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topjm [15]
4 years ago
11

If there are 40 boys and 20 girls in a room, fill out all possible ratios of boys to girls that could be made.

Mathematics
2 answers:
marysya [2.9K]4 years ago
7 0

Answer:


Step-by-step explanation:

A ratio compares two quantities in terms of multiplication. For example, suppose that there are 10 boys and 15 girls in a classroom. One way, not using ratios, to compare these quantities would be to say that there are 5 more girls than boys in the classroom. To employ a ratio, we say that the ratio of boys to girls is 10 to 15, or 2 to 3. In other words, for every 2 boys in the classroom, there are 3 girls.


In your case, for every 40 boys, there are 20 girls. if we simplify this all the way down, we can get 2 to 1.

All of the following are ratios that apply to your question:

40 to 20

20 to 10

10 to 5

4 to 2

2 to 1

I hope I helped you out! <3

Jalaina3 years ago
0 0

40:20 \rightarrow→ \frac{40\text{:}20}{2}
2
40:20
​
\rightarrow→ 20\text{:}1020:10
40\text{:}2040:20 \rightarrow→ \frac{40\text{:}20}{4}
4
40:20
​
\rightarrow→ 10\text{:}510:5
40\text{:}2040:20 \rightarrow→ \frac{40\text{:}20}{5}
5
40:20
​
\rightarrow→ 8\text{:}48:4
40\text{:}2040:20 \rightarrow→ \frac{40\text{:}20}{10}
10
40:20
​
\rightarrow→ 4\text{:}24:2
40\text{:}2040:20 \rightarrow→ \frac{40\text{:}20}{20}
20
40:20
​
\rightarrow→ 2\text{:}12:1

Jalaina
3 years ago
8:4
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Luba_88 [7]

Step-by-step explanation:

So, there is something known as a removable discontinuity, and it's essentially where you can define f(x) using the most simplified fraction, where you could normally not define f(x).

So we have the following equation:

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

As you may know, we cannot divide a number by the value of zero. When the denominator is equal to zero, on the graph this will appear as a vertical asymptote, where x approaches the value that makes the denominator zero, but never actually reaches it.

If you look at each denominator, you can set them equal to zero to find the vertical asymptotes

x+1 = 0

x=-1

There should be a vertical asymptote at x=-1, since it would make two of the denominators equal to -1, but let's divide the two fractions first.

Original Equation

f(x) = (\frac{x+5}{x+1}\div\frac{(x+3)(x-2)}{(x-4)(x+1)})-\frac{1}{x-2}

Keep, change, flip

f(x) = (\frac{x+5}{x+1}*\frac{(x-4)(x+1)}{(x+3)(x-2)})-\frac{1}{x-2}

Multiply the two fractions

f(x) = (\frac{(x+5)(x-4)(x+1)}{(x+1)(x+3)(x-2)})-\frac{1}{x-2}

Notice how the x+1 is in the numerator and fraction? That means we can cancel it out!

f(x) = (\frac{(x+5)(x-4)}{(x+3)(x-2)})-\frac{1}{x-2}

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2 years ago
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