Pregunta
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Simplify the radical expression. \( \sqrt{28} \)

Ask by Ellis Luna. in the United States
Jan 22,2025

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\( \sqrt{28} = 2\sqrt{7} \)

Solución

To simplify the radical expression \( \sqrt{28} \), follow these steps: 1. **Factor the Number Inside the Radical:** Start by finding the prime factors of 28. \[ 28 = 4 \times 7 \] Here, 4 is a perfect square since \( 4 = 2^2 \). 2. **Apply the Square Root to Each Factor:** Use the property of square roots that \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \). \[ \sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} \] 3. **Simplify the Square Roots:** Since \( \sqrt{4} = 2 \), substitute this value back into the expression. \[ \sqrt{4} \times \sqrt{7} = 2 \times \sqrt{7} \] 4. **Final Simplified Form:** \[ \sqrt{28} = 2\sqrt{7} \] **Therefore, the simplified form of \( \sqrt{28} \) is \( 2\sqrt{7} \).**

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Bonus Knowledge

To simplify \( \sqrt{28} \), we start by finding the prime factorization of 28. It can be broken down as \( 28 = 4 \times 7 \), where 4 is a perfect square. So, we can express the square root as follows: \[ \sqrt{28} = \sqrt{4 \times 7} = \sqrt{4} \times \sqrt{7} = 2\sqrt{7}. \] Thus, the simplified form of \( \sqrt{28} \) is \( 2\sqrt{7} \). Now, isn’t it neat how we can break down numbers and uncover their simpler forms? Next time you encounter a radical, just remember you can factor it to make your life easier!

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Write down the next two terms for each geometric sequence. a \( 2 ; 4 ; 8 ; 16 ; \ldots \) b \( 5 ; 15 ; 45 ; \ldots \) c \( 3 ; 6 ; 12 ; 24 ; \ldots \) d \( 18 ; 6 ; 2 ; \ldots \) e \( 20 ; 10 ; 5 ; \ldots \) f \( 4 ;-12 ; 36 ; \ldots \) g \( 7 ;-14 ; 28 ; \ldots \) h \( 8 ; 4 ; 2 ; \ldots \) i \( \frac{1}{9} ; \frac{1}{3} ; 1 ; \ldots \) j \( 500(1,25) ; 500(1,25)^{2} ; 500(1,25)^{3} ; \ldots \) k \( 1000(1,8) ; 1000(1,8)^{2} ; 1000(1,8)^{3} ; \ldots \) i \( 6000(1,1) ; 6000(1,1)^{2} ; 6000(1,1)^{3} ; \ldots \) m \( 400\left(1+\frac{0,09}{12}\right) ; 400\left(1+\frac{0,09}{12}\right)^{2} ; 400\left(1+\frac{0,09}{12}\right)^{3} ; \ldots \) n \( 300\left(1+\frac{0,1125}{4}\right) ; 300\left(1+\frac{0,1125}{4}\right)^{2} ; 300\left(1+\frac{0,1125}{4}\right)^{3} ; \ldots \) o \( x\left(1+\frac{0,092}{2}\right) ; x\left(1+\frac{0,092}{2}\right)^{2} ; x\left(1+\frac{0,092}{2}\right)^{3} ; \ldots \) 2 Find the first three terms for each geometric sequence. a \( \mathrm{T}_{1}=2 \) and \( r=3 \) b \( \mathrm{T}_{1}=4 \) and \( r=\frac{1}{2} \) c \( \mathrm{T}_{1}=12 \) and \( r=\frac{-1}{3} \) d \( T_{1}=500 \) and \( r=1,1 \) e. \( \mathrm{T}_{1}=8000 \) and \( r=\left(1+\frac{0,09}{4}\right) \) f \( T_{1}=3 \) and \( T_{6}=96 \) g \( \quad T_{1}=7 \) and \( T_{5}=\frac{7}{81} \) h \( T_{2}=6 \) and \( T_{7}=192 \) \( T_{3}=18 \) and \( T_{5}=162 \) -d \( T_{3}=16 \) and \( T_{7}=256 \) k \( T_{2}=10 \) and \( T_{5}=80 \) I \( T_{2}=3 \) and \( T_{6}=\frac{1}{27} \) Determine: a which term is equal to 1280 in the sequence \( 5 ; 10 ; 20 ; \ldots \) b which term is equal to 1536 in the sequence \( 3 ; 6 ; 12 ; \ldots \) c which term is equal to 6561 in the sequence \( 3 ; 9 ; 27 ; \ldots \) d which term is equal to \( \frac{5}{16384} \) in the sequence \( \frac{5}{4} ; \frac{5}{8} ; \frac{5}{16} ; \ldots \) e which term is equal to \( \frac{1}{8192} \) in the sequence \( 16 ; 8 ; 4 ; \ldots \)
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