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Andru [333]
3 years ago
11

The product of two consecutive numbers is 4 more than three times the larger. find the integer​

Mathematics
1 answer:
Stella [2.4K]3 years ago
7 0

Answer:

Step-by-step explanation:

Let the first integer be x, then the next is x+1.

x(x + 1) = 3(x + 1) + 4

x^2 + x = 3x+ 3 + 4

x^2 + x - 3x - 7 = 0

x^2 - 2x - 7 = 0

Solving this does not give an integer so maybe there's something wrong with the question??

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Given,

Side of a square = 4xy

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Packaging By cutting away identical squares from each corner of a rectangular piece of cardboard and folding up the resulting fl
Kobotan [32]

Answer:

The dimension of the box is 18.66 in by 2.66 in by 1.17 in.

Step-by-step explanation:

Given that, The cardboard is 21 in. long and 5 in. wide.

Assume,x be the length of the each sides of the square .

Then the length of the box is = (21-2x) in.

The breadth of the box is =(5-2x) in.

The height of the box is  = length of side of the square

                                         = x in.

The volume of the box = length × wide × height

                                      =(21-2x)(5-2x)x

                                      =(105-52x+4x²)x

                                      =(105 x-52x^2+4x^3)

Let

V=105 x-52x^2+4x^3

Differentiating with respect to x

V'=105-104x+12x^2

Again differentiating with respect to x

V''=-104+24x

To find the dimension of the box, we set V'=0

105-104x+12x^2=0

Applying the quadratic formula x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} , here a=12, b= -104 and c =105

\Rightarrow x=\frac{-(-104)\pm\sqrt{(-104)^2-4.12.105}}{2.12}

      =\frac{104\pm 76}{24}

      =7.5, 1.17

For x= 7.5 , the wide of the box will be negative.

∴x=1.17 in.

V''_{x=1.17}=-104+(24\times 1.17)

Since at x= 1.17, V''<0 , therefore at x=1.17 in. the volume of the given box will be maximum.

The length of the box is = [21-(1.17×2)]=18.66 in.

The wide of the box is = [5-(1.17×2)]=2.66 in.

The height of the box is = 1.17 in.

The dimension of the box is 18.66 in by 2.66 in by 1.17 in.

5 0
4 years ago
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