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igor_vitrenko [27]
3 years ago
7

If all four sides of a quadrilateral measure the same length, and one corner angle measures 90°, circle that shape the quadrilat

eral can be classified as.
A.) Rhombus
B.) Rectangle
C.) Square
D.) Parallelogram
E.) All of these
Mathematics
2 answers:
emmasim [6.3K]3 years ago
5 0
Its B for sure!!!!
hope this help

Andrei [34K]3 years ago
3 0
The answer is e. all of these because a square has all four sides the same length and has 90 degree angles and rectangles and rhombuses can be squares. And all of these shapes are parallelograms 
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(a + b - c )(a + b + c )
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You are constructing a cardboard box from a piece of cardboard with the dimensions 2 m by 4 m. You then cut equal-size squares f
postnew [5]

Answer:

Dimensions:

l =3,15 m

w=1,15 m

x= 0,42 m   (height)

V(max) = 1,52 m³

Step-by-step explanation:

The cardboard is L = 4 m   and   W = 2 m

Let call x the length of the square to cut in each corner, then, volume of open box is:

For the side L       is  L - 2*x               l =  4 - 2*x

For the side W      is  W - 2*x             w=  2 - 2*x

The height              is  x

Volume of the open box, as function of x is:

V(x) = ( 4 -2x) * ( 2 - 2x) *x    ⇒  V(x) = ( 8 - 8x -4x + 4x²) *x

V(x) = ( 8 -12x + 4x² )*x     V(x) = 8x - 12x² + 4x³

V(x) = 8x - 12x² + 4x³

Taking derivatives on both sides of the equation

V´(x) = 8 - 24x + 12x²

V´(x) = 0       8  - 24x + 12x² = 0     reordering    12x² - 24x + 8  = 0

or    3x² - 6x + 2 = 0

A second degree equation. Solving for x

x₁,₂  = ( 6 ± √36 - 24 ) /6

x₁,₂  = ( 6 ± 3.46) /6

x₁  =  6 + 3,46 /6     x₁  = 1.58  we dismiss such solution because 1,58 * 2 = 3,15 and is bigger than 2 one of the side of the cardboard

x₂  =( 6 - 3,46 ) / 6

x₂  = 0,42 m

Dimensions of the open box

l  = 4 - 2*x     l  = 4 - 0,85      l  =  3,15 m

w = 2 -2*x    w  = 2 - 0,85     w = 1,15 m

x = 0,42 m

V(max) =3,15*1,15*0,42

V(max) = 1,52 m³

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Answer 1,2 and 3 and challenge thank you
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