Loi sur les exemples de Questions partie 2
fr
All Modules View all topics Close topics
Mathématiques de la Loi sur l'examen partie 2

### Ce cours a été révisé!

Pour une expérience d'apprentissage plus agréable, nous vous recommandons d'étudier la version mobile relogée de ce cours.

Emmenez-moi au cours révisé.

- or -

Continue studying this course

# Loi sur les exemples de Questions partie 2

• Study Reminders

We'll email you at these times to remind you to study

You can set up to 7 reminders per week

#### You're all set

We'll email you at these times to remind you to study

Monday

Tuesday

Wednesday

Thursday

Friday

Saturday

Sunday

• Notes d'étude
• Révisions du sujet
 Petra S. 0 0 I never knew how to do ratio between a circle and a square. Helpful to now know the formula. One side of square is x and r is the radius of the circle. Formula to get r squared is x/2 squared plus x/2 squared. Figure out r. Square area equals length times width and circle equals Pi times r squared. Divide area of square by area of circle and the answer is 2/Pi.
 Jorge Carlos V. 0 0 There is a simpler solution: The square is two triangles with base 2r and height r, that is 2*squared(r). The area of circle is pi*squared(r).The requested ratio is the square area divided by the circle area, that is: 2/pi
 Khalid B. 1 0 5/2π
 Khalid B. 0 0 3
 Khalid B. 0 0 1
 Khalid B. 0 0 0.25
 Khalid B. 0 0 0.50
 Linda E. 0 0 It was very helpful
 Candance S. 0 0 I guess I was the only one who was lost by that whole equation
 Hafid A. 0 0 G
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}