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brilliants [131]
3 years ago
12

A. x-int:-0.5,y-int:1B. x-int:0.5,y-int:1C. x-int:-0.5,y-int:-1D. x-int:1,y-int:0.5

Mathematics
1 answer:
hodyreva [135]3 years ago
5 0
The correct answer is:  [D]:  " <span>x-int : 1 ,  y-int:  0.5  " .
_____________________________________________________
Note:
_____________________________________________________
The "x-intercept" refers to the point(s) at which the the graph of a function (which is a "line", in this case) cross(es) the "y-axis".  

In other words, what is (are) the point(s) of the graph at which "x = 0<span>" ?
</span>
By examining the graph, we see that when " x = 0" ; y is equal to:  "1<span>" .
</span>
So; the "x-intercept" is at point:  "(0, 1)" ; or, we can simply say that the 
   "x-intercept" is:  "1" .
_________________________________________________________</span>  Note:
_____________________________________________________
The "y-intercept" refers to the point(s) at which the the graph of a function (which is a line, in this case) cross(es) the "x-axis".  

In other words, what is (are) the point(s) of the graph at which " y = 0 <span>" ?
</span>
By examining the graph, we see that when " y = 0 " ; x  is equal to:  "0.5<span>" .
</span>
So; the "x-intercept" is at point:  "(0.5, 0)" ; or, we can simply say that the 
   "y-intercept" is:  "0.5 " .<span>
______</span>_________________________________________________
This would correspond to:<span>
_______________________________________________________
           Answer choice:  [D]:  </span>" x-int: 1 , y-int: 0.5  " .
_______________________________________________________
     {that is;  The "x-intercept" is: "0" ; and the "y-intercept" is:  "0.5 ".} .
_______________________________________________________
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What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{130}$ and $n^{-1}\pmod{231}$ are both defined?
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First of all, the modular inverse of n modulo k can only exist if GCD(n, k) = 1.

We have

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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:

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231 = 13 • 17 + 10

17 = 1 • 10 + 7

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Then

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

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