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3 years ago

What attributes help you tell the difference between a rhombus and a rectangle? Explain.

2 answers:

4 0

Angles and sides are the main differences! In a rectangle, all angles will be 90 degrees. In a rhombus, opposite angles will always be congruent. And, all sides in a rhombus are congruent, while in a rectangle, opposite sides are congruent.

An easy way to think of it is that, if a rhombus and a rectangle had a baby, it would be a square! A square gets its 90 degree angles from the rectangle, and the congruent sides from the rhombus. Hope that helped!

goblinko [34]3 years ago

3 0

The angles will help you tell the difference. In a rectangle, all angles are 90 degrees. In a rhombus, the 2 opposite facing angles are the same and all side lengths are the same.
Hope I helped. :P

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a local high schools cheerleading squad is selling snow cones during lunch to raise money for a national cheerleading competitio

svlad2 [7]

We have:

Volume of cone = 535 cm³ at the maximum
Height of cone= 8cm

We know the formula to find volume of cone is $\frac{1}{3} r^{2} \pi h$

We need to find the radius of the base of the cone
$535= \frac{\pi r^{2}8} {3}$
$535*8= \pi r^{2}$
$\frac{4280}{\pi}= r^{2}$
$r= \sqrt{ \frac{4280}{ \pi} }$
$r=36.9$ to the nearest one decimal place

The width of the opening of the cone is the diameter of the circle. Diameter is twice the radius, hence 2×36.9=73.8 cm

8 0

3 years ago

My favorite songstress gives away 25% of the cells of her signature fragrance collection to her charity. last month her cells we

Pachacha [2.7K]

Answer:

$300

Step-by-step explanation:

Multiply 1,200 by .25 and that is your answer

3 0

3 years ago

Someone please help me I only have until January 18th to turn this in. Question: Mr Lopez picks up 2 pallets of dog food to deli

miv72 [106K]

Answer:

Mr. Lopez delivers 1.5 tons of dog food to the animal shelter.

Step-by-step explanation:

Mr Lopez picks up 2 pallets of dog food to deliver to the animal shelter. Each pallet contains 40 bags of dog food, and the dog food weighs 37 1/2 pounds pet bag.

The weight of 1 bag of dog food = $37\frac{1}{2}\text{ pounds }$

The weight of 40 bags of dog food = $40 \times 37\frac{1}{2} =1500 \text{ pounds }$

Each pallet weighs = 1500 pounds

2 pallets weighs = 1500×2=3000 pounds.

We know that 2000 pounds = 1 ton

$\frac{3000}{2000} \text{ pounds}=1.5 \text{ tons}$

Hence, Mr. Lopez delivers 1.5 tons of dog food to the animal shelter.

6 0

3 years ago

Explain why the graph of a linear equation in the form of y=c is a horizontal line

xeze [42]

Let's say "c" is a constant, hmmmm any constant, for any value whatsoever of "x", "y" is always that constant, for example, say c = 3, thus y = 3, so a table for it will look like

$\bf \begin{array}{ccll}
x&y\\
\text{\textemdash\textemdash\textemdash}&\text{\textemdash\textemdash\textemdash}\\
-\infty&3\\
-6&3\\
-5&3\\
-4&3\\
-3&3\\
-2&3\\
-1&3\\
0&3\\
1&3\\
2&3\\
3&3\\
4&3\\
5&3\\
6&3\\
+\infty&3
\end{array}$

now, if you plot those points, it'd looks like the picture below.

4 0

3 years ago

A right triangle has one vertex on the graph of y = 9 - x^2 , x > 0, at ( x , y ), another at the origin, and the third on th

jenyasd209 [6]

Answer:

a. A(x) = (1/2)x(9 -x^2)

b. x > 0 . . . or . . . 0 < x < 3 (see below)

c. A(2) = 5

d. x = √3; A(√3) = 3√3

Step-by-step explanation:

a. The area is computed in the usual way, as half the product of the base and height of the triangle. Here, the base is x, and the height is y, so the area is ...

A(x) = (1/2)(x)(y)

A(x) = (1/2)(x)(9-x^2)

b. The problem statement defines two of the triangle vertices only for x > 0. However, we note that for x > 3, the y-coordinate of one of the vertices is negative. Straightforward application of the area formula in Part A will result in negative areas for x > 3, so a reasonable domain might be (0, 3).

On the other hand, the geometrical concept of a line segment and of a triangle does not admit negative line lengths. Hence the area for a triangle with its vertex below the x-axis (green in the figure) will also be considered to be positive. In that event, the domain of A(x) = (1/2)(x)|9 -x^2| will be (0, ∞).

c. A(2) = (1/2)(2)(9 -2^2) = 5

The area is 5 when x=2.

d. On the interval (0, 3), the value of x that maximizes area is x=√3. If we consider the domain to be all positive real numbers, then there is no maximum area (blue dashed curve on the graph).

5 0

3 years ago