If x = - 1.5, 2x + 3 = 0 and in that case g( - 1.5) is undefined. 1 2x + 3 1010 next

Preface to Problem #2 There is an inherent danger in using generic symbolism. © 2007 Herbert I. Gross For example, up to now in our course we have used x to denote the input of the function. In other words, when we write f(x), it is understood that x is in the domain of f. next

In a similar way when we write f -1 (x) it is understood that x is in the domain of f -1. © 2007 Herbert I. Gross The problem is that the domain of f -1 is the image of f (that is, the output of f), and we generically use y to denote a member of the image of f. next Thus, the danger of this generic symbolism is that we may not realize that the “y” in the notation “y = f(x)” is the “x” in the notation “y = f -1 (x)”and correspondingly the “x” in the notation “y = f(x)” is the “y” in the notation “y = f -1 (x)”.

The key to avoiding this danger is to recognize that in the same way that subtraction “undoes” addition, f -1 “undoes” f. In other words, if we are more comfortable working with f than with f -1, we may rewrite… © 2007 Herbert I. Gross …in the equivalent form… next y = f -1 (x) x = f(y)

Rewriting the equation y = f -1 (x) in the form of the equation x = f(y) is just another way of saying that to find the inverse of a function we interchange x and y. next © 2007 Herbert I. Gross next This will be emphasized a bit more in our solution for Problem #4; but for now we will apply the above discussion to the solution for Problem #2.

Solution for #2a: Remember that the input of f -1 is the output of f. Thus, this problem is asking us to find what the input is if the output is 0. To avoid the type of confusion described in the above preface, let’s not use x to denote f -1 (0). Instead, let’s denote f -1 (0) by c. next © 2007 Herbert I. Gross next Thus, we are being asked to find the value of c for which… f(c) = 0

Solution for #2a: Since f(x) = 2x + 3, next © 2007 Herbert I. Gross next If we replace f(c) in the equation f(c) = 0 by its value in the equation f(c) = 2c + 3, we obtain the linear equation… f(c) = 2c + 3 2c + 3 = 0 …and to find the value of c, we subtract 3 from both sides of the equation above and then divide by 2. Doing this we see that c = - 1.5; and since c = f -1 (0), it means that next f -1 (0) = - 1.5.

next Notes on #2a Be careful not to confuse f(0) with f -1 (0). © 2007 Herbert I. Gross next In the case of f -1 (0), 0 is in the image of f (that is, it’s an output of the f-program). In the case of f(0), 0 is in the domain of f (that is, it’s an input to the f-program”).

Notes on #2a © 2007 Herbert I. Gross To this end we see that… next Just as we can check every subtraction problem by using addition, we can check every inverse function problem by using the function. In the present problem, we need only check that f( - 1.5) = 0. f( x ) = 2( x ) + 3 - 1.5 = - 3 + 3 = 0 next

Solution for #2b: The solution here parallels our solution to Problem #2a. Namely, we are being asked to find the value of c for which… next © 2007 Herbert I. Gross next Since f(x) = 2x + 3, f(c) = 13 f(c) = 2c + 3

Solution for #2b: If we replace f(c) in the equation f(c) = 13 by its value in the equation f(c) = 2c + 3, we obtain the linear equation… next © 2007 Herbert I. Gross next 2c + 3 = 13 …and to solve the equation above we subtract 3 from both sides to obtain 2c = 10, and then we divide both sides by 2 to obtain c = 5. next As a check we see that… f(5) = 2(5) + 3 = 10 + 3 = 13

Solution for #2c: This problem is a generalization of Problems #1a and #1b. Namely, rather than look at f - 1 (x) for a specific value of x, we look at x as the generic name of any member, c, in the domain of f -1. next © 2007 Herbert I. Gross next In other words, we generalize what we did in the previous two parts of Problem #2 as follows…

Solution for #2c: We are being asked to find the value of c for which… next © 2007 Herbert I. Gross next Since f(x) = 2x + 3, f(c) = x f(c) = 2c + 3 If we replace f(c) in the equation f(c) = x by its value in the equation f(c) = 2c + 3, we obtain the linear equation… 2c + 3 = x next

