An island has some portion submerged in water. point x on the island is at negative 9 over 2 feet from the surface of the sea. point y on the island is at negative 4 over 3 feet from the surface of the sea. because negative 4 over 3 feet > negative 9 over 2 feet, what is true about the distance of the two points from the surface of the sea?

Point X is in more depth than the point Y.

Step-by-step explanation:

We are given that An island has some portion submerged in water.

Point X on the island is at distance from the surface of the sea=feet

Point Y on the island is at distance from the  surface of the sea=feet

We know if we move from left to right on the left of zero on number line then the value decreases.

We are given that

The distance of point X from the surface of the sea=

The distance of point Y from the surface of the sea=

The two distance of two points measure from the surface of the sea. It means the two points is present in water.

When we go in water then the distance measure in negative. Therefore, the point is more deepest than the point Y.

Hence, we can say point X is in more depth than the point Y in the sea.

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}

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