What is the square root of pie

2 answers:

7 0

Like Pi, The square root of Pi is irrational. The square root of pi can be displayed as 1.7725. It is irrational because it is an never ending flow of numbers.

givi [52]3 years ago

3 0

3.14 is the square root of pie.

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A playground is in the shape of a square with each side equal to 109 yards. It has skating rinks in the shape of the quadrants o

I am Lyosha [343]

Answer:

radius = 30 yd , cost of cement = $7065

Step-by-step explanation:

First we will find the area of the circular sectors. From that we can find the cost of cement and their radius.

The area of the playground is ...

playground area = (109 yd)^2 = 11,881 yd^2

Then the area of the skating rinks is:

rink area = playground area - remaining field

rink area = 11,881 yd^2 - 9,055 yd^2 = 2,826 yd^2

Then the cost of cement for the rink area is ...

(2826 yd^2)($2.50 yd^2) = $7065

The four quarter-circle skating rinks add to a total area of a full circle. That area is given by ...

A = πr^2

Put the values in the formula:

2826 = 3.14r^2

Divide both sides by 3.14

900 = r^2

By taking square root at both sides we get,

30 = r

The radius of each quadrant is 30 yards....

6 0

3 years ago

Write a decimal that is 1\10 of 0.9

exis [7]

I am pretty sure it would be 0.09

6 0

3 years ago

Read 2 more answers

Find sin 2a and cot 2a:

geniusboy [140]

Answer:

$sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cot(2\alpha ) = \frac{16511}{18720}$

Step-by-step explanation:

$\text{if } cos(\alpha)=\frac{12}{13}\\\text{That must mean we have a triangle with base 12, and hypotenuse 13.}\\\text{Using Pythagoras, we can determine the base of the triangle must be 5.}\\a^2+b^2=c^2 \text{, where c is the hypotenuse and a, b are the two other sides.}\\c^2-b^2=a^2\\\sqrt{c^2-b^2}=a\\\sqrt{13^2-12^2}=\sqrt{169-144}=\sqrt{25}=5\\\text{Therefore, }sin(\alpha) = \frac{5}{12}\\sin(2\alpha)=2sin(\alpha )cos(\alpha)\\\text{(From double angle formulae identities)}\\$

$sin(2\alpha )=2(\frac{5}{12})(\frac{12}{13})=\frac{10}{13}\\cos(2\alpha )=cos^2(\alpha)-sin^2(\alpha)\\cos(2\alpha )=(\frac{12}{13})^2-(\frac{5}{12})^2=\frac{16511}{24336}\\cot(2\alpha)=\frac{cos(2\alpha)}{sin(2\alpha)}=\frac{\frac{16511}{24336}}{\frac{10}{13}}=\frac{16511}{18720}$