Solution for #2c: And to solve the equation 2c + 3 = x, we subtract 3 from both sides to obtain… next © 2007 Herbert I. Gross next We then divide both sides of the equation 2c = x – 3 by 2 to obtain… c = x – 3 ÷ 2 = 1 / 2 (x – 3) 2c = x – 3 Recalling that c = f -1 (x), we see from the equation c = (x – 3) ÷ 2 = 1 / 2 (x – 3) that f -1 (x) = 1 / 2 (x – 3) next

next Notes on #2c And f -1 can be represented by the following program… © 2007 Herbert I. Gross next f -1 program Step 1Input xStep 2 Subtract 3 Step 3 Divide by 2 (Multiply by 1 / 2 ) Step 4 The output is f -1 (x)

next © 2007 Herbert I. Gross By the definition of the inverse function, if we input a number into the f-program and then use the output as the input for the f -1 -program, the resulting outcome should be the same as the original input. next Notes on #2c So for example, suppose we start with 6 as the input of the f-program. In this case we see that…

next © 2007 Herbert I. Gross next Notes on #2c f program Step 1Input x Step 2 Multiply by 2 Step 3 Add 3 Step 4 The output is f(x) 6 12 15 f -1 program Step 5Input answer from Step 4 Step 6 Subtract 3 Step 3 Divide by 2 Step 4 The output is (x) 15 12 6 6 6

next © 2007 Herbert I. Gross next Notes on #2c f program Step 1Input x Step 2 Multiply by 2 Step 3 Add 3 Step 4 The output is f(x) x 2x 2x + 3 f -1 program Step 5Input answer from Step 4 Step 6 Subtract 3 Step 3 Divide by 2 Step 4 The output is (x) 2x + 3 2x x x x Generalizing this result, we see that…

Solution for #3: A notation such as f(4x + 1) is often confusing to a beginning student. A good way to avoid this confusion is by omitting the x in the expression that defines f(x). next © 2007 Herbert I. Gross next For example, rather than to write f(x) = 2x + 3 write instead f( ) = 2( ) + 3

Solution for #3: In this way all, we have to do is to make sure that whatever we enclose within the parentheses on one side of the equation f( ) = 2( ) + 3, we also do on the other side. next © 2007 Herbert I. Gross next So with respect to the present problem, we want to work with f(4x + 1).

Solution for #3: Hence we place 4x + 1 into both sets of parentheses in the equation below to obtain… next © 2007 Herbert I. Gross next f( ) = 2( ) + 3 4x + 1 And we then use our “rules of the game” to obtain that… f(4x + 1) = 8x + 2 + 3 = next 8x + 5

next © 2007 Herbert I. Gross Notes on #3 next More specifically, since g(x) = 4x + 1, we may replace 4x + 1 by g(x) in which case f(4x + 1) = f(g(x)). In our “program” model, this says we input x into the g-program, and the output of the g-program becomes the input of the f-program. The notation f(4x + 1) is an abbreviation for indicating that the input of one function is the output of another function.

next © 2007 Herbert I. Gross In summary, the output of one function can be the input of another function. This leads to such notations as f(g(x)). We read f(g(x)) starting with x and working our way out. That is… Notes on #3 next We start with x. Run it through the g-program to obtain g(x). Then use g(x) as the input of the f-program to obtain f(g(x)).

© 2007 Herbert I. Gross next Notes on #3 g program Step 1Input x Step 2 Multiply by 4 Step 3 Add 1 Step 4 The output is g(x) f program Step 5Input answer from Step 4 Step 6 Multiply by 2 Step 3 Add 3 Step 4 The output is f(g(x)) For example, with respect to Problem #3 x 4x 4x + 1 8x + 2 8x + 5

next © 2007 Herbert I. Gross The notation f(g(x)), called the composition of f and g, is a bit tricky in the sense that f(g(x)) and g(f(x)) are usually different. Notes on #3 next For example, with f(x) = 2x + 3 and g(x) = 4x + 1 we see that… f(g(x)) = f(4x + 1) = 2(4x + 1) + 3 = 8x + 5 g(f(x)) = g(2x + 3) = 4(2x + 3) + 1 = 8x + 13 but…

next © 2007 Herbert I. Gross next Notes on #3 g program Step 5Input answer from Step 4 Step 6 Multiply by 4 Step 7 Add 1 Step 8 The output is gf((x)) f program Step 1x Step 2 Multiply by 2 Step 3 Add 3 Step 4 The output is f(x) Again in terms of the program model… x 2x 2x + 3 8x + 12 8x + 13

next Composition of Functions The concept of composition of functions was being used implicitly whenever we used the “program” model. Consider the following program… © 2007 Herbert I. Gross next Step 1Input xStep 2 Add 3 Step 3 Multiply by 4Step 4 Add 5 Program Step 5 Output is y

next For example, notice that the output of Step 2 is the input of Step 3; the output of Step 3 is the input of Step 4; and the output of Step 4 is the input of Step 5. © 2007 Herbert I. Gross next Step 1Input x Step 2 Add 3 Step 3 Multiply by 4 Step 4 Add 5 Program Step 5 Output is y Each step in the above program may be viewed as a “one step” function.

