I think the answer is.e=mc3
There will not be any specific name given to that image after translation
<u>Step-by-step explanation:</u>
- Whatever may be the transformation, either it may be rotation, reflection or translation, the size or shape of the image will not change no matter what the condition is and hence the name of that image will also not change and remains constant.
- Thus we can conclude that the name of the image will always tend to remain constant and not changes.
Answer:
a=41.8 BC=2.8
Step-by-step explanation:
sin30/6=sinx/8
8*sin30=6sinx
4=6sinx
sin^-1(4/6)
angle a
cosine rule
bc^2=3^2 +5^2-2(3)(5)*cos30
BC^2=
BC=2.83
2.8
All of them??? or just one
You need to use a ratio of height (H) to shadow length (L) to solve the first problem. It's basically a use of similar triangles, with two perpendicular sides, and with the shadow making the same angle with the vertical.
6 ft = 72 ins, so that rH/L = 72/16 = 9/2 for the player.
So the bleachers are 9/2 x 6 ft = 27 ft.
For the second problem, 9 ft = 108 in, so that the ratio of the actual linear dimensions to the plan's linear dimensions are 9ft/(1/2in) = 2 x 108 = 216.
So the stage will have dimensions 216 times larger than 1.75" by 3".
That would be 31ft 6ins x 54ft.
Live long and prosper.
