3 years ago

I need help pls if you can’t see let me know

1 answer:

4 0

Answer:

I cant see it

Step-by-step explanation:

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Stolb23 [73]

First some rewriting:

$\dfrac{n+3}{2n-6}\div\dfrac{n+3}{3n-9}=\dfrac{n+3}{2(n-3)}\div\dfrac{n+3}{3(n-3)}=\dfrac{\frac{n+3}{2(n-3)}}{\frac{n+3}{3(n-3)}}$

We see that the numerators of both fractions $(n+3)$ and a term in the denominators $(n-3)$ will cancel:

$\dfrac{n+3}{2(n-3)}\div\dfrac{n+3}{3(n-3)}=\dfrac{\frac12}{\frac13}$

Then

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pentagon [3]

Answer:

Step-by-step explanation:

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PolarNik [594]

Answer:

Step-by-step explanation:

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3 years ago