Use the spider tool located on page 1 of this activity to draw a 12-pointed star for the new logo. (Hint:If the spider rotates 3
60 degrees -- or 720 degrees or 1080 degrees -- she will be facing in the same direction in which she started. When the spider is done drawing, you want her to be facing in the same direction in which she started. She'll be making 12 rotations, all the same size, so each rotation must be some multiple of 360/12 = 30 degrees.)
Please help urgently. Been stuck on this problem for around 45 minutes now. Thanks.
PLEASE HELPPPPP! IT SHOULD BE EASY IF YOU'RE SMART ENOUGH
Please help urgently. Been stuck on this problem for around 45 minutes now. Thanks.
PLEASE HELPPPPP! IT SHOULD BE EASY IF YOU'RE SMART ENOUGH
1 answer:
Answer:
each of the 12 turns is 150°
Step-by-step explanation:
If we number the points of the star 1–12, in order for the star to be symmetrical, each point must connect to two points symmetrically located around the centerline.
That is, point 1 may connect to points {2, 12} or {3, 11} or {4, 10}, or {5, 9} or {6, 8} or {7, 7}. For each of these connections, the angle made at the point of the "star" is, respectively, 150°, 120°, 90°, 60°, 30° or 0°. For these angles, the figure obtained will be ...
150° - dodecagon, a 12-sided figure
120° - hexagon
90° - square
60° - equilateral triangle
30° - 12-pointed star
0° - straight line
The two 12-pointed figures are shown in the attachment.
__
We suspect the star you're interested in is the one with points that are 30°. In order to have that point angle, the spider must make a turn of 180° -30° = 150°.
The spider will make 12 turns of 150°, for a total of 1800°, for a total of 5 full turns of 360°.
You might be interested in
Check the picture below.
how do we know? well, notice h(t), starts off at 12, up up up reaches 47.84 then down down down, which is pretty much the trajectory of a flying object, by the time it gets to 44, is still going down.
now, let's look at g(t), starts off at 10, and goes up up up, never down, by the time it gets to 41, is still going up,
so at second 2, h(t) is 44 and going down, g(t) is 41 and going up, at 2.2 h(t) is 40.16, and g(t) is 44.1, between that lapse, h(t) became 44, 43, 42, 41, in the same lapse g(t) became 41, 42, 43, 44, so somewhere in those values h(t) = g(t).
what does the solution mean? It's the seconds or the instant lapse when the first cannon ball was at the same height as the second cannonball.
