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marysya [2.9K]
2 years ago
14

Race to get brainliest, need help!

Mathematics
2 answers:
Gemiola [76]2 years ago
7 0

Answer:

Original number is 86

Step-by-step explanation:

              Original:   10x + y

Double reversed:   2(10y + x)

Constraints:

  x + y = 14                                     ===>            x = 14 - y

  2(10y + x) + 10x + y = 222          ===>           21y + 12x = 222

Substitution to find y:

  21y + 12x = 222

  21y + 12(14 - y) = 222

  21y + 168 - 12y = 222

            9y + 168 = 222

                      9y = 54

                        y = 6

Substitution to find x:

  x = 14 - y

  x = 14 - 6

  x = 8

Original:   10x + y

                 10(8) + (6)

                 80 + 6

                 86

Sophie [7]2 years ago
4 0
X = Tens place
y = Ones place
x+y=14
y=14-x
2(10y+x)+(10x+y)=222
20y+2x+10x+y=222
21y+12x=222
21(14-x)+12x=222
294-21x+12x=222
-9x = -72
x = 8
y = 14-8 =6
ANSWER: 86
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Find the dimensions of the open rectangular box of maximum volume that can be made from a sheet of cardboard 23 in. by 13 in. by
faltersainse [42]

Answer:

The dimension of the box is 17.66 in by 7.66 in by 2.67 in.

Therefore the volume of the box is 361.19 in^3.

Step-by-step explanation:

Given that the dimensions of a cardboard is 23 in by 13 in.

Let the side of the square be x in.

Then the length of the box= (23-2x) in

and the width of the box =(13-2x) in

and height = x in.

The volume of the box is = length ×width × height

                                         =[(23-2x)(13-2x)x] in^3

                                         =(299x-72x² +4x^3) in^3

∴V=299x-72x² +4x³

Differentiating with respect x

V'= 299-144x+12x²

Again differentiating with respect x

V''= -144+24x

To find the dimensions, we set V'=0

∴299-144x+12x²=0

Applying Sridharacharya formula that is the solution of a quadratic equation ax²+bx+c is x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}

Here a=12, b=-144 , c=299

\therefore x=\frac{-(-144)\pm\sqrt{(-144)^2-4.12.299}}{2.12}

\Rightarrow x= 9.33, 2.67

If  we take x=9.33 in, then the width of the box [13-(2×9.33)] will negative.

∴x = 2.67 in

If at x = 2.67, V''<0 , then the volume of the box will be maximum or V''>0 then volume of the box will be minimum.

V''|_{x=2.67}=-144+(24\times 2.67)=-79.92

Therefore  at x = 2.67, the volume of the box maximum.

The length of the box =[23-(2×2.67)] in

                                    =17.66 in

The width of the box =[13-(2×2.67)] in

                                  =7.66 in

The height of the box= 2.67 in

The dimension of the box is 17.66 in by 7.66 in by 2.67 in.

Therefore the volume of the box is =(17.66×7.66×2.67) in^3

                                                            =361.19 in^3

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Answer:

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