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2 years ago

Write the numbers in the domain with commas between them (for example, {1, 2, 3}

Mathematics

1 answer:

zlopas [31]2 years ago

8 0

Answer:

{-5, -1, 2}

Step-by-step explanation:

domain is the inputs or x numbers

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Solve the equation x³ - 5x²+2x+8 = 0 given that - 1 is a zero of f(x)= x3 - 5x²+2x+8.

exis [7]

Answer:

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Step-by-step explanation:

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1 year ago

If y = x - 6 were changed to y = x + 8, how would the graph of the new function

Paraphin [41]

It would be shifted up

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2 years ago

Write the following number as ratios of integers

Grace [21]

We're going to "cut" the repeating part here in a few steps. First, we're going to put the number in a variable:

$x=-2.0\overline{42}$

Next, to get rid of the negative, we can multiply either side by -1 to get

$-x=2.0\overline{42}$

Now, we won't actually use this -x directly; instead, we want to create two new values, one by multiplying either side by 10:

$-10x=20.\overline{42}$

and the other by multiplying either side by 1000:

$-1000x=2042.\overline{42}$

Next, we can get rid of the repeated part of the number by subtracting -10x from -1000x:

$-1000x-(-10x)=2042.\overline{42}-20.\overline{42}\\ -1000x+10x=2022\\ -990x=2022$

And finally, we can divide either side of the equation by -990 to find that

$x= \frac{2022}{-990}=-\frac{2022}{990}\div \frac{6}{6}=-\frac{337}{165}$

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3 years ago

Suppose the graph of a cubic polynomial function

ivann1987 [24]

Answer:

(1/6)(x-2)(x-3)(x-5)

Step-by-step explanation:

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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