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Kaylis [27]
3 years ago
13

A gardener who wants a rectangular garden whose lot perimeter is 40 meters is suggested by a fellow gardener a plan that would m

ake it so that the build would have a maximum lot area. What could be the suggestion? You must answer in a mathematical solution. Show work.
Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
8 0

Answer:

a*(20-a)

Step-by-step explanation:

Let's assume that the length of the garden is a the width is b and the area is s.

b = 40/2 - a = 20-a

s =a*(20-a)=-a^2+20a=-(a-10)^2+100

then,

when a=10, max(-(a-10)^2) =0, max(s)=100

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4 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
Help Please! (8th Grade Math) :)
Yanka [14]

Answer:

c = 75 in

Step-by-step explanation:

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Plug in: <em>45² + 60² = c²</em>

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Add: √5625 = c

Square root: 75 in = c (no negatives because we are finding distance)

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Answer:

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Step-by-step explanation:

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3 years ago
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