Questions: Which expression is equivalent to b^(-2) / (a b^(-3)) ? Assume a ≠ 0, b ≠ 0. b/a a^3 b/1 1/(a b^5) a/(b^5)

Which expression is equivalent to b^(-2) / (a b^(-3)) ? Assume a ≠ 0, b ≠ 0. b/a a^3 b/1 1/(a b^5) a/(b^5)
Transcript text: Which expression is equivalent to $\frac{b^{-2}}{a b^{-3}}$ ? Assume $a \neq 0, b \neq 0$. $\frac{b}{a}$ $\frac{a^{3} b}{1}$ $\frac{1}{a b^{5}}$ $\frac{a}{b^{5}}$
failed

Solution

failed
failed

Solution Steps

Step 1: Simplify the Expression

The given expression is

\[ \frac{b^{-2}}{a b^{-3}} \]

We can simplify this by using the property of exponents: \(\frac{x^m}{x^n} = x^{m-n}\).

Step 2: Apply the Exponent Rule

First, simplify the expression by separating the terms:

\[ \frac{b^{-2}}{a b^{-3}} = \frac{b^{-2}}{b^{-3}} \cdot \frac{1}{a} \]

Now, apply the exponent rule to the \(b\) terms:

\[ \frac{b^{-2}}{b^{-3}} = b^{-2 - (-3)} = b^{1} = b \]

Step 3: Combine the Simplified Terms

Now, substitute back into the expression:

\[ b \cdot \frac{1}{a} = \frac{b}{a} \]

Final Answer

The expression equivalent to \(\frac{b^{-2}}{a b^{-3}}\) is

\[ \boxed{\frac{b}{a}} \]

Thus, the answer is the first option: \(\frac{b}{a}\).

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
Two hours after the start of a 100-kilometer bicycle race, a cyclist passes the 02-kilometer mark while riding at a velocity of 45 kilometers per hour. Complete parts (A) through (C) below. (A) Find the cyclist's average velocity during the first two hours of the race. kilometers per hour
failedAnswered
-1/3 x - 1/3 y = 1
failedAnswered
78.96 ÷ 6
failedAnswered
Find (f+g), (f-g), (f g) and (fracfg). Determine the domain for each function. (f(x)=4 x+7), (g(x)=x+1) ((f-g)(x)=3 x+6) (Simplify your answer.) What is the domain of (f-g) ? A. The domain of (f-g) is ((-∞, ∞)). (Type your answer in interval notation.) B. The domain of (f-g) is (). (Use a comma to separate answers as needed.) C. The domain of (f-g) is (varnothing). ((f g)(x)=) (Simplify your answer.)
failedAnswered
An admissions director wants to estimate the mean age of all students enrolled at a college. The estimate must be within 1.7 years of the population mean. Assume the population of ages is normally distributed. (a) Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 1.8 years. (b) The sample mean is 21 years of age. Using the minimum sample size with a 90% level of confidence, does it seem likely that the population mean could be within 10% of the sample mean? within 11% of the sample mean? Explain. Click here to view page 1 of the Standard Normal Table. Click here to view page 2 of the Standard Normal Table. (a) The minimum sample size required to construct a 90% confidence interval is students. (Round up to the nearest whole number.)