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Aleks [24]
2 years ago
5

1) The graph shows y as a function of x Suppose a point is added to this graph. Which choice gives a point that preserves the fu

nction?
A) (5,5)
B) (0,9)
919.-1)
D) (2.-4

Mathematics
1 answer:
barxatty [35]2 years ago
5 0
<h3>Answer:  B)  (0,9)</h3>

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

Explanation:

The points on the given graph are:

  • (-7,4)
  • (-3,-3)
  • (2,9)
  • (3,-8)
  • (5,7)
  • (9,3)

Notice that the <em>slightly</em> darker grid lines indicate 5 on either x or y axis. This helps visually spot where the points are fairly quickly. Though admittedly, it would be better if your teacher provided numbers on the x and y scales.

Now if we were to add something like point A(5,5) to the graph, then this function would cease to be a function. This is because the points (5,7) and (5,5) fail the vertical line test. The input x = 5 leads to the outputs y = 5 and y = 7 at the same time. A function is only possible if any x input leads to exactly one y output only.

So this is why we eliminate choice A from the possible answer pool.

Choice C is a similar story. The two points (9,3) and (9,-1) fail the vertical line test this time, i.e. the input x = 9 has multiple outputs y = 3 and y = -1 simultaneously. We rule out choice C.

Choice D is also ruled out because of (2,-4) and (2,9) failing the vertical line test.

We're left with choice B. Luckily, adding the point (0,9) to the mix does not break the function. The input x = 0 wasn't part of the original function.

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

In short, we eliminate choices A,C and D because they fail the vertical line test. Choice B does not fail the vertical line test so it's the final answer.

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3 years ago
8x^2+23=823<br><br> whats going to be the answer I got 0 am I right?
Naily [24]
1) No; you are incorrect.  "0" is NOT a solution.

Plug in "0" for "x" ;

8*0² + 23 =? 823? ; 

8*0  + 23 =? 823? ;

  0 + 23 = ? 823 ? ;  No! ;  "0 + 23 = 23 " ; NOT "823" .
_________________________________________________
2)  The answers:  x = 10, -10 ; or, write as:  x = ± 10 .
_________________________________________________

To solve:

GIven:  8x² + 23 = 823 ; 

Subtract "23" from EACH SIDE of the equation:
_____________________________________________
           8x² + 23 - 23 = 823 - 23 ; 
to get:  
           8x²  = 800 ; 

Now, divide EACH SIDE of the equation by "8" ; 

          8x² / 8 = 800 / 8 ;

to get:  x² = 100 ; 

        Take the "square root" of EACH SIDE of the equation; to isolate "x" on one side of the equation; and to solve for "x" ; 

          √(x²) = √(100) ;

             x  = <span>± 10 .

The answers:  x = 10, -10 ; or, write as:  x = </span><span>± 10 .
_____________________________________________________</span>
5 0
3 years ago
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