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1 year ago

7

The students in math class use square tiles to make arrays. Celia says they can make more arrays with 8 tiles than with 9 tiles

because 8 has more factors. Is Celia correct?

2 answers:

Marizza181 [45]1 year ago

6 0

Yes
8=1,2,4,8,
9=1,3,9

Cerrena [4.2K]1 year ago

5 0

The students in math class use square tiles to make arrays. If they

use 8 (8=1·2·2·2) tiles, then they can form arrays of the length in 1 tile, 2 tiles, 4 tiles and 8 tiles;
use 9 (9=1·3·3) tiles, they can form arrays of the length in 1 tile, 3 tiles and 9 tiles (one array less).

So, you can conclude that Celia is correct.

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Write two expressions where the solution is 4

Angelina_Jolie [31]

There are many expressions that have solution 4. some are:
3+1 = 4
2+2 = 4
5-1 = 4
7-3 =4
8-4 = 4
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1 year ago

George is comparing two different phones. The screen of phone A has a width of 5.3 cm. The width of the screen on phone B is 5 a

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Answer:

The answer is "Screen B is wider , because 5.3 is greater than 5 and one-third".

Step-by-step explanation:

The mobile display unit must be made similar. Phone A uses a decimal device, and telephone B uses a fraction in this case.

It's easier for you to use. Attempt using the decimal. The width of the phone B fraction could then be translated to decimal width. The computing is:

= 5 cm + $\frac{1}{3}$ cm

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= 5.333 cm

It is evident from here that the wider screen of Phone B is than that of Phone A.

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1 year ago

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Here are seven tiles: 1,1,3,3,3,5,5. Tom takes a tile at random. He does not replace the tile. Tom then takes a second tile. A)

zzz [600]

Answer:

a) 2/42

b)16/42

Step-by-step explanation:

a) 2/7 x 1/6 = 2/42

b) (1,2) (1,3) (2,3)

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P(1,3) = 2/7 x 2/6 = 4/42

P(2,3) = 3/7 x 2/6 = 6/42

Add all = 6/42 + 4/42 + 6/42 = 16/42

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1 year ago

A piece of paper is to display ~128~ 128 space, 128, space square inches of text. If there are to be one-inch margins on both si

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Answer:

The dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches

Step-by-step explanation:

We have that:

$Area = 128$

Let the dimension of the paper be x and y;

Such that:

$Length = x$

$Width = y$

So:

$Area = x * y$

Substitute 128 for Area

$128 = x * y$

Make x the subject

$x = \frac{128}{y}$

When 1 inch margin is at top and bottom

The length becomes:

$Length = x + 1 + 1$

$Length = x + 2$

When 2 inch margin is at both sides

The width becomes:

$Width = y + 2 + 2$

$Width = y + 4$

The New Area (A) is then calculated as:

$A = (x + 2) * (y + 4)$

Substitute $\frac{128}{y}$ for x

$A = (\frac{128}{y} + 2) * (y + 4)$

Open Brackets

$A = 128 + \frac{512}{y} + 2y + 8$

Collect Like Terms

$A = \frac{512}{y} + 2y + 8+128$

$A = \frac{512}{y} + 2y + 136$

$A= 512y^{-1} + 2y + 136$

To calculate the smallest possible value of y, we have to apply calculus.

Different A with respect to y

$A' = -512y^{-2} + 2$

Set

$A' = 0$

This gives:

$0 = -512y^{-2} + 2$

Collect Like Terms

$512y^{-2} = 2$

Multiply through by $y^2$

$y^2 * 512y^{-2} = 2 * y^2$

$512 = 2y^2$

Divide through by 2

$256=y^2$

Take square roots of both sides

$\sqrt{256=y^2$

$16=y$

$y = 16$

Recall that:

$x = \frac{128}{y}$

$x = \frac{128}{16}$

$x = 8$

Recall that the new dimensions are:

$Length = x + 2$

$Width = y + 4$

So:

$Length = 8 + 2$

$Length = 10$

$Width = 16 + 4$

$Width = 20$

To double-check;

Differentiate A'

$A' = -512y^{-2} + 2$

$A" = -2 * -512y^{-3}$

$A" = 1024y^{-3}$

$A" = \frac{1024}{y^3}$

The above value is:

$A" = \frac{1024}{y^3} > 0$

This means that the calculated values are at minimum.

<em>Hence, the dimensions of the smallest piece that can be used are: 10 by 20 and the area is 200 square inches</em>

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*Hint: In order to find the coordinates, you plug in the coordinates into the rule and solve.

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The coordinates of B' is (6, -13).

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2 years ago

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