For t!=0, which of the following expressions is equivalent to (81t^(2)+27 t)/(9t^(2)+9t) ? Choose 1 answer: (A) (9t+27)/(t+9) (B) (9t+3t)/(t+1t) (c) (9t+3)/(t+1) (D) (9t+9)/(t+3)

Factor out common terms: Simplify the numerator and the denominator by factoring out common terms. The numerator 81t^2 + 27t can be factored by taking out the common factor of 27t, which gives us 27t(3t + 1). The denominator 9t^2 + 9t can be factored by taking out the common factor of 9t, which gives us 9t(t + 1). So, the expression becomes (27t(3t + 1))/(9t(t + 1)).Cancel out common factors: Simplify the expression by canceling out common factors. We can cancel out the common factor of 9t from the numerator and the denominator. This leaves us with (3(3t + 1))/(t + 1).Expand remaining expression: Expand the remaining expression in the numerator. Multiplying 3 by each term in the parentheses gives us 9t + 3. So, the expression is now (9t + 3)/(t + 1).Match with answer choices: Match the simplified expression with the given answer choices. The expression (9t + 3)/(t + 1) matches with choice (C).

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For
t!=0, which of the following expressions is equivalent to
(81t^(2)+27 t)/(9t^(2)+9t) ?
Choose 1 answer:
(A)
(9t+27)/(t+9)
(B)
(9t+3t)/(t+1t)
(c)
(9t+3)/(t+1)
(D)
(9t+9)/(t+3)

For $t \neq 0$ , which of the following expressions is equivalent to $\frac{81 t^{2}+27 t}{9 t^{2}+9 t}$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\frac{9 t+27}{t+9}$ $\newline$ (B) $\frac{9 t+3 t}{t+1 t}$ $\newline$ (C) $\frac{9 t+3}{t+1}$ $\newline$ (D) $\frac{9 t+9}{t+3}$

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Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

Q. For

t \neq 0

, which of the following expressions is equivalent to

\frac{81 t^{2}+27 t}{9 t^{2}+9 t}

\newline

Choose

1

answer:

\newline

(A)

\frac{9 t+27}{t+9}

\newline

(B)

\frac{9 t+3 t}{t+1 t}

\newline

(C)

\frac{9 t+3}{t+1}

\newline

(D)

\frac{9 t+9}{t+3}

Factor out common terms: Simplify the numerator and the denominator by factoring out common terms. $\newline$ The numerator $81t^2 + 27t$ can be factored by taking out the common factor of $27t$ , which gives us $27t(3t + 1)$ . $\newline$ The denominator $9t^2 + 9t$ can be factored by taking out the common factor of $9t$ , which gives us $9t(t + 1)$ . $\newline$ So, the expression becomes $\frac{27t(3t + 1)}{9t(t + 1)}$ .
Cancel out common factors: Simplify the expression by canceling out common factors. $\newline$ We can cancel out the common factor of $9t$ from the numerator and the denominator. $\newline$ This leaves us with $\left(\frac{3(3t + 1)}{t + 1}\right)$ .
Expand remaining expression: Expand the remaining expression in the numerator. $\newline$ Multiplying $3$ by each term in the parentheses gives us $9t + 3$ . $\newline$ So, the expression is now $(9t + 3)/(t + 1)$ .
Match with answer choices: Match the simplified expression with the given answer choices. $\newline$ The expression $(9t + 3)/(t + 1)$ matches with choice (C).