Find an equation of the tangent to the curve at the given point by both eliminating the parameter and without eliminating the parameter. x = 6 + ln(t), y = t2 + 6, (6, 7)

2x-y=5

Step-by-step explanation:

Given

without eliminating parameter

=

=2t

at

6=6+

t=1

Equation of line is given by

2=

2x-12=y-7

2x-y=5

by eliminating parameter

differentiating we get

2x-y=5

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.