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3 years ago

(x)^3(-x^3y)^2please help

Mathematics

1 answer:

julsineya [31]3 years ago

4 0

Answer:

$\large\boxed{(x^3)(-x^3y)^2=x^9y^2}$

Step-by-step explanation:

$(x^3)(-x^3y)^2\qquad\text{use}\ (ab)^n=a^nb^n\\\\=(x^3)(-x^3)^2(y^2)\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=(x^3)(x^6)(y^2)\qquad\text{use}\ a^n\cdot a^m=a^{n+m}\\\\=x^{3+6}y^2\\\\=x^9y^2$

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3 years ago

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Step-by-step explanation:

$y=\log_3(x+4)$ only exists when $x+4$ is positive.

You can take the log of a negative or 0 number.

So $x+4>0$ implies $x>-4$ . (I just subtract 4 on both sides.)

So the domain is x>-4. You should see this also when you graph the curve that the curve only exist to the right of -4.

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Let's think about the inverse I found above a little more (I'm going to swap x and y).

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$y=3^x-4$ is an exponential function of 3^x that has been moved down 4 units.

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(x)^3(-x^3y)^2please help​

(x)^3(-x^3y)^2please help