next For example, with respect to Step 2, suppose we define the function f by f(x) = x + 3, the command “add 3” may be viewed as f(x). © 2007 Herbert I. Gross Step 1Input x Step 2 Add 3 Step 3 Multiply by 4 Step 4 Add 5 Program Step 5 Output is y Step 2 Add 3 f(x)

next Thus, f(x) is the output of Step 2 and the input of Step 3. © 2007 Herbert I. Gross next Step 1Input x Step 2 Add 3 Step 3 Multiply by 4 Step 4 Add 5 Program Step 5 Output is y Step 2 Add 3 Step 2 Add 3 Step 3 Multiply by 4 We may view the command “multiply by 4” as being the function g, where g(x) = 4x. In this way we may represent the output of Step 3 as g(f(x)). f(x) g( ) f(x)

next In other words the input of Step 4 is g(f(x)). And if we now define h by h(x) = x + 5, we may denote the output of Step 4 by h(g(f(x))). © 2007 Herbert I. Gross f(x) g(f(x)) h( ) Step 1Input x Step 2 Add 3 Step 3 Multiply by 4 Step 4 Add 5 Program Step 5 Output is y Step 2 Add 3 Step 3 Multiply by 4 Step 4 Add 5 g(f(x))

next So if the input of the above program is 2, we may represent the output by h(g(f(2))). That is… © 2007 Herbert I. Gross Step 1Input x2 Program Step 2 Add 3 Step 2 Add 3 5 Step 3 Multiply by 4 20 Step 4 Add 5 25 Step 5 Output is y 25 next f(2) g(f(2)) h(g(f(2)))

Solution for #4a: This is a repeat of Problem #2, only stated in the language of composition of functions. The function and its inverse undo one another. In terms of this example, we have already seen from our solution to Problem #2b that f -1 (13) = 5. Hence… next © 2007 Herbert I. Gross next f(f -1 (13)) = f(5) = 2(5) + 3 = 13

Answer: f(f -1 (13)) = 13 Solution for #4b: This is the “inverse” of Problem #4a. In general if f and g are two functions g(f(x)) and f(g(x)) are different functions. However, in the case of f and f -1, the order makes no difference. In other words, if f(f -1 (13)) = 13 then f -1 f(13)) is also equal to 13. As a check, we know from our previous result that… next © 2007 Herbert I. Gross next f -1 (13) = 5. Hence, f(f -1 (13)) = f(5) = 2(5) + 3 = 13.

Answer: f(f -1 (x)) = x Solution for #4c: This is a generalization of Problems #4a and #4b. That is, in Problems #4a and #4b we chose to let x = 13. However, this problem shows that the result holds true for all values of x. More specifically, we already have shown that… next © 2007 Herbert I. Gross next f -1 (x) = 1 / 2 (x – 3) f(x) = 2x + 3 and

next © 2007 Herbert I. Gross An easy way to visualize the results of Problem #4 is to think of the inverse of a function as being obtained by interchanging the input and the output of the given function. Notes on #4 next What we have shown is that if we interchange the input and the output of the function to obtain its inverse, and then we interchange the input and the output of the inverse function, we are back to the original function. With respect to the above note, keep in mind that f -1 is itself a function; and its inverse is f. next

© 2007 Herbert I. Gross A simple, but important, function (usually denoted by I ) is one that seems to “do nothing”. It’s called the identity function and it’s defined by I (x) = x for every x in the domain of I. Just as the number 1 plays a special role in numerical multiplication, the identity function plays a special role in the composition of functions. More specifically, just as n × 1 = n for every number n, the identity function composed with any function f is still f. That is, I (f(x)) = f( I (x)) = f(x), for all x Identity Function

next © 2007 Herbert I. Gross Recall the property of the multiplicative inverse; namely for any non-zero number n, there exists a number (called the multiplicative inverse of n) denoted by n -1 or 1 / n such that n × n -1 = n -1 × n. Identity Function

next © 2007 Herbert I. Gross A similar property exists for the inverse function of a given function. Namely, if f is any function for which f -1 exists… f(f -1 (x)) = f -1 (f(x)) = I (x) = x Identity Function We call I the identity function because the image of each number in the domain of I is the same as the number itself. In this problem, we chose the specific case for which f(x) = 2x + 3. next

