Ask question

Ask question

All categories

anyanavicka [17]

3 years ago

10

Coz and Amelia each make a tile design with 24 tiles. Coz puts his in rows of 2.Amelia puts hers in rows of 3.How many more tile

s are in each of Coz's rows than Amelia's?

1 answer:

Ivahew [28]3 years ago

4 0

Answer:

4 tiles

Step-by-step explanation:

Divide 24 by 2 (Coz): 12 tiles

Divide 24 by 3 (Amelia): 8 tiles

Subtract 8 from 12.

12 - 8 = 4

There are 4 more tiles in each of Coz's rows than Amelia's.

You might be interested in

How do you simplify this?<br>x²y+xy² / y²+2/5 × xy

polet [3.4K]

$\huge \boxed{\mathbb{QUESTION} \downarrow}$

How do you simplify this?
x²y+xy² / y²+2/5 × xy

$\large \boxed{\mathbb{ANSWER\: WITH\: EXPLANATION} \downarrow}$

$\sf\frac{ { x }^{ 2 } y+x { y }^{ 2 } }{ { y }^{ 2 } + \frac{ 2 }{ 5 } \times xy } \\$

Factor the expressions that are not already factored.

_____

<u>How </u><u>to</u><u> factorise</u><u> </u><u>:</u><u>-</u>

<u>NUMERATOR</u> $\downarrow$

$\sf \: x ^ { 2 } y + x y ^ { 2 }$

Factor out xy.

$\sf \: xy\left(x+y\right)$

<u>DENOMINATOR</u> $\downarrow$

$\sf{ y }^{ 2 } + \frac{ 2 }{ 5 } \times xy \\$

Factor out 1/5.

$\sf \: {\frac{1}{5}y\left(2x+5y\right)} \\$

_____

Continuing...

$\sf\frac{xy\left(x+y\right)}{\frac{1}{5}y\left(2x+5y\right)} \\$

Cancel out y in both the numerator and denominator.

$\sf\frac{x\left(x+y\right)}{\frac{1}{5}\left(2x+5y\right)} \\$

Expand the expression.

$\sf\frac{x^{2}+xy}{\frac{2}{5}x+y} \\$

This can further simplified to as $\downarrow$

$= \boxed{\boxed{\bf\frac{5x\left(x +y\right)}{2x+5y}}}$

3 0

2 years ago

Sam opened a money-market account that pays 3% simple interest. He started the account with $7,000 and made no further deposits.

Verizon [17]

He hold account for 5 years or 5 months. Depending on 3% per year or 3% per month

6 0

2 years ago

In general, the value measured by the horizontal axis is called the_______<br> variable.

azamat

Independent variable

which is the x axis the “horizontal axis

8 0

2 years ago

A tennis coach took his team out for lunch and bought 8 hamburgers and 5 fries for $24. The players were still hungry so the coa

g100num [7]

The cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8

<h3><u>Solution:</u></h3>

Let "f" be the cost of 1 fries

Let "h" be the cost of 1 hamburger

<em><u>Given that, tennis coach took his team out for lunch and bought 8 hamburgers and 5 fries for $24</u></em>

8 x cost of 1 hamburger + 5 x cost of 1 fries = 24

$8 \times h + 5 \times f = 24$

8h + 5f = 24 -------- eqn 1

<em><u>The players were still hungry so the coach bought six more hamburgers and two more fries for $16.60</u></em>

6 x cost of 1 hamburger + 2 x cost of 1 fries = 16.60

$6 \times h + 2 \times f = 16.60$

6h + 2f = 16.60 ------ eqn 2

<em><u>Let us solve eqn 1 and eqn 2</u></em>

Multiply eqn 1 by 2

16h + 10f = 48 ------ eqn 3

Multiply eqn 2 by 5

30h + 10f = 83 -------- eqn 4

<em><u>Subtract eqn 3 from eqn 4</u></em>

30h + 10f = 83

16h + 10f = 48

( - ) ----------------------

14h = 35

<h3>h = 2.5</h3>

Substitute h = 2.5 in eqn 1

8(2.5) + 5f = 24

20 + 5f = 24

5f = 4

<h3>f = 0.8</h3>

Thus cost of 1 hamburger is $ 2.5 and cost of 1 fries is $ 0.8

5 0

3 years ago

Given that 2x+7=27 and 3x+1=28, does 2x+7=3x+1? explain

diamong [38]

\left[x \right] = \left[ 6\right][x]=[6] totally answer

5 0

2 years ago