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Burka [1]
4 years ago
8

Which point is the midpoint of segment AB?

Mathematics
1 answer:
Tanzania [10]4 years ago
8 0

it is 0.

0 is the midpoint.

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A line segment has the endpoints (9, -16) and L (7,12) find the coordinates of its midpoints M
Anastasy [175]

Answer:

The answer to your question is: (8, -2)

Step-by-step explanation:

Data

A (9, -16)

L (7, 12)

Formula

                Xm = \frac{x1 + x2 }{2}              Ym = \frac{y1 + y2}{2}

                Xm = Xm = \frac{9 + 7}{2}         Ym = \frac{-16 + 12}{2}

               Xm = 8                                              Ym = -2

8 0
3 years ago
If (7^0)^x = 1, what are the possible values of x? Explain your answer.
Mandarinka [93]

Answer:

Endless

Step-by-step explanation:

There are endless possibilities for x, because 1 to the power of any x will always be 1.

8 0
3 years ago
20 points, please answer which option is correct
Natalija [7]

Answer:

A

Step-by-step explanation:

its the only negative whilst others are positive

5 0
2 years ago
Read 2 more answers
What is the change in temperature from -8°F to 14°F?
vova2212 [387]
The change in temperature from -8F to 14F would be 22F because -8 to 14 would require subtraction so 14 - (-8) is 22 and -8 + 22 = 14 so its true.

Answer: 1) 22F.
3 0
3 years ago
Read 2 more answers
Please put down the steps?
e-lub [12.9K]
The answer is:  " x = 0, 1 " .
__________________________________________
Explanation:
__________________________________________
Given:
__________________________________________
   " √(x + 1)  <span>− 1 = x "  ;   Solve for "x" ;

First, let us assume that "x </span>≥ -1 "
<span>
Add "1" to EACH SIDE of the equation:

 </span>→  √(x + 1)  − 1  + 1 = x + 1  ; 

to get:  

 →  √(x + 1)  = x + 1 .

Now, "square" EACH side of the equation:

 →  [√(x + 1) ]²  = (x + 1 )² ;

to get: 

   x + 1 = (x + 1)² 

↔  (x + 1)²  = (x + 1) .


Expand the "left-hand side" of the equation:

 → (x + 1)² = (x + 1)(x +1) ;

Note: (a+b)(c+d) = ac +ad + bc + bd ;

As such:  (x + 1)(x + 1) = (x*x) + (x*1) +(1(x) + (1*1) ;
                                   =  x² + 1x + 1x + 1 ; 
                                   =  x²  + 2x  +  1 ;

Now, substitute this "expanded" value, and bring down the "right-hand side" of the equation; and rewrite the equation:
__________________________________________
   " (x + 1)² = (x + 1)(x +1) " ; 

  →  Rewrite as:  " x² + 2x + 1 = x + 1 " ;

 Subtract "x" ;  and subtract "1" ; from EACH SIDE of the equation:

               →   x² + 2x + 1 - x - 1 = x + 1 - x - 1 ; 

to get:    →   x² <span>− x = 0  

Factor out an "x" on the "left-hand side" of the equation:
___________________________________________
         </span>x² − x = x(x − 1) ;

         →  x (x − 1) = 0 ;

We have:  "x" and "(x − 1)" ; when either of these two multiplicands are equal to zero, then the "right-hand side of the equation equals "zero" .

So, one value of "x" is "0" .

The other value for "x" ;

→  x − 1 = 0 ;

Add "1" to each side of the equation:

  → x − 1 + 1 = 0 + 1 ; 

 →  x = 1 ; 
__________________________________
So, the answers:

 " x = 0, 1 " .
_________________________________

4 0
4 years ago
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