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emmainna [20.7K]
3 years ago
10

The apothem of a regular hexagon measures 8 cm. Which are true of the regular hexagon? Check all that apply. // 1.The perimeter

of the hexagon is 48 cm. // 2.The measure of the angle formed by the radius and the apothem is 30°. // 3.The side length of the hexagon is about 4.6 cm. // 4.In a regular hexagon, the radius and side length are equal in length. // 5.The area of the hexagon is about 221.7 square cm
Mathematics
2 answers:
Julli [10]3 years ago
8 0

the answer would be choices 2, 4, and 5

olganol [36]3 years ago
5 0
The right answer would be.

The measure of the angle formed by the radius and the apothem is 30 degrees.

In a regular hexagon, the radius and side length are equal in length.

The area of the hexagon is about 221.7 square cm.
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barxatty [35]

Answer:

<h2>100, -50, 25, -12.5, 6.25, -3.125, 1.5625, -0.78125</h2>

Step-by-step explanation:

Each term is being multiplied by <u>-0.5</u>

How did I know?

Divide the second term by the first.

-50 divided by 100 = -1/2

Find the last four terms.

-12.5 x -1/2 =<em> </em><u><em>6.25</em></u>

6.25 x -1/2 = <u><em>-3.125</em></u>

-3.125 x -1/2 = <u>1.5625</u>

1.5625 x -1/2 =<em><u> -0.78125</u></em>

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3 years ago
If we sample from a small finite population without replacement, the binomial distribution should not be used because the events
liraira [26]

Answer:

0.0008

Step-by-step explanation:

The formula for a hypergeometric probability is

\frac{(_AC_x)(_BC_{n-x})}{_NC_n}

where N is the population, A is the number of objects of type A, B is the number of objects of type B, n is the sample size and x is the number of successes.

In this problem, N is 13.  A is 5, since there are 5 objects drawn, and B is 8, since there are 8 remaining objects.

The sample size, n, is 5, and x is 5:

\frac{(_5C_5)(_8C_0)}{_{13}C_5}\\\\=\frac{1(1)}{1287}=\frac{1}{1287}\approx 0.0008

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3 years ago
An object that weighs 1 pound on Earth weighs 1.19 pounds on Neptune. Suppose a dog weighs 9 pounds on Earth. How much would it
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Answer: 10.71


Step-by-step explanation:

1 pound on earth = 1.19 pounds on Neptune

9 pounds on earth means 9 x 1.19 = 10.71

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IgorC [24]

The speed of the object is 12121 m per week.

What does it mean to do speed?

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  • Some people also use various forms of speed to reduce their appetite. Types of speed include: Amphetamines (used to treat ADHD, narcolepsy, and depression)

The object is moving at a speed of 2 centimeters every 7 seconds.

We need to find the speed in m per week.

2 cm = 0.02 m

1 week = 604800 s

7 s =  1.65 * 10^{-6}   week

Speed = distance/time

So,

v = \frac{0.02 m}{1.65 * 10^{-6 } week}

v = 12121.21  m/ week

or v = 12121 m/ week

So, the speed of the object is 12121 m per week.

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8 0
1 year ago
\int\limits^0_\pi {x*sin^{m} (x)} \, dx
Ket [755]

Let

I(m) = \displaystyle \int_0^\pi x\sin^m(x)\,\mathrm dx

Integrate by parts, taking

<em>u</em> = <em>x</em>   ==>   d<em>u</em> = d<em>x</em>

d<em>v</em> = sin<em>ᵐ </em>(<em>x</em>) d<em>x</em>   ==>   <em>v</em> = ∫ sin<em>ᵐ </em>(<em>x</em>) d<em>x</em>

so that

I(m) = \displaystyle uv\bigg|_{x=0}^{x=\pi} - \int_0^\pi v\,\mathrm du = -\int_0^\pi \sin^m(x)\,\mathrm dx

There is a well-known power reduction formula for this integral. If you want to derive it for yourself, consider the cases where <em>m</em> is even or where <em>m</em> is odd.

If <em>m</em> is even, then <em>m</em> = 2<em>k</em> for some integer <em>k</em>, and we have

\sin^m(x) = \sin^{2k}(x) = \left(\sin^2(x)\right)^k = \left(\dfrac{1-\cos(2x)}2\right)^k

Expand the binomial, then use the half-angle identity

\cos^2(x)=\dfrac{1+\cos(2x)}2

as needed. The resulting integral can get messy for large <em>m</em> (or <em>k</em>).

If <em>m</em> is odd, then <em>m</em> = 2<em>k</em> + 1 for some integer <em>k</em>, and so

\sin^m(x) = \sin(x)\sin^{2k}(x) = \sin(x)\left(\sin^2(x)\right)^k = \sin(x)\left(1-\cos^2(x)\right)^k

and then substitute <em>u</em> = cos(<em>x</em>) and d<em>u</em> = -sin(<em>x</em>) d<em>x</em>, so that

I(2k+1) = \displaystyle -\int_0^\pi\sin(x)\left(1-\cos^2(x)\right)^k = \int_1^{-1}(1-u^2)^k\,\mathrm du = -\int_{-1}^1(1-u^2)^k\,\mathrm du

Expand the binomial, and so on.

8 0
2 years ago
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