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geniusboy [140]
3 years ago
9

Sometimes you to ___ some points to get a good approximation of the location of extreme values

Mathematics
1 answer:
Marianna [84]3 years ago
3 0

Answer:

Sometimes you to plot some points to get a good approximation of the location of extreme values.

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Somebody tell me this ...
Sloan [31]
I think y= -9/12x + .5
4 0
3 years ago
Which of the following numbers is a multiple of 8?
Andru [333]
16, 24, 32, 40, 48, 56, 64, 72, etc.
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3 years ago
This is just a random question don't answer im testing something out
notsponge [240]

Answer:

I think these should help.

Step-by-step explanation:

4 0
3 years ago
2a + 4b + c =5 <br> x =-2 <br><br> a - 4b = -6<br> y=1
cricket20 [7]
There is no solution ,<span>a+c=-10;b-c=15;a-2b+c=-5 </span>No solution System of Linear Equations entered : [1] 2a+c=-10 [2] b-c=15 [3] a-2b+c=-5 Equations Simplified or Rearranged :<span><span>  [1] 2a + c = -10 </span><span>  [2] - c + b = 15 </span><span>  [3] a + c - 2b = -5 </span></span>Solve by Substitution :

// Solve equation [3] for the variable  c  
 

<span> [3] c = -a + 2b - 5 </span>

// Plug this in for variable  c  in equation [1]

<span><span>  [1] 2a + (-a +2?-5) = -10 </span><span>  [1] a = -5 </span></span>

// Plug this in for variable  c  in equation [2]

<span><span>  [2] - (-? +2b-5) + b = 15 </span><span>  [2] - b = 10 </span></span>

// Solve equation [2] for the variable  ?  
 

<span> [2] ? = b + 10 </span>

// Plug this in for variable  ?  in equation [1]

<span><span>  [1] (? +10) = -5 </span><span>  [1] 0 = -15 => NO solution </span></span><span>No solution</span>
3 0
3 years ago
Find the coordinates of C' after a reflection across the line
iragen [17]
<h3>Answer:  (4,2)</h3>

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

Explanation:

C is at (0,0). Ignore the other points.

Reflecting over y = 1 lands the point on (0,2) because we move 1 unit up to arrive at the line of reflection, and then we keep going one more unit (same direction) to complete the full reflection transformation. I'll call this point P.

Then we reflect point P over the line x = 2 to arrive at the location Q = (4,2). Note how we moved 2 units to the right to get to the line of reflection, and then keep moving the same direction 2 more units, then we have applied the operation of "reflect over the line x = 2"

So we have started at C = (0,0), moved to P = (0,2) and then finally arrived at the destination Q = (4,2). This is the location of C' as well.

All of this is shown in the diagram below.

4 0
3 years ago
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