© 2007 Herbert I. Gross Keep in mind that even some very simple functions might not have an inverse. More specifically, if two (or more) numbers in the domain of f have the same image, then f cannot possess an inverse function. Notes on #4 As an example, consider the function f where f(x) = x 2 ; and let’s suppose that g is the rule that interchanges the image and domain of f. For example, if f(2) = 4, then g(4) = 2. next

© 2007 Herbert I. Gross However, f( - 2) is also equal to 4. Hence, g(4) is also equal to - 2. Thus, the rule g assigns to the number 4 both 2 and - 2. Notes on #4 The definition of a function requires that for each element in the domain there is one and only one number in the image. next Hence, in this case g is not a function. More generally, for every positive value of n there exist two numbers x for which x 2 = n.

© 2007 Herbert I. Gross If you have difficulty internalizing that f -1 (n) = m means the same thing as n = f(m), we may derive the result more formally by starting with… Notes on #4 f -1 (n) = m next and then using the fact that… f(f -1 (n)) = n

next © 2007 Herbert I. Gross Since f -1 (n) = m, we know that… Notes on #4 f(f -1 (n)) = f(m) If we then replace f(f -1 (n)) in the equation f -1 (n) = m by its value in the equation f(f -1 (n)) = n, we see that… next n = f(m)

Problem #5 © 2007 Herbert I. Gross next In the program shown, what number was the input if the output in Step 9 was 13? next Answer: The input was 5. Step 1Input x Step 2 Add 2 Step 3 Multiply by 3 Step 4 Subtract 5 Program Step 5 Add x Step 6Add 3 Step 7 Divide by 2 Step 8 Add 1 Step 9 The output is 13.

Answer: The output was 5. Solution for #5: The aim of this problem is to reinforce the idea that even if the steps in the “f-program” cannot be undone step-by-step; the inverse function might still exist. next © 2007 Herbert I. Gross In this example, if we start with Step 9, we can undo step 8 by subtracting 1 from 13 to obtain 12. We can then undo Step 7 by multiplying 12 by 2 to obtain 24. We can then undo Step 6 by subtracting 3 from 24 to obtain 21. However, we can’t undo Step 5 because we don’t know the value of x. next

Solution for #5: So the better approach is to paraphrase the program step-by-step and hope that the simpler form will supply us with the answer. To this end… next © 2007 Herbert I. Gross next Step 1Input xx Program Step 2 Add 2 x + 2 Step 3 Multiply by 33(x + 2)Step 4 Subtract 5 3x + 6 – 5Step 5 Add x (3x + 1) + xStep 6Add 34x + 1 + 3Step 7 Divide by 2 1 / 2 (4x + 4) Step 8 Add 12x + 2 + 1Step 9 The output is 13. 2x + 3 3x + 6 3x + 1 4x + 1 4x + 4 2x + 2 2x + 3

Solution for #5: next © 2007 Herbert I. Gross next Step 1Input xx Program Step 2 Add 2 x + 2 Step 3 Multiply by 33(x + 2) Step 4 Subtract 5 3x + 6 – 5 Step 5 Add x (3x + 1) + x Step 6Add 34x + 1 + 3 Step 7 Divide by 2 1 / 2 (4x + 4) Step 8 Add 12x + 2 + 1 Step 9 The output is 13. 2x + 3 3x + 6 3x + 1 4x + 1 4x + 4 2x + 2 2x + 3 Looking at Step 9 in the above program, we see that 2x + 3 = 13 from which it follows that 2x =10 and x = 5.

