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Ber [7]
2 years ago
5

Consider the following 8 numbers, where one labelled

Mathematics
1 answer:
AlladinOne [14]2 years ago
7 0
<h3>Answers:     x = -17  and   x = 64</h3>

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

Explanation

Consider three scenarios:

  • A) The value of x is the smallest of the set (aka the min)
  • B) The value of x is the largest of the set (aka the max)
  • C) The value of x is neither the min, nor the max. So 8 < x < 39.

These scenarios cover all the possible cases of what x could be. It's either the min, the max, or somewhere in between the min and max.

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

We'll start with scenario A.

If x is the min, then that must mean 39 is the max as it's the largest of the set {18, 36, 16, 39, 27, 8, 34}

The range is 56, so,

range = max - min

56 = 39 - x

56+x= 39

x = 39-56

x = -17  which is one possible answer

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

If instead we go with scenario B, then x is the max and 8 is the min

range = max - min

56 = x - 8

56+8 = x

64 = x

x = 64 is the other possible answer

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

Lastly, let's consider scenario C. If x is not the min or the max, then it's somewhere between the min 8 and max 39. in short, 8 < x < 39.

Note that range = max - min = 39-8 = 31 which is not the range of 56 that we want. So there's no way scenario C can be possible here.

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Write the slope-intercept form of the equation of each line 13x-2y=14
HACTEHA [7]

Answer:

Step-by-step explanation:

First, lets put this into y=mx+b form.

You can bring the 13x over to the 14, by subtracting and you get -2y=14-13x, or also -2y = -13x + 14

Then you can divide both sides by -2 to get y, and you get

-2y/-2 = -13x/-2 + 14/-2 which is simplified to y=13/2x + (-7) which is

y=13/2x-7

                                {[( IMPORTANT )]}

 THIS HAS THE SLOPE IN A IMPROPER FRACTION... CHECK IF YOU USE MIXED NUMBERS OR IMPROPER FRACTIONS

the mixed number form is y = 7     1/2 x -7

Hope this helps!

       

7 0
3 years ago
Mr. Paulsen challenges Olivia and Angelica to move figure ABCDE onto figure A"B"C"D"E" using a series of two different transform
Jobisdone [24]

Problem 1

One method Olivia could have used is to reflect over the x axis and then translate 10 units to the right.

Let's apply these set of steps to point C.

C = (-2,-4)

C' = (-2,4) after reflecting over the x axis; y coordinate flips in sign

C'' = (8,4) after shifting to the right 10 units; add 10 to the x coordinate

Point C(-2,-4) moves to C''(8,4) which is shown in the diagram. I'll let you check the other four points.

-----------

Another method Olivia could have used is to shift to the right 10 units first, then reflect over the x axis. In this case, the order doesn't matter. Though for some other combinations of transformations, order does matter.

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

Problem 2

Reflecting over the y axis will have the point (x,y) turn into (-x,y). Only the x coordinate changes, and we flip the sign here. The y coordinate stays the same.

A point like C(-2,-4) becomes C'(2,-4) after the y axis reflection. Shifting up 8 units means we add 8 to the y coordinate to arrive at C''(2,4) but this does not match what the diagram shows. Therefore, Angelica's method is incorrect.

You should find other inconsistencies for the other four points as well.

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

Problem 3

Michael's method isn't correct either. He does not apply an x axis reflection and instead relies entirely on translations to move ABCDE to A''B''C''D''E''

At a quick glance, it looks like Michael may be correct. But upon a much closer look, you'll see that things don't line up perfectly.

Instead of point C, we'll pick on point D this time.

D is located at (-6,-2). If we shift up 8 units, then we add 8 to the y coordinate to get to D'(-6,6). Then add 10 to the x coordinate to shift 10 units to the right. This leads us to D''(4,6).

However, the diagram shows D'' is actually at (4,2)

I'll let you check the other points. You should find that only points C and E go to the correct locations, but the other points do not. Again, this is because Michael did not apply an x axis reflection. His figure of A''B''C''D''E'' is upside down compared to what it should be.

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

To summarize, the methods described in problem 1 are the correct way to get ABCDE to move to A''B''C''D''E''. If Olivia followed either of those two methods, then she is correct. Both Angelica and Michael have incorrect methods.

5 0
3 years ago
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