Pangolin14

Favorable / possible

Possible would be 10 contestants, then choose 3 of them. 

There are 10 ways to choose first, 9 ways to choose second, 8 ways to choose 3rd.

720 possible.

For the favorable, we have 1 way to choose 1st, 1 way to choose 2nd, 1 way to choose 3rd.

So 1/720

Based on the information and because BD || GM, we have that ∆AMG ~ ∆ABD and ∆ANG ~ ∆ACD and ∆AMN ~ ∆ABC. 

By median ratios, 3AG = 2AD, so common ratio is 4:9 for all triangles' areas. So we have that GM = GN because they are both similar with the same ratio to BD=BC. Therefore, [AGN] = [AGM]. Thus, [AGN] = 144, and you can check this by finding area of AMN/2 and ABD then ABC... etc.

Inscribed arc = 1/2 * arc measure

170 * 2 = 340

340/360 = 34/36 = 17/18 of the whole circle --> 18^2 * 17/18 * pi = 17*18 = 306 pi

We have: (x-4)(2y+1)-1 = xy

So xy + x - 8y - 5 = 0

The numbers are 2 and 7 so 14.

Divide by 9 on both sides

x^2 + 6x - 80 = 0

So x^2 + 6x = 80

(6/2)^2 = 9

s = 80 + 9 = 89

:))))))

8. Domain = All possible X values and Range = All possible y values (withouth duplicates. Can you calculate it now with the given points?

9. Same as #8

10. Remove duplicates, you'll be good to go with #8.

11. Also #8

12. Sometimes. All functions are relations, but not all relations are functions, by the vertical line test.

13. Graph the points. Next, determine if any two x coordinates are the same. If they are, it's a no for being a function. If they aren't then yes.

14. Follow #13

15. Follow #13, or just look at the x and y coordinates.

16. Are any x coordinates the same? (#13 again)

17.

a) We must not have the same x coordinate, by the vertical line test. So, because age is the x coordinate, age cannot go down, so it must be a function!

b) Probably not, unless the person is super healthy. The reasoning behind this is that as the age goes higher, the old person's might get shorter and thus the graph will have overlapping x coordinates, as you must have reached a certain point to regress back to it, as the word implies.

18. No, as there are two values for y when x = -1.

19. Yes, by the vertical line test.

20. No, as there are two values for y when x = 1 or 2.

This would mean that $9m \equiv 1 \pmod m.$ Therefore, $0 \equiv 1 \pmod m,$ so $m = 1.$ But, if we plug this in to $2m \equiv 1 \pmod m,$ we get 2 = 1 mod 1 which is obviously true. Therefore, there is no solution.

Let the center of circle with diameter AB be O. Let the circle with diameter BC have center Q.

Let the tangent to circle O be M. Let the tangent to circle Q be N.

Then $\triangle QNA$ is right. The length QA is 3, and it is the hypotenuse, and QN is a radius, so it equals 1.

If we extend QN downward until it intersects MC at point X, we have that our desired distance is XN.

By AA similarity, $\triangle ANQ \sim CXQ,$ so $\frac{NQ}{AQ} = \frac{QX}{QC} \Longrightarrow \frac{1}{3} = QX \Longrightarrow QX + DQ = \frac{4}{3} !!!!$

Thanks!