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3 years ago

What do rectangles and parallelograms have in common?

Mathematics

2 answers:

IgorLugansk [536]3 years ago

7 0

Answer:

D. They are quadrilaterals that have opposite sides equal in length.

Step-by-step explanation:

Rectangles and parallelograms both have 2 sets of parallel sides. The parallel sides are of equal length.

Rectangles are special parallelograms that have 2 sets of corners that have 90 degree angles

Squares are special rectangles that have all 4 sides of equal lengths.

Aleks [24]3 years ago

3 0

Answer:

They are quadrilaterals that have opposite sides equal in length.

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Rectangle ABCD is graphed in the coordinate plane. The following are the vertices of the rectangle: A ( 2, -6) B ( 5, -6) C (5,-

Sliva [168]

Answer:

The answer is 14

Step-by-step explanation:

Point A to Point B is 3

Point C to Point D is 3

Point B to Point c is 4

Point D to point A is 4.

Add all those together to get 14

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4 years ago

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Mention the commutativity associative and distributive properties of rational number

evablogger [386]

Answer:

Associative property: a + (b + c), a – (b – c) ≠ (a – b) – c, a × (b × c) = (a × b) × c, and a ÷ (b ÷ c) ≠ (a ÷ b) ÷ c

When a = ½ and b = ¾

Now, for checking a × b = b × a, consider LHS and RHS.

LHS = a × b = ½ × ¾ = ⅜

RHS = b × a = ¾ × ½ = ⅜

Thus, LHS = RHS (Hence proved)

Step-by-step explanation:

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3 years ago

PLZ HELP ME GEOMETRY<br> question is below

konstantin123 [22]

The missing justification is for the statement that three angles add to a particular angle. The appropriate choice is ...

... c. Angle Addition Postulate

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3 years ago

Find the critical points of the surface f(x, y) = x3 - 6xy + y3 and determine their nature.

Vedmedyk [2.9K]

Compute the gradient of $f$ .

$\nabla f(x,y) = \left\langle 3x^2 - 6y, -6x + 3y^2\right\rangle$

Set this equal to the zero vector and solve for the critical points.

$3x^2-6y = 0 \implies x^2 = 2y$

$-6x+3y^2=0 \implies y^2 = 2x \implies y = \pm\sqrt{2x}$

$\implies x^2 = \pm2\sqrt{2x}$

$\implies x^4 = 8x$

$\implies x^4 - 8x = 0$

$\implies x (x-2) (x^2 + 2x + 4) = 0$

$\implies x = 0 \text{ or } x-2 = 0 \text{ or } x^2 + 2x + 4 = 0$

$\implies x = 0 \text{ or } x = 2 \text{ or } (x+1)^2 + 3 = 0$

The last case has no real solution, so we can ignore it.

Now,

$x=0 \implies 0^2 = 2y \implies y=0$

$x=2 \implies 2^2 = 2y \implies y=2$

so we have two critical points (0, 0) and (2, 2).

Compute the Hessian matrix (i.e. Jacobian of the gradient).

$H(x,y) = \begin{bmatrix} 6x & -6 \\ -6 & 6y \end{bmatrix}$

Check the sign of the determinant of the Hessian at each of the critical points.

$\det H(0,0) = \begin{vmatrix} 0 & -6 \\ -6 & 0 \end{vmatrix} = -36 < 0$

which indicates a saddle point at (0, 0);

$\det H(2,2) = \begin{vmatrix} 12 & -6 \\ -6 & 12 \end{vmatrix} = 108 > 0$

We also have $f_{xx}(2,2) = 12 > 0$ , which together indicate a local minimum at (2, 2).

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2 years ago

Help Plzzzzzzzzzzzzzzzzzzzzzzzzzzzzz

vladimir2022 [97]

Answer:

line number one is the answer

4 0

3 years ago