next © 2007 Herbert I. Gross As a check, we may replace x by 5 in the above program and see whether we obtain 13 as the output. Doing this we see that… Notes on #5 next Step 1Input x Step 2 Add 2 Step 3 Multiply by 3 Step 4 Subtract 5 Program Step 5 Add x (i.e.,5) Step 6Add 3 Step 7 Divide by 2 Step 8 Add 1 Step 9 The output is 13. 5 7 21 16 21 24 12 13 13 Check

next © 2007 Herbert I. Gross We have shown that the Given Program could be replaced by the simpler program above. Notes on #5 next Step 1 Input x Step 2 Add 2 Step 3 Multiply by 3 Step 4 Subtract 5 The Given Program Step 5 Add x Step 6Add 3 Step 7 Divide by 2 Step 8 Add 1 Step 9 The output is 13. Step 1 Input x Step 2 Multiply by 2 Step 3 Add 3 Step 4 The output is f(x) The Equivalent Program

next © 2007 Herbert I. Gross Notice that the above program represents the same f(x), namely f(x) = 2x + 3, that we used in Problems #1 through #4; during which time we showed that f -1 (x) = 1 / 2 (x – 3). Hence, f -1 is also the inverse of the given function. Notes on #5 Step 1 Input x Step 2 Multiply by 2 Step 3 Add 3 Step 4 The output is f(x) The Equivalent Program = f(x)

next © 2007 Herbert I. Gross We can now show that the given function possesses an inverse and in fact the inverse function is f -1 where f -1 (x) is shown above. By way of review… Step 1 Input x Step 2 Multiply by 2 Step 3 Add 3 Step 4 The output is y ( = f(x) ) f(x)Step 1Input xStep 2 Subtracxt 3 Step 3 Divide by 2Step 4 The output x ( = f -1 (y) ) f -1 (x) next

© 2007 Herbert I. Gross next Step 1Input xx The Given Program Step 2 Add 2 x + 2 Step 3 Multiply by 33(x + 2) Step 4 Subtract 5 3x + 6 – 5 Step 5 Add x (3x + 1) + x Step 6Add 34x + 1 + 3 Step 7 Divide by 2 1 / 2 (4x + 4) Step 8 Add 12x + 2 + 1 Step 9 The output is y ( = 2x + 3 ). 2x + 3 3x + 6 3x + 1 4x + 1 4x + 4 2x + 2 2x + 3 More specifically… The f -1 Program Step 10Input y2x + 3Step 11 Subtract 3 2x Step 12 Divide by 2xStep 13 The output is x ( = f -1 (y) ) x x

next © 2007 Herbert I. Gross Even with a good knowledge of the rules there are times when it is extremely difficult or maybe even impossible to paraphrase a problem into a form that we can solve by algebraic means. In such cases, trial-and- error can often be helpful. A Note on Trial- and-Error For example, suppose we wanted to solve Problem #5 but didn’t know how to use algebra to simplify the program. next

© 2007 Herbert I. Gross We could have made guesses for the value of the input such as shown below… next Step 1Input x Step 2 Add 2 Step 3 Multiply by 3 Step 4 Subtract 5 Program Step 5 Add x Step 6Add 3 Step 7 Divide by 2 Step 8 Add 1 Step 9 The output is 13. 3 5 15 10 13 16 8 9 10 Trial 8 10 30 25 33 36 18 19 19 Trial next

© 2007 Herbert I. Gross When we let x = 3, the output was 9, which was too small to be the correct answer. Step 1Input x (5) Step 2 Add 2 Step 3 Multiply by 3 Step 4 Subtract 5 Program Step 5 Add x (5) Step 6Add 3 Step 7 Divide by 2 Step 8 Add 1 Step 9 The output is 13. 3 5 15 10 13 16 8 9 9 Trial 8 10 30 25 33 36 18 19 19 Trial next And when we let x = 8, the output was 19, which was too big to be the correct answer. 3 9 8 19

next © 2007 Herbert I. Gross This tells us that the answer we seek is greater than 3 but less than 8. next We may continue in this way until we either arrive at the exact number or an approximation that is sufficient for our needs. Step 1Input x (5) Step 2 Add 2 Step 3 Multiply by 3 Step 4 Subtract 5 Program Step 5 Add x (5) Step 6Add 3 Step 7 Divide by 2 Step 8 Add 1 Step 9 The output is 13. 3 5 15 10 13 16 8 9 13 Trial 8 10 30 25 33 36 18 19 13 Trial 38

Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 2 By Herbert I. Gross and Richard A. Medeiros next.

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Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 2 By Herbert I. Gross and Richard A. Medeiros next.

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Presentation on theme: "Algebra Problems… Solutions Algebra Problems… Solutions © 2007 Herbert I. Gross Set 17 part 2 By Herbert I. Gross and Richard A. Medeiros next."— Presentation transcript:

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