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Hello Again, I'm going through a question bank and came across a question I missed which did not quite explain the problem in a way that I could fully understand. Here it is below:
A-----B-----C-----D-----E
In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = \(9^{13}\) and C = \(9^{11}\), what is the distance from point A to point D?
Here is the explanation: "consecutive exponents like \(9^{13}\),\(9^{12}\) ,\(9^{11}\) , and \(9^{10}\) are not evenly-spaced (think about 9, 81, and 243, the first three powers of 9...they're not evenly spaced at all). Your job for this question is to subtract to find the differences. And since you're given the distance for two increments (E to C) and need to find the distance for three increments (D to A), you can take the two-increment number and multiply by 1.5 to get the three-increment difference.
Your equation should look like:
\(1.5(9^{13}\) - \(9^{11})\). And here you can use one of the Guiding Principles of Exponents: Factor! Factor out the common \(9^{11}\) so that you have: \((1.5)(9^{11})(9^2\) - \(1)\) Which simplifies to:
\((1.5)(9^{11})(80)\) 1.5 multiplies with 80 to give you 120, so the correct answer is:
\((120)(9^{11})\)"
What I don't understand is why it is automatically assumed that when moving from increments of 2 to 3 that I should multiple the difference by 1.5. Would someone here be able to explain why this is so?
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Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Hello Again, I'm going through a question bank and came across a question I missed which did not quite explain the problem in a way that I could fully understand. Here it is below:
A-----B-----C-----D-----E
In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = \(9^{13}\) and C = \(9^{11}\), what is the distance from point A to point D?
Here is the explanation: "consecutive exponents like \(9^{13}\),\(9^{12}\) ,\(9^{11}\) , and \(9^{10}\) are not evenly-spaced (think about 9, 81, and 243, the first three powers of 9...they're not evenly spaced at all). Your job for this question is to subtract to find the differences. And since you're given the distance for two increments (E to C) and need to find the distance for three increments (D to A), you can take the two-increment number and multiply by 1.5 to get the three-increment difference.
Your equation should look like:
\(1.5(9^{13}\) - \(9^{11})\). And here you can use one of the Guiding Principles of Exponents: Factor! Factor out the common \(9^{11}\) so that you have: \((1.5)(9^{11})(9^2\) - \(1)\) Which simplifies to:
\((1.5)(9^{11})(80)\) 1.5 multiplies with 80 to give you 120, so the correct answer is:
\((120)(9^{11})\)"
What I don't understand is why it is automatically assumed that when moving from increments of 2 to 3 that I should multiple the difference by 1.5. Would someone here be able to explain why this is so?
Show more
The distance between the points C and E is 2 steps. This distance is given by \(9^{13} - 9^{11}\). So length of every step is \(\frac{9^{13} - 9^{11}}{2}\) We are asked the distance between A and D. This distance is 3 steps. So this distance will be given by \(3*\frac{9^{13} - 9^{11}}{2} = (3/2)*(9^{13} - 9^{11})\)
Hello Again, I'm going through a question bank and came across a question I missed which did not quite explain the problem in a way that I could fully understand. Here it is below:
A-----B-----C-----D-----E
In the figure above, points A, B, C, D, and E are evenly spaced along the number line. If E = \(9^{13}\) and C = \(9^{11}\), what is the distance from point A to point D?
Here is the explanation: "consecutive exponents like \(9^{13}\),\(9^{12}\) ,\(9^{11}\) , and \(9^{10}\) are not evenly-spaced (think about 9, 81, and 243, the first three powers of 9...they're not evenly spaced at all). Your job for this question is to subtract to find the differences. And since you're given the distance for two increments (E to C) and need to find the distance for three increments (D to A), you can take the two-increment number and multiply by 1.5 to get the three-increment difference.
Your equation should look like:
\(1.5(9^{13}\) - \(9^{11})\). And here you can use one of the Guiding Principles of Exponents: Factor! Factor out the common \(9^{11}\) so that you have: \((1.5)(9^{11})(9^2\) - \(1)\) Which simplifies to:
\((1.5)(9^{11})(80)\) 1.5 multiplies with 80 to give you 120, so the correct answer is:
\((120)(9^{11})\)"
What I don't understand is why it is automatically assumed that when moving from increments of 2 to 3 that I should multiple the difference by 1.5. Would someone here be able to explain why this is so?
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.