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UkoKoshka [18]
3 years ago
6

Rewrite without absolute value for the given conditions: y=|x-3|+|x+2|-|x-5|, if x>5

Mathematics
2 answers:
Wittaler [7]3 years ago
5 0

Answer:

  y = x + 4

Step-by-step explanation:

Since x > 5, x-3 > 0, x + 2 > 0 and x - 5 > 0

So Ix - 3I = x - 3,   Ix + 2I = x + 2   and x - 5 = x - 5

Therefore, y = x - 3 + x + 2 - x + 5

                  y = x + 4

Svetlanka [38]3 years ago
5 0

Answer:

Step-by-step explanation:

It might interest you to know what this question means without the restriction. See graph 36b. This shows exactly what happens to the absolute value. There are all sorts of turns and twists in it. Notice all the changes in direction and the minimum at -2,-2. The y intercept is at 0,0.

0,0 is also one of the x intercepts as well. The other is -4,0. The whole point of showing you this is to make you aware of what the restriction does.

The graph is reduced to a line starting at (5,9)

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Step-by-step explanation:

first rewrite  remove all (  )Parentheses

2nd   collect all like terms and calculate

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so she skipped the second step

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2 years ago
Explain how you would solve for the value of x when you are given the solution to a function.
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Answer:

Substitute

Step-by-step explanation:

Substitute the values given in the function

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3 years ago
What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
olasank [31]

First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

130 = 2 • 5 • 13

231 = 3 • 7 • 11

so n must be free of 2, 3, 5, 7, 11, and 13, which are the first six primes. It follows that n = 17 must the least integer that satisfies the conditions.

To verify the claim, we try to solve the system of congruences

\begin{cases} 17x \equiv 1 \pmod{130} \\ 17y \equiv 1 \pmod{231} \end{cases}

Use the Euclidean algorithm to express 1 as a linear combination of 130 and 17:

130 = 7 • 17 + 11

17 = 1 • 11 + 6

11 = 1 • 6 + 5

6 = 1 • 5 + 1

⇒   1 = 23 • 17 - 3 • 130

Then

23 • 17 - 3 • 130 ≡ 23 • 17 ≡ 1 (mod 130)

so that x = 23.

Repeat for 231 and 17:

231 = 13 • 17 + 10

17 = 1 • 10 + 7

10 = 1 • 7 + 3

7 = 2 • 3 + 1

⇒   1 = 68 • 17 - 5 • 231

Then

68 • 17 - 5 • 231 ≡ = 68 • 17 ≡ 1 (mod 231)

so that y = 68.

3 0
2 years ago
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