Answer
a) \(\frac{x^{5}}{x^{3}} = x^{2}\)
b) What is the result of dividing \(m^{5}\) by \(m^{2}\)? (Answer: \(m^{3}\))
c) \(\frac{1}{\text{a.a.a.a.a}} = a^{1} \div a^{6}\)
3) a) \(x^{6} \times x^{3} = x^{9}\)
b) \(x^{9} \div 6 = x^{3}\)
c) \(x^{4} y^{2} \div x^{3} = x y^{2}\)
d) \(x^{6} \times 2 = 2x^{6}\)
Solution
Alright, I need to tackle these math problems step by step. Let's start with the first part:
**a) Expand \(\frac{x^{5}}{x^{3}}\) and then cancel like factors to show that \(\frac{x^{5}}{x^{3}} = x^{2}\).**
Okay, so I have the expression \(\frac{x^{5}}{x^{3}}\). To simplify this, I can use the properties of exponents. Specifically, when dividing like bases, I subtract the exponents. So, \(x^{5} \div x^{3} = x^{5-3} = x^{2}\). That seems straightforward. But let me double-check to make sure I'm not missing anything. If I write it out, \(x^{5}\) is \(x \times x \times x \times x \times x\), and \(x^{3}\) is \(x \times x \times x\). Dividing them, I cancel out three \(x\)s from the numerator and the denominator, leaving me with \(x \times x = x^{2}\). Yep, that checks out.
**b) Write your own question that will give an answer \(m^{3}\) when two powers are divided.**
Alright, I need to create a division problem involving powers of \(m\) that results in \(m^{3}\). Let's think about it. If I have \(m^{a} \div m^{b} = m^{3}\), then according to the exponent rules, \(a - b = 3\). So, if I choose \(a = 5\) and \(b = 2\), then \(5 - 2 = 3\), which satisfies the condition. Therefore, the question could be: "What is the result of dividing \(m^{5}\) by \(m^{2}\)?" The answer would be \(m^{3}\).
**c) Show that \(\frac{1}{\text{a.a.a.a.a}}\) is the same as \(a^{1} \div a^{6}\).**
Hmm, this one is a bit tricky. Let's break it down. The expression \(\frac{1}{\text{a.a.a.a.a}}\) has five \(a\)s in the denominator. So, it's \(a^{5}\) in the denominator. Now, \(a^{1} \div a^{6}\) can be rewritten using the exponent rules as \(a^{1-6} = a^{-5}\). Wait, that doesn't seem right. Let me think again. Actually, \(a^{1} \div a^{6} = a^{1-6} = a^{-5}\), which is the same as \(\frac{1}{a^{5}}\). So, \(\frac{1}{\text{a.a.a.a.a}} = a^{-5}\) and \(a^{1} \div a^{6} = a^{-5}\). Therefore, they are indeed the same.
**3) Fill in the blanks:**
a) \(x^{6} \times x^{\square} = x^{9}\)
To find the missing exponent, I can use the rule that when multiplying like bases, I add the exponents. So, \(6 + \square = 9\). Solving for \(\square\), I get \(\square = 9 - 6 = 3\). Therefore, the blank should be filled with 3.
b) \(x^{9} \div \square = x^{3}\)
Here, I need to find the divisor that, when \(x^{9}\) is divided by it, results in \(x^{3}\). Using the exponent rule for division, \(x^{9} \div x^{b} = x^{9 - b}\). Setting this equal to \(x^{3}\), I have \(9 - b = 3\), so \(b = 9 - 3 = 6\). Therefore, the blank should be filled with 6.
c) \(x^{4} y^{2} \div x^{3} = \square y^{2}\)
To solve this, I'll divide the like terms separately. For the \(x\) terms: \(x^{4} \div x^{3} = x^{4-3} = x^{1}\). For the \(y\) terms, since there's no exponent on \(y\) in the denominator, it's \(y^{2} \div y^{0} = y^{2}\) (since \(y^{0} = 1\)). Therefore, the expression simplifies to \(x y^{2}\). So, the blank should be filled with \(x\).
**d) \(x^{6} \times 2\)**
This one is a bit different. It's not an exponent problem but a multiplication by a constant. So, \(x^{6} \times 2 = 2x^{6}\). There's no need to simplify further unless specified.
Answered by UpStudy AI and reviewed by a Professional Tutor
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Simplify this solution