Rewrite the root expressions as fractional exponents:
![\dfrac{\sqrt[3]{7}}{\sqrt[5]{7}} = \dfrac{7^{1/3}}{7^{1/5}}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B7%7D%7D%7B%5Csqrt%5B5%5D%7B7%7D%7D%20%3D%20%5Cdfrac%7B7%5E%7B1%2F3%7D%7D%7B7%5E%7B1%2F5%7D%7D)
Recall that 
, so that
Simplify the exponent:
Then you end up with
![\dfrac{\sqrt[3]{7}}{\sqrt[5]{7}} = 7^{2/15}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Csqrt%5B3%5D%7B7%7D%7D%7B%5Csqrt%5B5%5D%7B7%7D%7D%20%3D%207%5E%7B2%2F15%7D)
9514 1404 393
Answer:
9 square units
Step-by-step explanation:
Pick's theorem is a useful relationship in circumstances like these. It tells you the area is ...
A = i + b/2 - 1
where i is the number of interior points (8), and b is the number of points on the border (4).
The area of this figure is ...
A = 8 + 4/2 -1 = 9
The area of the polygon is 9 square units.
_____
You can also get there by realizing the bounding rectangle is 4 units square. From that, corner triangles are cut. CW from upper left, those triangles have (base, height) dimensions of (3, 2), (1, 3), (3, 1), and (1, 2). So, the total of their areas is (1/2)(6 +3 + 3 +2) = 7 square units. The shaded area is then 16-7 = 9 square units, same as above.
Answer:
Step-by-step explanation:
I'm going to use a Pythagorean identity to help make this make sense.
We are given that cos(0) = 1, so
cos²θ = 1² which simplifies to
cos²θ = 1. So filling into our trig identity:
sin²θ + 1 = 1 and
sin²θ = 0 then
sinθ = √0 so
sinθ = 0
so
sin(0) = 0
We can solve this problem in different ways:
0. Statement: As angle 1 and 3 are vertical angles, and angle 1, 5, 4 form a straigth angle, so angle 2 and (algle 4 + angle 5) are vertcal angles too.
Reason:
