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Juli2301 [7.4K]
2 years ago
5

PLZZ BRAINLIEST TO ANYBODY THAT CAN SOLVE THIS. (wrong or not at least try)

Mathematics
1 answer:
Soloha48 [4]2 years ago
7 0

Answer: Part A= True, Part B= False, Part C= Line 1's slope is -2, Line 2's slope is 3, Part D: Line 1 is 3, Line 2 is 3, Part E; sorry I can't do that for you but use desmos if you need help, Part F: (0,3)

Step-by-step explanation:

Put the equations in slope-intercept form. This will help greatly. To check solutions, substitute x and y in. To graph lines, use the slope and y-intercept. Desmos graphing calculator is very helpful if you need help. Finding solutions, just find where the lines intersect. Good luck!

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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
xxMikexx [17]

Answer with explanation:

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

 f(x)=log (3 x) +5 x²

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

Initial Guess =0.5

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

                =\frac{\text{Actual value}}{\text{Given Value}}\\\\=\frac{0.205}{0.5}\\\\=0.41

 Approximate relative error in terms of Percentage

   =0.41 × 100

   = 41 %

7 0
2 years ago
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