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3 years ago

Determine if the statement is Sometimes/Always/Never true. A square is a rhombus

1 answer:

8 0

To determine first if A Square is a Rhombus.

Let us define a rhombus.

1) Has all four sides equal.

2) Has opposite sides parallel.

3) The diagonals bisect the angles and bisect each other at 90 degrees.

A Square also displays all of these properties.

So a Square is ALWAYS a Rhombus.

But a Rhombus is a not always a Square, except that Rhombus is a Square.

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AVprozaik [17]

Answer:

thats alot

Step-by-step explanation:

feels bad to be in that position, fallen soldier :P

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3 years ago

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PLZ HELP!!

White raven [17]

Answer:

The probablity is 11

Step-by-step explanation:

I am sure it's 11

Hope this answer helps you :)

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7 0

3 years ago

What does 40 divided by 1+3-(3x7)+7-5 equal?

Serjik [45]

It equal to 24 because you have to multiply or divide first then you can add or subtract

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3 years ago

Read 2 more answers

What is the area of the shaded portion of the circle?

SOVA2 [1]

Answer:

The first option is the correct one, the area of the shaded portion of the circle is

[/tex](5 \pi -11.6)ft^2[/tex]

Step-by-step explanation:

Let us first consider the triangle + the shadow.

The full area of the circle is the radius squared times pi, so

A= $(5 ft)^2 \cdot \pi \\25 ft^2 \cdot \pi$

Since $\frac{72^{\circ}}{360^{\circ}}=\frac{1}{5}$ , the area of the triangle + the shaded area is one fifth of the area of the whole circle, thus

$A_1=\frac{1}{5}25 ft^2 \cdot \pi\\ =5 ft^2 \cdot \pi$

If we want to know the area of the shaded part of the circle, we must subtract the area of the triangle from $A_1$ .

The area of the triangle is given by

$A_{triangle}=\frac{1}{2}\cdot (2.9+2.9)ft \cdot 4 ft\\= 11.6 ft^2$

Thus the area of the shaded portion of the circle is

$A_1-A_{triangle}=5 \pi ft^2-11.6ft^2\\= (5 \pi -11.6)ft^2$

3 0

3 years ago

Graph triangle RST with vertices R(3, 7), S(-5, -2), and T(3, -5) and its image after a reflection over x = -3.

kirill115 [55]

Given:

The vertices of a triangle are R(3, 7), S(-5, -2), and T(3, -5).

To find:

The vertices of the triangle after a reflection over x = -3 and plot the triangle and its image on the graph.

Solution:

If a figure reflected across the line x=a, then

$(x,y)\to (-(x-a)+a,y)$

$(x,y)\to (-x+a+a,y)$

$(x,y)\to (2a-x,y)$

The triangle after a reflection over x = -3. So, the rule of reflection is

$(x,y)\to (2(-3)-x,y)$

$(x,y)\to (-6-x,y)$

The vertices of triangle after reflection are

$R(3,7)\to R'(-6-3,7)$

$R(3,7)\to R'(-9,7)$

Similarly,

$S(-5,-2)\to S'(-6-(-5),-2)$

$S(-5,-2)\to S'(-6+5,-2)$

$S(-5,-2)\to S'(-1,-2)$

And,

$T(3,-5)\to T'(-6-3,-5)$

$T(3,-5)\to T'(-9,-5)$

Therefore, the vertices of triangle after reflection over x=-3 are R'(-9,7), S'(-1,-2) and T'(-3,-5).

3 0

3 years ago