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3 years ago

15

Which equation represents the yearly cost for a nonmeber at the same video game arcade based on the total number of game tokens

purchased in a year

1 answer:

olganol [36]3 years ago

3 0

Answer: c) y=1/5x

Step-by-step explanation: took da test

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What is the slope of (2,5) and (5,8)

murzikaleks [220]

Answer:

Slope = 1

Step-by-step explanation:

Slope = change in y/change in x

= (8-5)/(5-2)

=3/3

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3 years ago

Round each number to the nearest tenth.

andreev551 [17]

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2 years ago

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2 years ago

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2 years ago

Prove the following statement using a proof by contraposition. Yr EQ,s ER, if s is irrational, then r + 1 is irrational.

sp2606 [1]

Answer:

I think that what you are trying to show is: If $s$ is irrational and $r$ is rational, then $r+s$ is rational. If so, a proof can be as follows:

Step-by-step explanation:

Suppose that $r+s$ is a rational number. Then $r$ and $r+s$ can be written as follows

$r=\frac{p_{1}}{q_{1}}, \,p_{1}\in \mathbb{Z}, q_{1}\in \mathbb{Z}, q_{1}\neq 0$

$r+s=\frac{p_{2}}{q_{2}}, \,p_{2}\in \mathbb{Z}, q_{2}\in \mathbb{Z}, q_{2}\neq 0$

Hence we have that

$r+s=\frac{p_{1}}{q_{1}}+s=\frac{p_{2}}{q_{2}}$

Then

$s=\frac{p_{2}}{q_{2}}-\frac{p_{1}}{q_{1}}=\frac{p_{2}q_{1}-p_{1}q_{2}}{q_{1}q_{2}}\in \mathbb{Q}$

This is a contradiction because we assumed that $s$ is an irrational number.

Then $r+s$ must be an irrational number.

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2 years ago