Colin has a pad with x pieces of paper on it. For his first class, he wrote on 5 fewer than half of the pieces of paper in the pad. He used 2 more sheets in his second class than in his first. How many sheets are left for his third class? ill give brainliest to the first answer

Colin has 8 sheets left for his third class.

Step-by-step explanation:

Given that:

Total Number of pieces of papers =

Number of pieces of papers used for 1st class = 5 fewer than half of the pieces in the pad

Writing the equation:

Also, Given that number of pieces of papers used for the 2nd class are 2 more than that of papers used in the 1st class.

Now, number of pieces of papers left for the third class = Total number of pieces of papers in the pad - Number of pieces of papers used in the first class - Number of pieces of papers used in the first class

So, the answer is:

Colin has 8 sheets left for his third class.

for the first question it would be the second answer for the second question it would be the third answer and for the third question it would be the first answer

step-by-step explanation:

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}