C because when you add 2 in one column you keep adding three to the other
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.
Answer:
The answer would be 'B. perimeter'.
(In the picture, QRST looks like a rhombus, so I will assume that it is a rhombus.)
In a rhombus, opposite angles are congruent based on the alternate interior angle theorem. In other words, R and T along with Q and S are both alternate interior angle pairs.
(You can actually test this by printing out a rhombus, cutting out the angles, and matching up the opposite ones.)
We know that Q is equal to 4x + 10, but we need to solve for x. Since Q and S are congruent, we can set them equal to each other:
1) 4x + 10 = 5x - 3
2) 4x + 13 = 5x
3) 5x = 4x + 13
4) x = 13
Now let's plug 13 in for Q:
5) 4(13) + 10 = 52+10 = 62
I got m<Q = 62 degrees. Hope this helps!
