Mrs. Fox, the art teacher at Eastside Elementary school, has an empty space on her wall 6 ft.* 3 1/3 ft. She was given some white tiles that measure 1/3 ft. by 1/3 ft. She wants her 175 fifth grade students to each design and paint a tile to be hung in the empty space on the wall. Will there be enough room in the designed space to hang all 175 tiles?
What is the area of the space Mrs. Fox wants to fill? Justify your answer with words, diagrams, or an equation for both questions.

1. The space Mrs. Fox wants to fill is bigger than the space needed to hang 175 tiles , therefore it will be enough room in the designed space.

2. The area of the space Mrs. Fox wants to fill is 20 ft².

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Empty space Mrs. Fox has on her wall = 6 ft. * 3 1/3 ft.

White tiles measures Mrs. Fox was given = 1/3 ft. * 1/3 ft.

2. She wants her 175 fifth grade students to each design and paint a tile to be hung in the empty space on the wall. Will there be enough room in the designed space to hang all 175 tiles?  What is the area of the space Mrs. Fox wants to fill? Justify your answer with words, diagrams, or an equation for both questions.

Let's recall the formula for the area of a rectangle:

Area = Length * Width

Area = 6 ft * 3 1/3 ft.

Area = 6 * 3.33 = 20 ft²

The area of the space Mrs. Fox wants to fill is 20 ft²

Now, let's find the area of each white tile and let's multiply that value 175 times, as follows:

Area of a white tile = 1/3 * 1/3

Area of a white tile = 0.33 * 0.33 = 0.1089 ft²

Area of 175 tiles = 0.1089 * 175 = 19.06 ft² (rounding to two decimal places)

20 ft² > 19.06 ft² or the space Mrs. Fox wants to fill is bigger than the space needed to hang 175 tiles , therefore it will be enough room in the designed space.

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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}