Use the Euclidean algorithm to express 1 as a linear combination of 
and 
.
a. 
because
77 = 1*52 + 25
52 = 2*25 + 2
25 = 12*2 + 1
so we can write
1 = 25 - 12*2 = 25*25 - 12*52 = (77 - 52)(77 - 52) - 12*52 = 77^2 - 2*52*77 + 52^2 - 12*52
Taken modulo 77 leaves us with
b. First, 
, so really we're looking for the inverse of 25 mod 52. We've basically done the work in part (a) already:
1 = 25*25 - 12*52
Taken modulo 52, we're left with
c. The EA gives
71 = 1*53 + 18
53 = 2*18 + 17
18 = 1*17 + 1
so we get
1 = 18 - 17 = 3*18 - 53 = 3*71 - 4*53
so that taken module 71, we find
d. Same process as with (b). First we have 
, and we've already shown that
1 = 3*18 - 53
which means, taken modulo 53, that
GCF - What is the highest number possible that the two given terms have in common.
In the given problem, one can clearly see that the two share
.
The factors of 14 are 1, 2, 7 and 14 while the factors of 42 are 1, 2, 3, 6, 7, 14, 21 and 42.
Out of the two, 14 is the highest number that they both share.
So our answer will be 14
Answer:
Player 1's position is Player 2's position reflected across the y-axis; only the signs of the y-coordinates of Player 1 and Player 2 are different
Step-by-step explanation:
Answer:
4 and 6
Step-by-step explanation:
Answer:
Q1
cos 59° = x/16
x = 16 cos 59°
x = 8.24
Q2
BC is given 23 mi
Maybe AB is needed
AB = √34² + 23² = 41 (rounded)
Q3
BC² = AB² - AC²
BC = √(37² - 12²) = 35
Q4
Let the angle is x
cos x = 19/20
x = arccos (19/20)
x = 18.2° (rounded)
Q5
See attached
Added point D and segments AD and DC to help with calculation
BC² = BD² + DC² = (AB + AD)² + DC²
Find the length of added red segments
AD = AC cos 65° = 14 cos 65° = 5.9
DC = AC sin 65° = 14 sin 65° = 12.7
Now we can find the value of BC
BC² = (19 + 5.9)² + 12.7²
BC = √781.3
BC = 28.0 yd
All calculations are rounded
