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3 years ago

Which equation does not define a function?

Mathematics

1 answer:

stira [4]3 years ago

5 0

The first relation gives two y-values for any given x-value (except x=6).

y² +x = 6 . . . . . is not a function

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shepuryov [24]

Answer:

it is 612 I think

Step-by-step explanation:

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2 years ago

What is 36 pieces of candy for 6 children as a unit rate?

Leni [432]

$\frac{36}{6} =\frac{6}{1}$

7 0

3 years ago

PLEASE HELP I WILL MARK YOU BRAINLIEST

Delvig [45]

Answer:

Step-by-step explanation:

I counted each of the cubes and got 64.

Then I divided it by 4 since you need to divide it be 1/4.

once I divided 4 for 64 I got 16.

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3 years ago

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What is not a factor of six 2,3,4,or1

Andre45 [30]

4 because 2, 3, and 1 can all fit into 6.

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3 years ago

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(−5)3x7(yz)4 / (3)2x2y8z2

g100num [7]

Answer:

$\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}$

Step-by-step explanation:

Given:

The expression to simplify is given as:

$\frac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}$

In order to simplify this, we have to use the law of indices.

1. $(ab)^m=a^mb^m$

So, $(yz)^4=y^4z^4$

Substitute this value in the above expression. This gives,

$=\dfrac{(-5)3x^7y^4z^4}{(3)2x^2y^8z^2}\\\\\\=\dfrac{-15x^7y^4z^4}{6x^2y^8z^2}......(-5\times 3=15\ and\ 3\times 2=6)$

Now, we use another law of indices.

2. $\frac{a^m}{a^n}=a^{m-n}$

So, $\frac{x^7}{x^2}=x^{7-2}=x^5,\frac{y^4}{y^8}=y^{4-8}=y^{-4}, \frac{z^4}{z^2}=z^{4-2}=z^2$

Substitute these values in the above expression. This gives,

$=\frac{-15}{6}\times x^5\times y^{-4}\times z^2\\\\=\frac{-5x^5y^{-4}z^2}{2}$

Finally, we further simplify it using the law $a^{-m}=\frac{1}{a^m}$

So, $y^{-4}=\frac{1}{y^4}$

Therefore, the given expression is simplified as:

$\dfrac{(-5)3x^7(yz)^4}{(3)2x^2y^8z^2}=\dfrac{-5x^5z^2}{2y^4}$

5 0

3 years ago