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2 years ago

What is the mode? 0-0-2-1-3-2-1-0

Mathematics

2 answers:

exis [7]2 years ago

4 0

The mode is 0 since it is the most frequent.

Tanya [424]2 years ago

4 0

The mode is 0 shjsjs

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What polygon is formed when 2 equilateral triangles that are congruent and share a side

zalisa [80]

You form a rhombus.

Draw an equilateral triangle, and then mirror it around its base. All the sides are clearly equal, because they are all sides of the same equilateral triangle.

Moreover, the top and bottom angle are 60°, because they are angles of an equilateral triangle, while the left and right angles are 120, because they're 60+60 angles.

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3 years ago

2x2 + 2x-4?

nexus9112 [7]

Answer:

2x^2 +2x-4
——————
2x^2-4x+2

Factor out 2 from the expression

2(x^2+x-2)
—————-
2(x^2-2x+1)

Write x as a difference

2(x^2x-x-2)
—————-
2(x^2-2x+1)

Use a^2-2ab+b^2=(ab)^2

2(x^2x-x-2)
—————-
2(x-1)^2

Reduce the fraction with 2

x^2x-x-2
—————-
(x-1)^2

Factor out x from the expression

X*(x^2)-x-2
—————-
(x-1)^2

Factor out negative sign from the expression

X*(x+2)-(x-2)
—————-
(x-1)^2

Factor out x+2 from the expression

(x+2)(x-1)
—————-
(x-1)^2

Simplify the expression

x+2
——
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6 0

3 years ago

Study the graphs, then select the best answers from the drop-down menus.

-Dominant- [34]

1. b 2. a 3. a

100% correct edge 2021

Step-by-step explanation:

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2 years ago

Kay added 253 + 14 and got a sum of 393. What did she do wrong? Check all that apply.

SashulF [63]

Answer:

She did not correctly add up the numbers

Step-by-step explanation:

253 + 14 = 267

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3 years ago

Arrange the cones in order from lease volume to greatest volume

bearhunter [10]

Answer:

Volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

cone with DIAMETER of 18 & height of 10

cone with RADIUS of 10 & height of 9

cone with RADIUS of 11 & height of 9

cone with DIAMETER of 20 & height of 12

Step-by-step explanation:

Let $V_{2}. V_{3}. and\ V_{4}.$ be the volume of the cone.

Let d, r and h be the diameter, radius and height of the cone.

Given:

$d_{1} = 20\ and\ h_{1}=12$

$d_{2} = 18\ and\ h_{2}=10$

$r_{3} = 10\ and\ h_{3}=9$

$r_{4} = 11\ and\ h_{14}=9$

Arrange the cones in order from lease volume to greatest volume.

Solution:

The volume of the cone is given below.

$V=\pi r^{2} \frac{h}{3}$ ----------------(1)

where: r is radius of the base of cone.

and h is height of the cone.

The volume of the cone for $d_{1} = 20\ and\ h_{1}=12$

$r_{1} = \frac{d_{1}}{2}$

$r_{1} = \frac{20}{2}=10\ units$

$V_{1}=\pi (r_{1})^{2} \frac{h_{1}}{3}$

$V_{1}=\pi (10)^{2} \frac{12}{3}$

$V_{1}=\pi\times 100\times 4$

$V_{1}=400\pi\ units^{3}$

Similarly, for volume of the cone for $d_{2} = 18\ and\ h_{2}=10$

$r_{2} = \frac{d_{2}}{2}$

$r_{2} = \frac{18}{2}=9\ units$

$V_{2}=\pi (r_{2})^{2} \frac{h_{2}}{3}$

$V_{2}=\pi (9)^{2} \frac{10}{3}$

$V_{2}=\pi\times 81\times \frac{10}{3}$

$V_{2}=\pi\times 27\times 10$

$V_{2}=270\pi\ units^{3}$

Similarly, for volume of the cone for $r_{3} = 10\ and\ h_{3}=9$

$V_{3}=\pi (r_{3})^{2} \frac{h_{3}}{3}$

$V_{3}=\pi (10)^{2} \frac{9}{3}$

$V_{3}=\pi\times 100\times 3$

$V_{3}=\pi\times 300$

$V_{3}=300\pi\ units^{3}$

Similarly, for volume of the cone for $r_{4} = 11\ and\ h_{4}=9$

$V_{4}=\pi (r_{4})^{2} \frac{h_{4}}{3}$

$V_{4}=\pi (11)^{2} \frac{9}{3}$

$V_{4}=\pi\times 121\times 3$

$V_{4}=\pi\times 363$

$V_{4}=363\pi\ units^{3}$

So, the volume of the cone in ascending order.

$V_{2}=270\pi\ units^{3}$

7 0

3 years ago