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Lady_Fox [76]
3 years ago
15

I need to do this now

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
8 0

Answer:

4.55294117. If you want it rounded look below.

Step-by-step explanation:

-Rounded-

Hundredths:4.55

Tenth: 4.6

Thousandth: 4.553

Whole Number: 5

Hope this helps. :D

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If 3 + 4 7 and 6 + 1 = 7, then 3 + 4 = 6 + 1a. Distributive property . Associative property Transitive property . Commutative pr
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If 3+4 = 7 and 6+7 = 7 leads to 3+4 = 6+1, then you are using the transitive property. In general the property says that if a = b and b = c, then a = c.
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A 9cm wife cross section pipe has a height of 13cm . What is the volume of the pipe
rjkz [21]
I think you meant "wide." :) Since it is a pipe, you should know that it is the shape of a cylinder. Therefore, the formula would be: πr^2h. Substitute the height for h. (π(4.5)^2(13). Since it's 9 cm wide total, divide it by 2 to get the radius (part of formula), which is 4.5 So multiply π times 4.5^2 (which equals 20.25) times 13. This all will equal 826.605. You can simplify this decimal to 827. Therefore, the volume is 827 cm^3.
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If you flip 11 coins, the probability of not getting all 11 tails is
Marta_Voda [28]
The probability of *NOT* getting all 11 tails is 91/100
4 0
2 years ago
You took a 5-question multiple choice quiz with 4 choices for each question. If you guess at random on each question, what is th
Arturiano [62]

Answer:

80%

Step-by-step explanation:

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5 0
3 years ago
The polar curve $r = 1 + \cos \theta$ is rotated once around the point with polar coordinates $(2,0).$ What is the area of the r
mash [69]

Answer:

Area = -2.3147

Step-by-step explanation:

Given

$r = 1 + \cos \theta$

Required

Determine the area with coordinates (2,0)

The area is represented as:

Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta

Where

$r = 1 + \cos \theta$

and

(a,b) = (2,0)

Substitute values for r, a and b in

Area = \frac{1}{2}\int\limits^b_a {r^2} \, d\theta

Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)^2} \, d\theta

Expand

Area = \frac{1}{2}\int\limits^0_2 {(1 + cos\theta)(1 + cos\theta)} \, d\theta

Area = \frac{1}{2}\int\limits^0_2 {(1 + 2cos\theta+cos^2\theta} )\, d\theta

By integratin the above, we get:

Area = \frac{1}{2}*\frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{2}[0,2]

Area = \frac{(cos(\theta) + 4)sin(\theta) + 3\theta}{4}[0,2]

Substitute 0 and 2 for \theta one after the other

Area = \frac{(cos(0) + 4)sin(0) + 3*0}{4} - \frac{(cos(2) + 4)sin(2) + 3*2}{4}

Area = \frac{(cos(0) + 4)sin(0)}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area = \frac{(1 + 4)*0}{4} - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area =  - \frac{(cos(2) + 4)sin(2) + 6}{4}

Area =  \frac{-sin(2)(cos(2) + 4) - 6}{4}

Get sin(2) and cos(2) in radians

Area = \frac{-0.9093 * (-0.4161 + 4) - 6}{4}

Area = \frac{-9.2588}{4}

Area = -2.3147

3 0
2 years ago
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