(x^(3)+9x^(2))/(x^(3)) Which expression is equivalent to the given expression for all x > 1 ? Choose 1 answer: (A) 9x^(2) (B) 1+9x^(2) (c) (x+9)/(x) (D) (1+9x)/(x)

Divide x^(3): We have the expression (x^(3)+9x^(2))/(x^(3)). To simplify this expression, we can divide each term in the numerator by the term in the denominator.Divide 9x^(2): Divide x^(3) by x^(3) to get 1.Combine results: Divide 9x^(2) by x^(3) to get 9/x.Match with option: Combine the results from the previous steps to get the simplified expression: 1 + 9/x.Now we need to match our simplified expression with one of the given options. The expression 1 + 9/x is equivalent to (1+9x)/(x), which is option (D).

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(x^(3)+9x^(2))/(x^(3))
Which expression is equivalent to the given expression for all
x > 1 ?
Choose 1 answer:
(A)
9x^(2)
(B)
1+9x^(2)
(c)
(x+9)/(x)
(D)
(1+9x)/(x)

$\frac{x^{3}+9 x^{2}}{x^{3}}$ $\newline$ Which expression is equivalent to the given expression for all x>1 ? $\newline$ Choose $1$ answer: $\newline$ (A) $9 x^{2}$ $\newline$ (B) $1+9 x^{2}$ $\newline$ (C) $\frac{x+9}{x}$ $\newline$ (D) $\frac{1+9 x}{x}$

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Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

\frac{x^{3}+9 x^{2}}{x^{3}}

\newline

Which expression is equivalent to the given expression for all

x>1

\newline

Choose

1

answer:

\newline

(A)

9 x^{2}

\newline

(B)

1+9 x^{2}

\newline

(C)

\frac{x+9}{x}

\newline

(D)

\frac{1+9 x}{x}

Divide $x^{3}$ : We have the expression $\frac{x^{3}+9x^{2}}{x^{3}}$ . To simplify this expression, we can divide each term in the numerator by the term in the denominator.
Divide $9x^{2}$ : Divide $x^{3}$ by $x^{3}$ to get $1$ .
Combine results: Divide $9x^{2}$ by $x^{3}$ to get $\frac{9}{x}$ .
Match with option: Combine the results from the previous steps to get the simplified expression: $1 + \frac{9}{x}.$
Match with option: Combine the results from the previous steps to get the simplified expression: $1 + \frac{9}{x}$ .Now we need to match our simplified expression with one of the given options. The expression $1 + \frac{9}{x}$ is equivalent to $\frac{(1+9)x}{x}$ , which is option (D).