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Answer :
Sure, let's solve this step-by-step.
The given expression is:
[tex]\[ 8 x^6 \sqrt{200 x^{13}} \div 2 x^5 \sqrt{32 x^7} \][/tex]
Step 1: Break down inside the square roots
For the numerator:
[tex]\[ \sqrt{200 x^{13}} = \sqrt{4 \cdot 25 \cdot 2 \cdot (x^6)^2 \cdot x} \][/tex]
For the denominator:
[tex]\[ \sqrt{32 x^7} = \sqrt{16 \cdot 2 \cdot (x^3)^2 \cdot x} \][/tex]
Step 2: Simplify the square roots and rewrite the expression
Now we rewrite the expression:
[tex]\[ 8 x^6 \cdot \sqrt{4 \cdot 25 \cdot 2 \cdot (x^6)^2 \cdot x} \div 2 x^5 \cdot \sqrt{16 \cdot 2 \cdot (x^3)^2 \cdot x} \][/tex]
Breaking down the square roots:
[tex]\[ \sqrt{4} = 2 \][/tex]
[tex]\[ \sqrt{25} = 5 \][/tex]
[tex]\[ \sqrt{16} = 4 \][/tex]
So:
[tex]\[ 8 \cdot 2 \cdot 5 \cdot x^6 \cdot x^6 \sqrt{2x} \div 2 \cdot 4 \cdot 4 \cdot x^5 \cdot x^3 \sqrt{2x} \][/tex]
Which simplifies to:
[tex]\[ 80 x^{12} \sqrt{2x} \div 32 x^8 \sqrt{2x} \][/tex]
Step 3: Cancel common factors
Next, we simplify by canceling common terms:
[tex]\[ \frac{80 x^{12} \sqrt{2x}}{32 x^8 \sqrt{2x}} \][/tex]
The [tex]\(\sqrt{2x}\)[/tex] terms cancel out:
[tex]\[ \frac{80 x^{12}}{32 x^8} \][/tex]
Step 4: Simplify the fraction and exponents
Simplify the numeric fraction:
[tex]\[ \frac{80}{32} = \frac{5}{2} \][/tex]
Simplify the exponents of [tex]\(x\)[/tex]:
[tex]\[ x^{12} / x^8 = x^{12 - 8} = x^4 \][/tex]
So the final simplified expression is:
[tex]\[ \frac{5}{2} x^4 \][/tex]
The mistake occurred in Step 4, where Seth mistakenly introduced the variable [tex]\(z\)[/tex] instead of continuing with [tex]\(x\)[/tex].
The given expression is:
[tex]\[ 8 x^6 \sqrt{200 x^{13}} \div 2 x^5 \sqrt{32 x^7} \][/tex]
Step 1: Break down inside the square roots
For the numerator:
[tex]\[ \sqrt{200 x^{13}} = \sqrt{4 \cdot 25 \cdot 2 \cdot (x^6)^2 \cdot x} \][/tex]
For the denominator:
[tex]\[ \sqrt{32 x^7} = \sqrt{16 \cdot 2 \cdot (x^3)^2 \cdot x} \][/tex]
Step 2: Simplify the square roots and rewrite the expression
Now we rewrite the expression:
[tex]\[ 8 x^6 \cdot \sqrt{4 \cdot 25 \cdot 2 \cdot (x^6)^2 \cdot x} \div 2 x^5 \cdot \sqrt{16 \cdot 2 \cdot (x^3)^2 \cdot x} \][/tex]
Breaking down the square roots:
[tex]\[ \sqrt{4} = 2 \][/tex]
[tex]\[ \sqrt{25} = 5 \][/tex]
[tex]\[ \sqrt{16} = 4 \][/tex]
So:
[tex]\[ 8 \cdot 2 \cdot 5 \cdot x^6 \cdot x^6 \sqrt{2x} \div 2 \cdot 4 \cdot 4 \cdot x^5 \cdot x^3 \sqrt{2x} \][/tex]
Which simplifies to:
[tex]\[ 80 x^{12} \sqrt{2x} \div 32 x^8 \sqrt{2x} \][/tex]
Step 3: Cancel common factors
Next, we simplify by canceling common terms:
[tex]\[ \frac{80 x^{12} \sqrt{2x}}{32 x^8 \sqrt{2x}} \][/tex]
The [tex]\(\sqrt{2x}\)[/tex] terms cancel out:
[tex]\[ \frac{80 x^{12}}{32 x^8} \][/tex]
Step 4: Simplify the fraction and exponents
Simplify the numeric fraction:
[tex]\[ \frac{80}{32} = \frac{5}{2} \][/tex]
Simplify the exponents of [tex]\(x\)[/tex]:
[tex]\[ x^{12} / x^8 = x^{12 - 8} = x^4 \][/tex]
So the final simplified expression is:
[tex]\[ \frac{5}{2} x^4 \][/tex]
The mistake occurred in Step 4, where Seth mistakenly introduced the variable [tex]\(z\)[/tex] instead of continuing with [tex]\(x\)[/tex].
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