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scoundrel [369]
3 years ago
15

Mr. Cannon has a piece of land that is 3/7 of an acre. He wants to build 10 houses on his land. How much land does each house ge

t?
Show me the division statement
explain why you chose the dividend and the divisor
give me the quotient.
Mathematics
1 answer:
sashaice [31]3 years ago
4 0

Answer:

Step-by-step explanation:

3/7 divided by 10 houses= 0.042 acres per house

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I have the formula <br><br> A fraction • π r^2
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Step-by-step explanation:

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4 0
4 years ago
What is the answer explain please
eduard

Answer:

Pattern B

<h3> Explain:  </h3>

A quadratic relationship is characterized by constant second differences.

<em><u>Pattern A </u></em>

Sequence: 0, 2, 4, 6

First Differences: 2, 2, 2 . . . . constant indicates a 1st-degree (linear, arithmetic) sequence

__________________________________________________________

<em><u>Pattern B</u></em>

Sequence: 1, 2, 5, 10

First Differences: 1, 3, 5

Second Differences: 2, 2 . . . . constant indicates a 2nd-degree (quadratic) sequence

__________________________________________________________

<em><u>Pattern C</u></em>

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Pattern B shows a geometric relationship between step number and dot count.

6 0
2 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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