Answer:
5
Step-by-step explanation:
Answer:
12870ways
Step-by-step explanation:
Combination has to do with selection
Total members in a tennis club = 15
number of men = 8
number of women = 7
Number of team consisting of women will be expressed as 15C7
15C7 = 15!/(15-7)!7!
15C7 = 15!/8!7!
15C7 = 15*14*13*12*11*10*9*8!/8!7!
15C7 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2
15C7 = 15*14*13*12*11/56
15C7 = 6,435 ways
Number of team consisting of men will be expressed as 15C8
15C8 = 15!/8!7!
15C8 = 15*14*13*12*11*10*9*8!/8!7!
15C8 = 15*14*13*12*11*10*9/7 * 6 * 5 * 4 * 3 * 2
15C8 = 6,435 ways
Adding both
Total ways = 6,435 ways + 6,435 ways
Total ways = 12870ways
Hence the required number of ways is 12870ways
When given a system of equations, the "solutions" are defined where two equations intersect, or meet.
A. The point where the lines p(x) and g(x) meet is (3, -1), and thus this is considered the solution set.
B. Because there are three lines in total, g(x) is able to intersect both lines one time, and so it has two pairs of solutions.
The first is (3, -1), which has already been established with p(x).
The second is (0, 5), and this is where it intersects with f(x).
C. The solution to f(x) = g(x) is 0, as this is the only x value where both equations are equal.
Hope my answer helped!
The third term would be 5h2 because the descending order would be
-2h4 - 8h3 + 5h2 - 2h - 3
Answer:
56.39 nm
Step-by-step explanation:
In order to have constructive interference total optical path difference should be an integral number of wavelengths (crest and crest should be interfered). Therefore the constructive interference condition for soap film can be written as,
where λ is the wavelength of light and n is the refractive index of soap film, t is the thickness of the film, and m=0,1,2 ...
Please note that here we include an additional 1/2λ phase shift due to reflection from air-soap interface, because refractive index of latter is higher.
In order to have its longest constructive reflection at the red end (700 nm)
Here we take m=0.
Similarly for the constructive reflection at the blue end (400 nm)
Hence the thickness difference should be