20 + 5 + 2 + 3 + 1 minus 1 =30
find the perimeter of a triangle with sides 15 inches, 15 inches, and 21 inches length
To find the perimeter of a triangle we add all the sides of the triangle
The length of the sides of the triangle are given as 15 inches, 15 inches, and 21 inches
Perimeter of a triangle =
15 inches + 15 inches + 21 inches = 51 inches
So 51 inches is the perimeter
Answer:
Let 'a' be the first term, 'r' be the common ratio and 'n' be the number of terms
Series = 2+6+18.......= 2+2•3¹+ 2•3².......= 728
Now,

So,

Therefore, number of terms is 6
Answer: option d. x = 3π/2Solution:function y = sec(x)
1) y = 1 / cos(x)
2) When cos(x) = 0, 1 / cos(x) is not defined
3) cos(x) = 0 when x = π/2, 3π/2, 5π/2, 7π/2, ...
4) limit of sec(x) = lim of 1 / cos(x).
When x approaches π/2, 3π/2, 5π/2, 7π/2, ... the limit →+/- ∞.
So, x = π/2, x = 3π/2, x = 5π/2, ... are vertical asymptotes of sec(x).
Answer: 3π/2
The figures attached will help you to understand the graph and the existence of multiple asymptotes for y = sec(x).
Pretty sure the answer is 29/20 or 1 9/20. You get this by find the common denominator , 60, and then adding the number all together and simplifying