How would I at least start out those two problems?
It looks to me like you already did. Now just follow through with it.
If the length of each side of the frame's inner dimension (i.e. the glass) is x, and the length of each side of the frame's outer dimension is x^2 - 3, then the area of the frame is:
A = (outer dimension)^2 - (inner dimension)^2
27 = (x^2 - 3)^2 - (x)^2
27 = x^4 - 6x^2 + 9 - x^2
27 = x^4 - 7x^2 + 9 <----- This is where you got to already.
0 = x^4 - 7x^2 - 18
0 = (x^2 - 9)(x^2 + 2)
Therefore, either x^2 = 9 or x^2 = -2... and you know it can't be the latter.
Thus, x^2 = 9 and therefore x = 3. The glass is 3" on each side.
<span>Each (outer) side of the frame is x^2 - 3 = 9 - 3 = 6" </span>
Answer: a:b:c:d = 16:24:30:40
You can also divide the ratio by 2 because all terms have 2 in common. Which would be a:b:c:d = 8:12:15:20
The cosine of an angle is the x-coordinate of the point where its terminal ray intersects the unit circle. So, we can draw a line at x=-1/2 and see where it intersects the unit circle. That will tell us possible values of θ/2.
We find that vertical line intersects the unit circle at points where the rays make an angle of ±120° with the positive x-axis. If you consider only positive angles, these angles are 120° = 2π/3 radians, or 240° = 4π/3 radians. Since these are values of θ/2, the corresponding values of θ are double these values.
a) The cosine values repeat every 2π, so the general form of the smallest angle will be
... θ = 2(2π/3 + 2kπ) = 4π/3 + 4kπ
b) Similarly, the values repeat for the larger angle every 2π, so the general form of that is
... θ = 2(4π/3 + 2kπ) = 8π/3 + 4kπ
c) Using these expressions with k=0, 1, 2, we get
... θ = {4π/3, 8π/3, 16π/3, 20π/3, 28π/3, 32π/3}
The answer is no, because the Rank Theorem tells us that rank A + dim Nul A = 12 . Since the rank can be at most 10, the null space has to have dimension at least two.