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Grace [21]
3 years ago
12

How many ways are there to arrange 7 distinct red and 5 distinct blue balls in a row such that:?

Mathematics
1 answer:
BARSIC [14]3 years ago
5 0
You can change them around up to 144 times.
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Rationalize the numerator <br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B3%20%5Ctimes%20a%20%5Ctimes%20b%20%5Cfrac%3Cbr%3E%7
Eva8 [605]

Answer:

rationalize my skill capability

Step-by-step explanation:

7 0
3 years ago
Solve the system. <br> y= -2x + 1<br><br> y= 2x - 3
Neporo4naja [7]

Equation 1: y = -2x + 1

Equation 2: y = 2x - 3

Since both equations already have y isolated, we are able to simply set the right side of both equations equal to each other. Since we know that the value of y must be the same, we can do this.

-2x + 1 = 2x - 3

1 = 4x - 3

4 = 4x

x = 1

Then, we need to plug our value of x back into either of the original two equations and solve for y. I will be plugging x back into equation 2 above.

y = 2x - 3

y = 2(1) - 3

y = 2 - 3

y = -1

Hope this helps!! :)

5 0
3 years ago
Read 2 more answers
What number complete the fallowing equation? 8×(40+7)=(8×_)+(8×7) A=40 B=47 C=320 D =376
11Alexandr11 [23.1K]

8x40=320 (+) 8x7=56 (=) 376.

A=40

hope this helped


3 0
3 years ago
Read 2 more answers
Are these equations linear or non-linear?​
never [62]

Answer:

Linear

Step-by-step explanation:

Find the degree of the equation to determine if linear.

7 0
3 years ago
One positive number is one-fifth of another number. the difference of the two numbers is 84. find the numbers.
seraphim [82]
Answer:  The numbers are:  " 21 " and " 105 " .
___________________________________________________
Explanation:
___________________________________________________
Let "x" be the "one positive number:

Let "y" be the "[an]othyer number".

x = 1/5 (y)
___________________________________________________
Given that the difference of the two number is "84" ;  and that "x" is (1/5) of  "y" ;  we determine that "x" is smaller than "y".

So, y − x = 84 .

Add "x" to each side of this equation; to solve for "y" in terms of "x" ;

y − x + x = 84 + x  ;

 y = 84 + x ;
___________________________________________________
So, we have: 

 x = (1/5) y ;

and:  y = 84 + x  ;

Substitute "(1/5)y" for "x" ;  in  "y = 84 + x " ;  to solve for "y" ;

 y = 84 + [ (1/5)y ]

Subtract  " [ (1/5)y ] " from EACH SIDE of the equation ;

y − [ (1/5)y ] = 84 + [ (1/5)y ] −  [ (1/5)y ]  ;

to get:

  [ (4/5)y ] = 84 ;


       ↔    (4y) / 5 = 84  ;
      
        →  4y = 5 * 84  ;

      Divide EACH SIDE of the equation by "4" ; 
to isolate "y" on one side of the equation; and to solve for "y" ;

           4y / 4 = (5 * 84) / 4 ;

                 y =  5 * (84/4) = 5 * 21 = 105 .

   y = 105 .
___________________________________________________
Now, plug "105" for "y" into:
___________________________________________________
Either:
___________________________________________________
 x = (1/5) y ;

OR:

  y = 84 + x  ;
___________________________________________________
to solve for "x" ;
___________________________________________________
Let us do so in BOTH equations; to see if we get the same value for "x" ; which is a method to "double check" our answer ;
___________________________________________________
Start with:

x = (1/5)y 

    →  (1/5)*(105) = 105 / 5 = 21 ;  x = 21 ; 

___________________________________________________
So, x = 21;  y = 105 .
___________________________________________________
Now, let us see if this values hold true in the other equation:
___________________________________________________
y = 84 + x ;

105 = ? 84 + 21 ?
 
105 = ? 105 ? Yes!
___________________________________________________
The numbers are:  " 21 " and  "105 " .
___________________________________________________

6 0
3 years ago
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