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solmaris [256]
3 years ago
14

The diameter of a circular garden is 22 feet. What is the approximate area of the shaded region

Mathematics
2 answers:
SOVA2 [1]3 years ago
7 0

Answer:

Step-by-step explanation:

The radius it .5 of the diameter so in this case the radius is 11.  The formula for area is pi*r^2= pi*11^2=pi*121=380.13

andrew-mc [135]3 years ago
4 0
<h2>Answer:</h2>

<u>Area = </u><u>3454 feet²</u>

<h2>Step-by-step explanation:</h2>

Area of a circle is given by

A = π r ²

Area = 3.14 * 11²

Area = 3454 feet²

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Tell which property is illustrated by the statement. 3+z=z+3
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Answer:

communitive property of addition

Step-by-step explanation:

The community property of addition states: a+b=b+a

This basically means that when you add two numbers, the order of when you add them does not occur.

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Answer:

No

Step-by-step explanation:

By multiplying 1/3 4 times, you would get 1/81 because you are multiplying the numerators by the numerators and the denominators by the denominators. However Priya is adding the fractions instead which gave her 4/3

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4 years ago
2. What is the volume of a box<br>of cereal with the dimensions<br>Shown?<br>2 in<br>I Join<br>rin​
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Answer:

140 in

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3 years ago
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Which expression is a fourth root of -1+isqrt3?
aleksklad [387]

Answer:

Step-by-step explanation:

\sf n^{th} roots of a complex number is given by DeMoivre's formula.

   \sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}

Here, k lies between 0 and (n -1) ; n is the exponent.

\sf -1 + i\sqrt{3}

a = -1 and b = √3

\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}

\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}

                   \sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})

                   \sf = \dfrac{-\pi }{3}

n = 4

For k = 0,

          \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin  \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi  }{12}+iSin  \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]

For k =1,

         \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{5\pi}{12}+i \ Sin \ \dfrac{5\pi}{12}\right]

For k =2,

       z = \sqrt[4]{10}\left[Cos \ \dfrac{11\pi}{12}+i \ Sin \ \dfrac{11\pi}{12}\right]

For k = 3,

      \sf z = \sqrt[4]{10}\left[Cos \ \dfrac{17\pi}{12}+i \ Sin \ \dfrac{17\pi}{12}\right]

For k = 4,

      \sf z =\sqrt[4]{10}\left[Cos \ \dfrac{23\pi}{12}+i \ Sin \ \dfrac{23\pi}{12}\right]

4 0
2 years ago
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