Answer:
<h2>
-0.3913 </h2>
Step-by-step explanation:
Given the initial value of X0 = -1, we can determine the solution of the equation x³ + 5x +2 = 0 using the Newton's method. According to newton's approximation formula;
![y = f(x_0) + f'(x_0)(x-x_0)](https://tex.z-dn.net/?f=y%20%3D%20f%28x_0%29%20%2B%20f%27%28x_0%29%28x-x_0%29)
![x_n = x_n_-_1 - \frac{f(x_n_-_1 )}{f'(x_n_-_1 )}](https://tex.z-dn.net/?f=x_n%20%3D%20x_n_-_1%20%20-%20%5Cfrac%7Bf%28x_n_-_1%20%29%7D%7Bf%27%28x_n_-_1%20%29%7D)
If ![x_0 = 1\\](https://tex.z-dn.net/?f=x_0%20%3D%201%5C%5C)
We will iterate using the formula;
![x_1 = x_0 - \frac{f(x_0 )}{f'(x_0 )}](https://tex.z-dn.net/?f=x_1%20%3D%20x_0%20%20-%20%5Cfrac%7Bf%28x_0%20%29%7D%7Bf%27%28x_0%20%29%7D)
Given f(x) = x³ + 5x +2
f(x0) = f(-1) = (-1)³ + 5(-1) +2
f(-1) = -1 -5 +2
f(-1) = -4
f'(x) = 3x²+5
f'(-1) = 3(-1)²+5
f'(-1) = 8
![x_1 = -1+4/8\\x_1 = -1+0.5\\x_1 = -0.5\\\\x_2 = x_1 - \frac{f(x_1)}{f'(x_1)}\\x_2 = -0.5 - \frac{f(-0.5)}{f'(-0.5)}](https://tex.z-dn.net/?f=x_1%20%3D%20-1%2B4%2F8%5C%5Cx_1%20%3D%20-1%2B0.5%5C%5Cx_1%20%3D%20-0.5%5C%5C%5C%5Cx_2%20%3D%20x_1%20-%20%5Cfrac%7Bf%28x_1%29%7D%7Bf%27%28x_1%29%7D%5C%5Cx_2%20%3D%20-0.5%20-%20%5Cfrac%7Bf%28-0.5%29%7D%7Bf%27%28-0.5%29%7D)
f(-0.5) = (-0.5)³ + 5(-0.5) +2
f(-0.5) = -0.125-2.5+2
f(-0.5) = -0.625
f'(-0.5) = 3(-0.5)²+5
f'(-0.5) = 3(0.25)+5
f'(-0.5) = 0.75+5
f'(-0.5) = 5.75
![x_2 = -0.5 - \frac{(-0.625)}{5.75}\\x_2 = -0.5 + \frac{(0.625)}{5.75}\\x_2 = -0.5 + 0.1086957\\x_2 = -0.3913](https://tex.z-dn.net/?f=x_2%20%3D%20-0.5%20-%20%5Cfrac%7B%28-0.625%29%7D%7B5.75%7D%5C%5Cx_2%20%3D%20-0.5%20%2B%20%5Cfrac%7B%280.625%29%7D%7B5.75%7D%5C%5Cx_2%20%3D%20-0.5%20%2B%200.1086957%5C%5Cx_2%20%3D%20-0.3913)
The estimated solution is -0.3913 (to 4dp)
I think the correct answer is c
Answer:
A
Step-by-step explanation:
A decrease by 30%, means that the price is now 70% of the original price ( 100% - 30% = 70%)
To find the original price divide the new price by the percent of the original price it is now.
147 / 0.70 = 210
The original price was £210
240 and can i get brainlist?
and hope this help you(-;