1.
a.11/12=less than one
b. 77/72= greater than one
c.23/60= less than one
d. 31/56= less than one
2. <
Answer:
16
Step-by-step explanation:
The triangle AEC is a right triangle and 2 sides are already filled in.
The hypotenuse is 10 and one side is 6.
So, this is a 3-4-5 right triangle.
Thus, the remaining side EC is 8.
Since EC is congruent to ED, then segment CD is 8+8=16
These are the steps, with their explanations and conclusions:
1) Draw two triangles: ΔRSP and ΔQSP.
2) Since PS is perpendicular to the segment RQ, ∠ RSP and ∠ QSP are equal to 90° (congruent).
3) Since S is the midpoint of the segment RQ, the two segments RS and SQ are congruent.
4) The segment SP is common to both ΔRSP and Δ QSP.
5) You have shown that the two triangles have two pair of equal sides and their angles included also equal, which is the postulate SAS: triangles are congruent if any pair of corresponding sides and their included angles are equal in both triangles.
Then, now you conclude that, since the two triangles are congruent, every pair of corresponding sides are congruent, and so the segments RP and PQ are congruent, which means that the distance from P to R is the same distance from P to Q, i.e. P is equidistant from points R and Q
Answer:
A) 
B) 
C) for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Step-by-step explanation:
A) A recurrence relation for the number of bit strings of length n that contain a pair of consecutive Os can be represented below
if a string (n ) ends with 00 for n-2 positions there are a pair of consecutive Os therefore there will be : 
strings
therefore for n ≥ 2
The recurrence relation for the number of bit strings of length 'n' that contains consecutive Os
b ) The initial conditions
The initial conditions are :
C) The number of bit strings of length seven containing two consecutive 0s
here we apply the re occurrence relation and the initial conditions
for n = 2
= 1
for n = 3
= 3
for n = 4
= 8
for n = 5
= 19
Answer:
y=2/3-4x/9
Step-by-step explanation:
