Which mathematicians first used the symbol pi Why?

Mathematics

1 answer:

gavmur [86]3 years ago

8 0

<h2>Answer:</h2>

William Jones

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

The number pi (π) is a platonic concept and has been used for 4000 years. The number pi can be approached but never reached. In general, this number means the constant ratio of the circumference to the diameter of any circle. Ancient Babylonians, Egyptians, and even Archimedes, one of the greatest mathematicians of the ancient world tried to approximate the value of pi, but in the 1700s, mathematicians began using the Greek letter π that was introduced by William Jones, in his second book <em>Synopsis Palmariorum Matheseos</em> or <em>A New Introduction to the Mathematics base</em>. Then, this symbol was popularized and adopted in 1737 by the greatest mathematician Leonhard Euler.

You might be interested in

Find the area of ABC

shutvik [7]

Answer:

24 ft sq.

Step-by-step explanation:

formula for area of triangle is: bh/2

8(6)/2 = area

48/2 = area

24 = area

6 0

3 years ago

DUE TODAY NO LINKS OR FILES

kupik [55]

64 dollars is the answer

4 0

2 years ago

Read 2 more answers

Let $f(x) = x^2$ and $g(x) = \sqrt{x}$. Find the area bounded by $f(x)$ and $g(x).$

Anna [14]

Answer:

$\large\boxed{1\dfrac{1}{3}\ u^2}$

Step-by-step explanation:

Let's sketch graphs of functions f(x) and g(x) on one coordinate system (attachment).

Let's calculate the common points:

$x^2=\sqrt{x}\qquad\text{square of both sides}\\\\(x^2)^2=\left(\sqrt{x}\right)^2\\\\x^4=x\qquad\text{subtract}\ x\ \text{from both sides}\\\\x^4-x=0\qquad\text{distribute}\\\\x(x^3-1)=0\iff x=0\ \vee\ x^3-1=0\\\\x^3-1=0\qquad\text{add 1 to both sides}\\\\x^3=1\to x=\sqrt[3]1\to x=1$

The area to be calculated is the area in the interval [0, 1] bounded by the graph g(x) and the axis x minus the area bounded by the graph f(x) and the axis x.

We have integrals:

$\int\limits_{0}^1(\sqrt{x})dx-\int\limits_{0}^1(x^2)dx=(*)\\\\\int(\sqrt{x})dx=\int\left(x^\frac{1}{2}\right)dx=\dfrac{2}{3}x^\frac{3}{2}=\dfrac{2x\sqrt{x}}{3}\\\\\int(x^2)dx=\dfrac{1}{3}x^3\\\\(*)=\left(\dfrac{2x\sqrt{x}}{2}\right]^1_0-\left(\dfrac{1}{3}x^3\right]^1_0=\dfrac{2(1)\sqrt{1}}{2}-\dfrac{2(0)\sqrt{0}}{2}-\left(\dfrac{1}{3}(1)^3-\dfrac{1}{3}(0)^3\right)\\\\=\dfrac{2(1)(1)}{2}-\dfrac{2(0)(0)}{2}-\dfrac{1}{3}(1)}+\dfrac{1}{3}(0)=2-0-\dfrac{1}{3}+0=1\dfrac{1}{3}$