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lawyer [7]
2 years ago
6

Help help help help please pleaseeeee

Mathematics
1 answer:
Lerok [7]2 years ago
3 0

Answer:

\sqrt{29}

Step-by-step explanation:

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8090 [49]

Answer:

Okay?

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I think m a week or 2 late

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Use Newton’s Method to find the solution to x^3+1=2x+3 use x_1=2 and find x_4 accurate to six decimal places. Hint use x^3-2x-2=
luda_lava [24]

Let f(x) = x^3 - 2x - 2. Then differentiating, we get

f'(x) = 3x^2 - 2

We approximate f(x) at x_1=2 with the tangent line,

f(x) \approx f(x_1) + f'(x_1) (x - x_1) = 10x - 18

The x-intercept for this approximation will be our next approximation for the root,

10x - 18 = 0 \implies x_2 = \dfrac95

Repeat this process. Approximate f(x) at x_2 = \frac95.

f(x) \approx f(x_2) + f'(x_2) (x-x_2) = \dfrac{193}{25}x - \dfrac{1708}{125}

Then

\dfrac{193}{25}x - \dfrac{1708}{125} = 0 \implies x_3 = \dfrac{1708}{965}

Once more. Approximate f(x) at x_3.

f(x) \approx f(x_3) + f'(x_3) (x - x_3) = \dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125}

Then

\dfrac{6,889,342}{931,225}x - \dfrac{11,762,638,074}{898,632,125} = 0 \\\\ \implies x_4 = \dfrac{5,881,319,037}{3,324,107,515} \approx 1.769292663 \approx \boxed{1.769293}

Compare this to the actual root of f(x), which is approximately <u>1.76929</u>2354, matching up to the first 5 digits after the decimal place.

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1 year ago
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dolphi86 [110]
The midpoint is (-2, -15)

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If 18% of the beads in a bag are green, 25% are red, and 40% are purple, what is the probability of choosing a green bead and th
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the answer is a)22 %

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