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iogann1982 [59]
3 years ago
8

Add. 7 2 15+5 2 3+9 13 15 plz help btw its a fraction

Mathematics
1 answer:
LUCKY_DIMON [66]3 years ago
8 0
How could it be a fraction? All the numbers are whole & no division? May I assume the #s listed next to each other are multiplied, i.e. 7 2? If so,...

multiply first...
14 * 15 + 10*3 + 1755 (9*13*15)
210 + 30 + 1755
1,995
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Elsa tries to solve the following equation, and determines there is no solution. Is she correct? Explain.
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Elsa's answer is incorrect since there is a solution of the given equation. In the given logarithmic problem, we need to simplify the problem by transposing log2(3x+5) in the opposite side. The equation will now be log2x-log2(3x+5)=4. Using properties of logarithm, we further simplify the problem into a new form log (2x/6x+10)=4.  Then transform the equation into base form 10^4=(2x/6x+10) and proceed in solving for x value which is equal to 1.667. 
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Use the Newton-Raphson method to find the root of the equation f(x) = In(3x) + 5x2, using an initial guess of x = 0.5 and a stop
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Answer with explanation:

The equation which we have to solve by Newton-Raphson Method is,

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f'(x)=\frac{1}{3x}+10 x

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Formula to find Iteration by Newton-Raphson method

  x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\\\\x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}\\\\ x_{1}=0.5-\frac{\log(1.5)+1.25}{\frac{1}{1.5}+10 \times 0.5}\\\\x_{1}=0.5- \frac{0.1760+1.25}{0.67+5}\\\\x_{1}=0.5-\frac{1.426}{5.67}\\\\x_{1}=0.5-0.25149\\\\x_{1}=0.248

x_{2}=0.248-\frac{\log(0.744)+0.30752}{\frac{1}{0.744}+10 \times 0.248}\\\\x_{2}=0.248- \frac{-0.128+0.30752}{1.35+2.48}\\\\x_{2}=0.248-\frac{0.17952}{3.83}\\\\x_{2}=0.248-0.0468\\\\x_{2}=0.2012

x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057

x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052

So, root of the equation =0.205 (Approx)

Approximate relative error

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 Approximate relative error in terms of Percentage

   =0.41 × 100

   = 41 %

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