1113.

]]>Thanks for another interesting puzzle.

Bob

]]>Thanks for your latest puzzle.

Bob

]]>As far as I know there is no totally algebraic way to solve this.

What I would do is to sketch two graphs, y = 2^x and y = 2x to see where these cross.

You can try this at https://www.mathsisfun.com/data/function-grapher.php

In this case they look like they cross at (1,2) and at (2,4); and it's easy to check by substitution that these are solutions **. But are they the only ones?

y = 2x is an increasing function, negative when x<0.

y = 2^x is also increasing but never negative. So we can rule out any negative solutions for x.

2x increases at a steady rate (constant gradient) whereas 2^x gets steeper as x goes up. So they will never cross again after (2,4) when the 2^x curve crosses y = 2x with an ever increasing gradient. So x = 1 and x = 2 are the only solutions.

Does it matter that I spotted the answer without complicated algebraic work? Well no actually. If you have shown a solution works and found any others and can prove you've got them all, then that's ok as a way to answer the question.

Bob

** You shouldn't assume x= 1 is the answer just from the graph. The 'correct' answer might be x = 0.9999997. From a graph alone you only know the answer is roughly 1 as graphs are only as accurate as your ability to draw them (thickness of the pencil; degree of accuracy with the calculator etc)

]]>]]>

IQR = Q3 - Q1. The interquartile range [link removed by moderator] shows how the data is spread about the median. It is less susceptible than the range to outliers and can, therefore, be more helpful.

]]>functions from A to B? Explain.

9. A = {0, 1, 2, 3} and B = {−2, −1, 0, 1, 2}

(a) {(0, 1), (1, −2), (2, 0), (3, 2)}

(b) {(0, −1), (2, 2), (1, −2), (3, 0), (1, 1)}

(c) {(0, 0), (1, 0), (2, 0), (3, 0)}

(d) {(0, 2), (3, 0), (1, 1)}

10. A = {a, b, c} and B = {0, 1, 2, 3}

(a) {(a, 1), (c, 2), (c, 3), (b, 3)}

(b) {(a, 1), (b, 2), (c, 3)}

(c) {(1, a), (0, a), (2, c), (3, b)}

(d) {(c, 0), (b, 0), (a, 3)}

NOTE: THIS IS EXTRA PRACTICE FOR MEMBERS. I AM NOT ASKING FOR HELP WITH 9 AND 10.

]]>mathland wrote:Find the distance between the points.

17. (−2, 6), (3, −6)

18. (8, 5), (0, 20)

19. (1, 4), (−5, −1)

20. (1, 3), (3, −2)

21. (1/2, 4/3) (2, −1)

22. (9.5, −2.6), (−3.9, 8.2)

___________________

would the formula be √ (x1 - x2)² + (y1 -y2)² ? edit: oop the link already said so

I posted the problems as extra practice for members.

]]>in which (x, y) could be located.

9. x > 0 and y < 0

10. x < 0 and y < 0

11. x = −4 and y > 0

12. x < 0 and y = 7

13. x + y = 0, x ≠ 0, y ≠ 0

14. xy > 0

]]>In the formulas...

r = the radius of the circumscribed circle

i = the radius of the respective inscribed circle

s = the sagitta of the respective circular segment

**Method A** (Area of circular segments minus area of inscribed circles)

1. Total area of the 3 circular segments = 481.49203......respective segment areas = r²ArcCos((r - s)/r) - (r - s)√(2rs - s²)

2. Total area of the 3 inscribed circles R, M & L = 175.92919......respective circle areas = πi²

481.49203

- 175.92919

------------

305.56284 = shaded region area

=======

**Method B** (Area of circumscribed circle minus areas of inscribed objects)

1. Area of circumscribed circle = 756.47691......πr²

2. Total area of the 3 inscribed circles R, M & L = 175.92919......respective circle areas = πi²

3. Area of triangle ABC, using Heron's Formula on the 3 circular segment chord lengths AB, AC & BC = 274.98488......respective chord lengths = 2√(2sr - s²)

756.47691

- 175.92919

- 274.98488

------------

305.56284 = shaded region area

=======

*Edit: I first had Method B as my only method (same as the one in my post #5), but then discovered Method A.*

http://www.mathisfunforum.com/misc.php?action=rules]]>

Hi,

I wasn't able to solve this using substitution, but here's a solution I came up with:

]]>

#4

Hi,

]]>