The first 5 terms in a pattern are shown below.
25 13 27
-6,-4' 2' 4-7
If the pattern continues, which
equatior. could be used to find the nth
term?

Option B. m(n) = –5.75 –¼n

Step-by-step explanation:

The sequence is given below:

–6, –25/4, –13/2, –27/4, –7

Next, we shall determine the common difference.

Common difference (d) = 2nd – First

2nd term = –25/4

First term = –6

d = –25/4 – – 6

d = –25/4 + 6

d = –6.25 + 6

d = –0.25 = –¼

Finally, we shall determine the nth term

Tₙ = a + (n –1)d

First term (a) = –6

Common difference (d) = –¼

Tₙ = a + (n –1)d

Tₙ = –6 + (n –1)–¼

Tₙ = –6 –¼n + ¼

Tₙ = –6 + ¼ –¼n

Tₙ = –6 + 0.25 –¼n

Tₙ = –5.75 –¼n

Thus, the nth term is given by:

m(n) = –5.75 –¼n

step-by-step explanation:

through just a quick inspection, one can see that the graph of this cubing function starts in quadrant iii and ends in quadrant i.   when x = 0, y = 3, so the y-intercept is (0, 3).

if graphs are part of the problem statement, share them.   .

a.

step-by-step explanation:

step 1 distribute 2 to every term in the parentheses

step 2 add 10 on both sides of the equal sign