Let A, B and C be sets such that |A| = 12, |B| = 7, |C| = 10. (a) If |A ∩ B| = 0, how many ways can we choose two elements, one from
A and one from B?

(b) If |A ∩ B| = 4, what is |A ∪ B|?

(c) If |A ∩ B| = 0, |A ∩ C| = 0, and |B ∩ C| = 1, how many ways can
we choose three distinct elements, one from A and one from B and one

from C?​

cant open file sorry!

step-by-step explanation:

answer: the angel abc is 30°,and the abc is a triangle where the angel bac is 90°

so ac is the half of 24=12ft,and from the pythagorean theorem we find ba

ba=√bc^2-ac^2=√576-144=√432=12√3ft

so bac is 12+12√3=12(1+√3)ft

i’m sorry but i don’t know sorry

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.