I need help with solving this with synthetic division

Mathematics

1 answer:

Anvisha [2.4K]2 years ago

3 0

Answer:

Synthetic division is another way to divide a polynomial by the binomial x - c , where c is a constant.

Step-by-step explanation:

Set up the synthetic division.

Bring down the leading coefficient to the bottom row.

Multiply c by the value just written on the bottom row.

Add the column created in step 3.

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Given the function f(x) x^2-2x/x^3 +9x^2-10x Find any holes, vertical asymptotes, and horizontal asymptotes there may be.

yulyashka [42]

Answer:

See below

Step-by-step explanation:

I assume you mean $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ :

Holes: Since $f(x)=\frac{x^2-2x}{x^3+9x^2-10x}$ reduces to $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ , then there is a hole at $x=0$ as $x$ exists in both the numerator and denominator (however, its limit as x approaches 0 is 1/5).

Vertical Asymptotes: If we further reduce $f(x)=\frac{x(x-2)}{x(x^2+9x-10)}$ to $f(x)=\frac{x-2}{(x+10)(x-1)}$ , then we see that there are vertical asymptotes at $x=-10$ and $x=1$

Horizontal Asymptotes: As the degree of the numerator is less than the degree of the numerator ( $2$ ), then there is a horizontal asymptote at $y=0$

4 0

2 years ago

15 POINTS!!!!!!!!!!!!!!!!!!!!!!!!!!!!

dem82 [27]

Answer:

The given expression is

$3 + (\frac{(2+8)2}{4} \times(\frac{1}{2} )^4)= \hspace{0.3cm}3 +( \frac{10 \times 2}{4}\times \frac{1}{2^4} ) \hspace{0.3cm} = \hspace{0.2cm} 3 + ( \frac{5}{16})$ = 3.3125

Step-by-step explanation:

The given expression is

$3 + (\frac{(2+8)2}{4} \times(\frac{1}{2} )^4)= \hspace{0.3cm}3 +( \frac{10 \times 2}{4}\times \frac{1}{2^4} ) \hspace{0.3cm} = \hspace{0.2cm} 3 + ( \frac{5}{16})$ = 3.3125

4 0

3 years ago

Read 2 more answers

Suppose that x=6 and y=8. If y is inversely related to x, what is y when x=3?

andrey2020 [161]

$\bf \qquad \qquad \textit{inverse proportional variation}\\\\
\textit{\underline{y} varies inversely with \underline{x}}\qquad \qquad y=\cfrac{k}{x}\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
\textit{we also know that }
\begin{cases}
x=6\\
y=8
\end{cases}\implies 8=\cfrac{k}{6}\implies 48=k
\\\\\\
therefore\qquad \boxed{y=\cfrac{48}{x}}
\\\\\\
\textit{when x = 3, what is \underline{y}?}\qquad y=\cfrac{48}{3}$