Answer:
6y?
It's probably wrong but off the top of my head, 6y
Answer:
Step-by-step explanation:
A binary string with 2n+1 number of zeros, then you can get a binary string with 2n(+1)+1 = 2n+3 number of zeros either by adding 2 zeros or 2 1's at any of the available 2n+2 positions. Way of making each of these two choices are (2n+2)22. So, basically if b2n+12n+1 is the number of binary string with 2n+1 zeros then your
b2n+32n+3 = 2 (2n+2)22 b2n+12n+1
your second case is basically the fact that if you have string of length n ending with zero than you can the string of length n+1 ending with zero by:
1. Either placing a 1 in available n places (because you can't place it at the end)
2. or by placing a zero in available n+1 places.
0 ϵ P
x ϵ P → 1x ϵ P , x1 ϵ P
x' ϵ P,x'' ϵ P → xx'x''ϵ P
86 and 87 utilize the Pythagorean theorem. Where those tangent lines meet the circle forms a right angle. So you have a side length of 6 and a hypotenuse of 22. So the missing side length is 21.16. For 87, its the same thing, but you have the 2 legs but are missing the hypotenuse.
Answer:
D
Step-by-step explanation:
Angle 5 is equal to angle 8 because the two angles are an alternate interior angle. We already know that angle 7 and 8 are supplementary because it's adjacent to each other on a straight line. So, since angle 5 and 8 is equal to each other, angle 5 is also supplementary to angle 7.
2(4z-9-11)=166-46
8z-18-22=120
8z-40=120
Add 40 from both side
8z-40+40=120+40
8z=160
Divided 8 from both side
8z/8=160/8
z=20. Hope it help!
