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Serjik [45]
3 years ago
10

The ratio of boys to girls in the entire 6th grade is 3 to 2. Given this ratio, which two class compositions are possible?

Mathematics
1 answer:
gtnhenbr [62]3 years ago
7 0

Answer:

<em>There's an infinite number of possible answers because there might be any multiplication of 3 for boys and any multiplication of 2 for girls. So, for example:</em>

<em>There might be 3 boys and 2 girls.</em>

<em>There might be 6 boys and 4 girls.</em>

<em>There might be 9 boys and 6 girls.</em>

<em>There might be 12 boys and 8 girls.</em>

<em>etc...</em>

<em>In each case the ratio stays the same - 3 : 2.</em>

<em>Of course there probably can't be more than a few dozens of people in one class, but theoretically we can raise the numbers up to infinite.</em>

Step-by-step explanation:<em>hope this help.</em>

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4 (w+1)-4w


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<h2>4</h2>

Step-by-step explanation:


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4 years ago
Short Response To earn a varsity letter on the baseball team, you must play in 60% of the games. Derek played in 11 out of 18 ba
Alla [95]

Answer:

Yes, he earned a Varsity letter because he played in 61% of the games.

Step-by-step explanation:

To know the percent of games he played out of the total, we can do it by dividing the games he played by the total of games and multiply this by 100% to get the percent of games he played:

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0.611111*100% = 61.11%

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3 years ago
What is 10+10+10+10+10+10?
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Step-by-step explanation:

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Find the perimeter of the triangle. Recall that the perimeter of a figure is the distance around a figure.
Svet_ta [14]

Answer:

1 inch

Step-by-step explanation:

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8 0
2 years ago
Prove that $5^{3^n} + 1$ is divisible by $3^{n + 1}$ for all nonnegative integers $n.$
Viktor [21]

When n=0, we have

5^{3^0} + 1 = 5^1 + 1 = 6

3^{0 + 1} = 3^1 = 3

and of course 3 | 6. ("3 divides 6", in case the notation is unfamiliar.)

Suppose this is true for n=k, that

3^{k + 1} \mid 5^{3^k} + 1

Now for n=k+1, we have

5^{3^{k+1}} + 1 = 5^{3^k \times 3} + 1 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k}\right)^3 + 1^3 \\\\ ~~~~~~~~~~~~~ = \left(5^{3^k} + 1\right) \left(\left(5^{3^k}\right)^2 - 5^{3^k} + 1\right)

so we know the left side is at least divisible by 3^{k+1} by our assumption.

It remains to show that

3 \mid \left(5^{3^k}\right)^2 - 5^{3^k} + 1

which is easily done with Fermat's little theorem. It says

a^p \equiv a \pmod p

where p is prime and a is any integer. Then for any positive integer x,

5^3 \equiv 5 \pmod 3 \implies (5^3)^x \equiv 5^x \pmod 3

Furthermore,

5^{3^k} \equiv 5^{3\times3^{k-1}} \equiv \left(5^{3^{k-1}}\right)^3 \equiv 5^{3^{k-1}} \pmod 3

which goes all the way down to

5^{3^k} \equiv 5 \pmod 3

So, we find that

\left(5^{3^k}\right)^2 - 5^{3^k} + 1 \equiv 5^2 - 5 + 1 \equiv 21 \equiv 0 \pmod3

QED

5 0
2 years ago
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