The second one is correct
The answer is 7/9.Hope it helps
Answer:
First term (a) =8
Common difference (d)= t2-t1
=12-8
=4
Now, sum of first 31th term (tn31) =n/2{2a+(n-1)d}
= 31/2{2×8+(31-1)4}
=31/2{16+(30×4)
=31/2(16+120)
=31/2×126
=31×63
Step-by-step explanation:
Similarly use 19 as (n) for the 19th term
Answer:
z= 13
x= 36
y= 18
Step-by-step explanation:
Solving for z.
We know that 108 degrees and (8z+4) are alternate exterior angles, so they are equal to each other.
We can set both of those angles equal to each other, and solve for our missing side, z.
108= 8z+4
104=8z
z= 13
Solving for x.
We know that 108 degrees and (3x) are alternate interior angles, so they would equal each other.
We can set both of these angles equal to each other, and solve for our missing side, x.
3x=108
x= 36
Solve for y.
We know that (4y) and (3x) are same side interior angles, so they would make 180 degrees. We know that (3x) would equal 108.
4y+108=180
4y= 72
y= 18
Answer is (A). For a term with an irreducible
-degree polynomial in the denominator, its numerator must have at most a polynomial of degree
.
