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attashe74 [19]
1 year ago
7

The logo for a company is a regular hexagon inscribed inside a circle. The logo will be painted on the side of the company's off

ice building. The radius of the circle will be 8 ft. Find the area of the hexagon to the nearest whole foot.
Mathematics
1 answer:
stiks02 [169]1 year ago
6 0

The radius becomes the side length of six isosceles triangles. Assuming each triangle is equilateral, the height of each triangle is 4*(sqrt3).

The total area is 6 times .5*4(4sqrt3)=41.56121938 or 42 square feet
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Anton [14]

This is very spread out just to show properly but you could get it smaller haha

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2 years ago
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The table represents the equation y = 2 – 4x. A 2-column table with 5 rows. The first column is labeled x with entries negative
kherson [118]

The missing value for x=-1 is y=6

Explanation:

It is given that the equation for the table is y=2-4x

The table has 2 column with 5 rows.

Thus, we have,

x       y

-2      10

-1       ---

0        2

1        -2

2       -6

We need to determine the value of y when x=-1

The value of y can be determined by substituting x=-1 in the equation y=2-4x

Thus, we have,

y=2-4(-1)

Multiplying the term within the bracket, we have,

y=2+4

Adding the terms, we have,

y=6

Thus, the value of y when x=-1 is 6.

Hence, the missing value for x=-1 is y=6

7 0
1 year ago
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Match each three-dimensional figure to its volume based on the given dimensions. (Assume π = 3.14.)
LekaFEV [45]

Answer:

The volume of the cylinder is 150.72 cm³ ⇒ last answer

The volume of the cone is 314 cm³ ⇒ 1st answer

The volume of the pyramid is 160 cm³ ⇒ 2nd answer

The volume of the pyramid is 48 cm³ ⇒ 3rd answer

Step-by-step explanation:

* Lets revise the volumes of some shapes

- The volume of the cylinder of radius r and height h is:

 V = π r² h

- The volume of the cone of radius r and height h is:

 V = 1/3 π r² h

- The volume of the pyramid is:

 V = 1/3 × its base area × its height

* Lets solve the problem

# A cylinder with radius 4 cm and height 3 cm

∵ V = π r² h

∵ π = 3.14

∵ r = 4 cm , h = 3 cm

∴ v = 3.14 (4)² (3) = 150.72 cm³

* The volume of the cylinder is 150.72 cm³

# A cone with radius 5 cm and height 12 cm

∵ V = 1/3 π r² h

∵ π = 3.14

∵ r = 5 cm , h = 12 cm

∴ V = 1/3 (3.14) (5)² (12) = 314 cm³

* The volume of the cone is 314 cm³

# A pyramid with base area 16 cm² and height 30 cm

∵  V = 1/3 × its base area × its height

∵ The area of the base is 16 cm²

∵ The height = 30 cm

∴ V = 1/3 (16) (30) = 160 cm³

* The volume of the pyramid is 160 cm³

# A pyramid with square base of length 3 cm and height 16 cm

∵  V = 1/3 × its base area × its height

∵ The area of the square = s²

∵ The area of the base = 3² = 9 cm²

∵ The height = 16 cm

∴ V = 1/3 (9) (16) = 48 cm³

* The volume of the pyramid is 48 cm³

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On the first swing, a pendulum swings through an arc of length 65 cm. On the successive swing, the length of the arc is 85% of t
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you mutliply 65 to 85%/0.85 and you get 55.25cm as your answer



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The vertex of this parabola is at (2,-4). When the y value is -3, the x-value is
marusya05 [52]

\bf ~~~~~~\textit{parabola vertex form} \\\\ \begin{array}{llll} \stackrel{\textit{we'll use this one}}{y=a(x- h)^2+ k}\\\\ x=a(y- k)^2+ h \end{array} \qquad\qquad vertex~~(\stackrel{}{ h},\stackrel{}{ k}) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \textit{we know that } \begin{cases} h=2\\ k=-4 \end{cases}\implies y=a(x-2)^2-4 \\\\\\ \textit{we also know that } \begin{cases} y = -3\\ x = -3 \end{cases}\implies -3=a(-3-2)^2-4\implies 1=a(-5)^2 \\\\\\ 1=25a\implies \boxed{\cfrac{1}{25}=a}

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