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3 years ago

I don’t know what I’m doing please help!!

Mathematics

1 answer:

elixir [45]3 years ago

5 0

Answer:

(n -13)/(n -7)

Step-by-step explanation:

Simplify the fraction on the left, then add the two fractions.

$\displaystyle\frac{n^2-10n+24}{n^2-13n+42}-\frac{9}{n-7}=\frac{(n-6)(n-4)}{(n-6)(n-7)}-\frac{9}{n-7}\\\\=\frac{n-4}{n-7}-\frac{9}{n-7}\\\\=\frac{n-4-9}{n-7}\\\\=\frac{n-13}{n-7}$

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<em>Comment on the graph</em>

The vertical asymptote tells you the simplified form has one zero in the denominator at x=7. That is, the denominator is x-7.

The x-intercept at 13 tells you that x-13 is a factor of the numerator.

The horizontal asymptote at y=1 tells you there is no vertical scaling, so the simplest form is ...

(n -13)/(n -7)

The hole at x=6 is a result of the factor (x-6) that is cancelled from the first fraction in the original expression. At that value of x, the fraction is undefined. So, the above solution should come with the restriction x ≠ 6.

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Step-by-step explanation:

Considering the expression

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$\mathrm{Apply\:binomial\:theorem}:\quad \left(a+b\right)^n=\sum _{i=0}^n\binom{n}{i}a^{\left(n-i\right)}b^i$

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$i=0\quad :\quad \frac{4!}{0!\left(4-0\right)!}3^4x^0$

$i=1\quad :\quad \frac{4!}{1!\left(4-1\right)!}3^3x^1$

$i=2\quad :\quad \frac{4!}{2!\left(4-2\right)!}3^2x^2$

$i=3\quad :\quad \frac{4!}{3!\left(4-3\right)!}3^1x^3$

$i=4\quad :\quad \frac{4!}{4!\left(4-4\right)!}3^0x^4$

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