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Simplify the product. \[ \sqrt{3} \cdot \sqrt{12} \]

Ask by Curry Bradley. in the United States
Jan 22,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The product simplifies to 6.

Solution

To simplify the product \(\sqrt{3} \cdot \sqrt{12}\), you can use the property of square roots that states: \[ \sqrt{a} \cdot \sqrt{b} = \sqrt{a \times b} \] Let's apply this property step by step: 1. **Multiply the radicands (the numbers inside the square roots):** \[ \sqrt{3} \cdot \sqrt{12} = \sqrt{3 \times 12} \] 2. **Calculate the product inside the square root:** \[ 3 \times 12 = 36 \] So, \[ \sqrt{3 \times 12} = \sqrt{36} \] 3. **Simplify the square root:** \[ \sqrt{36} = 6 \] **Therefore,** \[ \sqrt{3} \cdot \sqrt{12} = 6 \]

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Beyond the Answer

To simplify the product \(\sqrt{3} \cdot \sqrt{12}\), you can use the property of square roots that states \(\sqrt{a} \cdot \sqrt{b} = \sqrt{a \cdot b}\). First, combine the square roots: \[ \sqrt{3} \cdot \sqrt{12} = \sqrt{3 \cdot 12} = \sqrt{36} \] Next, simplify \(\sqrt{36}\): \[ \sqrt{36} = 6 \] Thus, the simplified product is: \[ \sqrt{3} \cdot \sqrt{12} = 6 \]

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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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